Chapter 8 Exponents & Powers

The multiplicative inverse of a non-zero number $a$ is a number $b$ such that the product of $a$ and $b$ is 1. That is, $a \times b = 1$. The multiplicative inverse of $a$ is typically written as $\frac{1}{a}$ or $a^{-1}$.

The given number is $2^7$.

We need to find a number $b$ such that $2^7 \times b = 1$.

This number $b$ is $\frac{1}{2^7}$.

Using the property of exponents which states that $\frac{1}{a^n} = a^{-n}$, we can rewrite $\frac{1}{2^7}$ as $2^{-7}$.

So, the multiplicative inverse of $2^7$ is $2^{-7}$.

Comparing this result with the given options:

(a) $2^{-7}$

(b) $7^2$

(c) $-2^7$

(d) $-2^7$

The correct option is (a).

The correct answer is (a) $2^{-7}$.

We are given the number of cells in the human body as about 100 billion.

First, let's write the number "100 billion" in standard numerical form.

A billion is $1,000,000,000$, which is $10^9$.

100 billion means $100 \times 1 \text{ billion}$.

$100 \text{ billion} = 100 \times 1,000,000,000$.

$100 \text{ billion} = 100,000,000,000$.

Now, we need to write this number in exponential form, specifically as a power of 10.

The number $100,000,000,000$ is 1 followed by 11 zeros.

The number of zeros after 1 indicates the power of 10.

So, $100,000,000,000 = 10^{11}$.

Thus, 100 billion can be written in exponential form as $10^{11}$.

Comparing this result with the given options:

(a) $10^{-11}$

(b) $10^{11}$

(c) $10^{9}$

(d) $10^{-9}$

The correct option is (b).

The correct answer is (b) $10^{11}$.

We are asked to fill in the blank for the expression $(-4)^4 \times \left( \frac{5}{4} \right)^4$.

We can use the property of exponents that states for any non-zero numbers $a$ and $b$, and any integer $m$, $(a \times b)^m = a^m \times b^m$.

Applying this property in reverse, $a^m \times b^m = (a \times b)^m$.

In the given expression, $a = -4$, $b = \frac{5}{4}$, and $m = 4$.

So, we have:

$(-4)^4 \times \left( \frac{5}{4} \right)^4 = \left( -4 \times \frac{5}{4} \right)^4$

Now, we perform the multiplication inside the parentheses:

$-4 \times \frac{5}{4} = -\cancel{4} \times \frac{5}{\cancel{4}} = -5$

The expression becomes:

$(-5)^4$

Now, we calculate the value of $(-5)^4$. Since the exponent is an even number (4), the result will be positive.

$(-5)^4 = (-5) \times (-5) \times (-5) \times (-5) = 25 \times 25 = 625$

Thus, $(-4)^4 \times \left( \frac{5}{4} \right)^4 = 625$.

The statement becomes $(-4)^4 \times \left( \frac{5}{4} \right)^4 = \textbf{625}$.

We are asked to evaluate the expression $(2^{-3})^2 \times (3^{-2})^3$ and fill in the blank.

We will use the property of exponents that states for any non-zero number $a$ and integers $m$ and $n$, $(a^m)^n = a^{m \times n}$.

Apply this rule to the first term $(2^{-3})^2$:

$(2^{-3})^2 = 2^{-3 \times 2} = 2^{-6}$

Apply the same rule to the second term $(3^{-2})^3$:

$(3^{-2})^3 = 3^{-2 \times 3} = 3^{-6}$

Now substitute these results back into the original expression:

$(2^{-3})^2 \times (3^{-2})^3 = 2^{-6} \times 3^{-6}$

We can use another property of exponents that states for any non-zero numbers $a$ and $b$, and any integer $m$, $a^m \times b^m = (a \times b)^m$.

Apply this rule to combine the terms $2^{-6} \times 3^{-6}$:

$2^{-6} \times 3^{-6} = (2 \times 3)^{-6}$

Simplify the expression inside the parentheses:

$(2 \times 3)^{-6} = 6^{-6}$

Thus, the simplified form of the expression is $6^{-6}$. We can also write this with a positive exponent using the rule $a^{-n} = \frac{1}{a^n}$.

$6^{-6} = \frac{1}{6^6}$

The blank should be filled with $\textbf{6}^{\textbf{-6}}$ or $\textbf{\frac{1}{6^6}}$.

We are given the distance between the Earth and the Sun as 150 million kilometres.

First, let's write 150 million in standard numerical form.

A million is $1,000,000$, which is $10^6$.

So, 150 million is $150 \times 1,000,000$.

$150 \times 1,000,000 = 150,000,000$.

Now, we need to write this number in exponential form, specifically in scientific notation.

Scientific notation is a way of writing numbers that are too large or too small to be conveniently written in standard decimal form. It is written as a number between 1 and 10 (inclusive of 1 but exclusive of 10), multiplied by a power of 10.

Take the number $150,000,000$. The decimal point is currently at the end of the number.

To get a number between 1 and 10, we move the decimal point to the left until it is after the first digit (1).

$1.50000000$

We moved the decimal point 8 places to the left.

The number of places the decimal point is moved gives the power of 10. Since we moved it to the left, the power is positive.

So, $150,000,000 = 1.5 \times 10^8$.

The distance between Earth and Sun is $1.5 \times 10^8$ kilometres.

The blank should be filled with $\textbf{1.5} \times \textbf{10}^{\textbf{8}}$ km.

The statement is about expressing very small numbers in standard form using positive exponents.

Standard form (or scientific notation) for a number is written as $a \times 10^n$, where $1 \leq a < 10$ and $n$ is an integer.

For very small numbers (numbers between 0 and 1), the decimal point is moved to the right to obtain the digit $a$, which is between 1 and 10.

When the decimal point is moved to the right, the exponent $n$ is negative.

For example, $0.0000000012$ is a very small number. To write it in standard form, we move the decimal point 9 places to the right: $1.2$. Since we moved the decimal to the right, the exponent is negative 9.

$0.0000000012 = 1.2 \times 10^{-9}$. Here, the exponent is $-9$, which is a negative exponent.

Very large numbers are expressed in standard form using positive exponents (e.g., $1,200,000,000 = 1.2 \times 10^9$).

Therefore, very small numbers are expressed in standard form using negative exponents, not positive exponents.

The statement "Very small numbers can be expressed in standard form using positive exponents" is false.

The correct answer is F.

We need to determine if the statement $(-10) \times (-10) \times (-10) \times (-10) = 10^{-4}$ is true or false.

First, let's evaluate the left side of the equation:

$(-10) \times (-10) \times (-10) \times (-10)$

This is the product of four $-10$s. We can write this using an exponent:

$(-10) \times (-10) \times (-10) \times (-10) = (-10)^4$

Now, let's calculate the value of $(-10)^4$. Since the base is negative and the exponent is an even number (4), the result will be positive.

$(-10)^4 = (-10) \times (-10) \times (-10) \times (-10) = 100 \times 100 = 10000$

So, the left side of the equation is $10000$.

Next, let's evaluate the right side of the equation:

$10^{-4}$

Using the property of negative exponents, $a^{-n} = \frac{1}{a^n}$, we have:

$10^{-4} = \frac{1}{10^4}$

Now, let's calculate the value of $10^4$:

$10^4 = 10 \times 10 \times 10 \times 10 = 10000$

So, the right side of the equation is $\frac{1}{10000}$.

Now we compare the left side and the right side:

Left side = $10000$

Right side = $\frac{1}{10000}$

Clearly, $10000 \neq \frac{1}{10000}$.

Therefore, the statement $(-10) \times (-10) \times (-10) \times (-10) = 10^{-4}$ is false.

The correct answer is F.

We are asked to simplify the expression $\frac{(-2)^3 \;\times\; (-2)^7}{3 \;\times\; 4^6}$.

Let's simplify the numerator first: $(-2)^3 \times (-2)^7$.

Using the property $a^m \times a^n = a^{m+n}$ for multiplication of exponents with the same base, we have:

$(-2)^3 \times (-2)^7 = (-2)^{3+7} = (-2)^{10}$

Since the base is negative and the exponent is even (10), the result is positive:

$(-2)^{10} = 2^{10}$

So, the numerator is $2^{10}$.

Now, let's simplify the denominator: $3 \times 4^6$.

We can write the base 4 as $2^2$. So, $4^6 = (2^2)^6$.

Using the property $(a^m)^n = a^{m \times n}$, we have:

$(2^2)^6 = 2^{2 \times 6} = 2^{12}$

So, the denominator is $3 \times 2^{12}$.

Substitute the simplified numerator and denominator back into the original expression:

$\frac{2^{10}}{3 \times 2^{12}}$

Now, we can simplify the powers of 2 using the property $\frac{a^m}{a^n} = a^{m-n}$:

$\frac{2^{10}}{2^{12}} = 2^{10 - 12} = 2^{-2}$

The expression becomes:

$\frac{2^{-2}}{3}$

We can express $2^{-2}$ with a positive exponent using the property $a^{-n} = \frac{1}{a^n}$:

$2^{-2} = \frac{1}{2^2} = \frac{1}{4}$

Substitute this back into the expression:

$\frac{\frac{1}{4}}{3} = \frac{1}{4 \times 3} = \frac{1}{12}$

Thus, the simplified form of the expression is $\frac{1}{12}$.

The simplified expression is $\frac{1}{12}$.

We are given the equation $(–5)^{x+1} \times (–5)^{5} = (–5)^{7}$.

We can use the property of exponents that states for any non-zero number $a$ and integers $m$ and $n$, $a^m \times a^n = a^{m+n}$.

Applying this property to the left side of the given equation, where the base is $(-5)$, we have:

$(-5)^{x+1} \times (-5)^{5} = (-5)^{(x+1) + 5}$

$(-5)^{x+1} \times (-5)^{5} = (-5)^{x+6}$

So, the given equation becomes:

$(-5)^{x+6} = (-5)^{7}$

When the bases are equal on both sides of an equation and the base is not 0, 1, or -1 (in this case the base is -5), the exponents must also be equal.

Therefore, we can equate the exponents:

$x+6 = 7$

Now, we solve for $x$ by subtracting 6 from both sides of the equation:

$x = 7 - 6$

$x = 1$

Thus, the value of $x$ is 1.

The value of x is 1.

In an exponential expression of the form $a^n$, $a$ is called the base and $n$ is called the exponent or power.

In the given expression $2^n$, the base is 2 and the exponent is $n$.

A variable is a symbol that represents a quantity in a mathematical expression, which may change or be assigned different values.

A constant is a value that does not change.

In the expression $2^n$, the number 2 is a fixed value, so it is a constant. The symbol $n$ is used to represent the exponent, and its value can change. For example, $n$ could be 1, 2, 3, -1, 0, etc., resulting in different values for $2^n$ ($2^1=2$, $2^2=4$, $2^{-1}=0.5$, $2^0=1$).

Therefore, $n$ in the expression $2^n$ is known as a variable.

Comparing this with the given options:

(a) Base - Incorrect (2 is the base)

(b) Constant - Incorrect (n can vary)

(c) x - Incorrect (x is a different symbol)

(d) Variable - Correct

The correct answer is (d) Variable.

Let the fixed base be $a$ and the initial exponent be $n$. The original number is $a^n$.

If the exponent decreases by 1, the new exponent is $n-1$. The new number is $a^{n-1}$.

We need to find the relationship between the new number ($a^{n-1}$) and the original number ($a^n$).

Using the property of exponents, $a^{m-n} = a^m \times a^{-n}$, we can write $a^{n-1}$ as:

$a^{n-1} = a^{n} \times a^{-1}$

Using the property $a^{-1} = \frac{1}{a}$, we have:

$a^{n-1} = a^n \times \frac{1}{a} = \frac{a^n}{a}$

This shows that the new number ($a^{n-1}$) is equal to the previous number ($a^n$) divided by the base ($a$). In other words, the new number is $\frac{1}{a}$ times the previous number.

The options are given in terms of factors of 10. Let's consider the case where the fixed base $a$ is 10, as this is the base related to the options provided.

If the base $a = 10$, then the relationship is:

$10^{n-1} = \frac{10^n}{10}$

This means that when the base is 10, the new number is $\frac{1}{10}$ times the previous number, which is "One-tenth of the previous number".

Comparing this with the given options, option (a) matches this finding for base 10.

The correct answer is (a) One-tenth of the previous number.

We are asked to write $3^{-2}$ in an equivalent form.

We can use the property of exponents that states for any non-zero number $a$ and any integer $n$, $a^{-n} = \frac{1}{a^n}$.

In the given expression $3^{-2}$, the base is $a = 3$ and the positive value of the exponent is $n = 2$.

Applying the property $a^{-n} = \frac{1}{a^n}$, we get:

$3^{-2} = \frac{1}{3^2}$

Comparing this result with the given options:

(a) $3^2$ - Incorrect

(b) $\frac{1}{3^2}$ - Correct

(c) $\frac{1}{3^{-2}}$ - Incorrect (This would be $3^{-(-2)} = 3^2$)

(d) $-\frac{2}{3}$ - Incorrect

The correct answer is (b) $\frac{1}{3^2}$.

We are asked to find the value of the expression $\frac{1}{4^{-2}}$.

We can simplify the term in the denominator, $4^{-2}$, using the property of exponents that states for any non-zero number $a$ and any integer $n$, $a^{-n} = \frac{1}{a^n}$.

Applying this property to $4^{-2}$, where $a=4$ and $n=2$, we get:

$4^{-2} = \frac{1}{4^2}$

Now substitute this back into the original expression:

$\frac{1}{4^{-2}} = \frac{1}{\frac{1}{4^2}}$

When we have a fraction in the denominator, we can multiply the numerator by the reciprocal of the denominator. The reciprocal of $\frac{1}{4^2}$ is $4^2$.

So, $\frac{1}{\frac{1}{4^2}} = 1 \times 4^2 = 4^2$.

Now, calculate the value of $4^2$:

$4^2 = 4 \times 4 = 16$

Thus, the value of $\frac{1}{4^{-2}}$ is 16.

Comparing this result with the given options:

(a) 16 - Correct

(b) 8 - Incorrect

(c) $\frac{1}{16}$ - Incorrect

(d) $\frac{1}{8}$ - Incorrect

The correct answer is (a) 16.

We are asked to find the value of the expression $3^5 \div 3^{-6}$.

We can write the division as a fraction: $\frac{3^5}{3^{-6}}$.

We use the property of exponents for division with the same base: $\frac{a^m}{a^n} = a^{m-n}$, where $a$ is a non-zero number and $m$ and $n$ are integers.

In this expression, the base is $a = 3$, the exponent in the numerator is $m = 5$, and the exponent in the denominator is $n = -6$.

Applying the property, we get:

$3^5 \div 3^{-6} = 3^{5 - (-6)}$

Simplify the exponent:

$5 - (-6) = 5 + 6 = 11$

So, the expression simplifies to $3^{11}$.

$3^5 \div 3^{-6} = 3^{11}$

Comparing this result with the given options:

(a) $3^5$ - Incorrect

(b) $3^{-6}$ - Incorrect

(c) $3^{11}$ - Correct

(d) $3^{-11}$ - Incorrect

The correct answer is (c) $3^{11}$.

We are asked to find the value of the expression $\left( \frac{2}{5} \right)^{-2}$.

We can use the property of exponents that states for any non-zero rational number $\frac{a}{b}$ and any integer $n$, $\left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^n$.

Applying this property to the given expression $\left( \frac{2}{5} \right)^{-2}$, where $a=2$, $b=5$, and $n=2$, we have:

$\left( \frac{2}{5} \right)^{-2} = \left( \frac{5}{2} \right)^2$

Now, we use the property that states for any rational number $\frac{a}{b}$ and any integer $n$, $\left( \frac{a}{b} \right)^n = \frac{a^n}{b^n}$.

Applying this property to $\left( \frac{5}{2} \right)^2$, we get:

$\left( \frac{5}{2} \right)^2 = \frac{5^2}{2^2}$

Calculate the squares:

$5^2 = 5 \times 5 = 25$

$2^2 = 2 \times 2 = 4$

So, the value of the expression is $\frac{25}{4}$.

Comparing this result with the given options:

(a) $\frac{4}{5}$ - Incorrect

(b) $\frac{4}{25}$ - Incorrect

(c) $\frac{25}{4}$ - Correct

(d) $\frac{5}{2}$ - Incorrect

The correct answer is (c) $\frac{25}{4}$.

We need to find the reciprocal of the expression $\left( \frac{2}{5} \right)^{-1}$.

First, let's simplify the expression $\left( \frac{2}{5} \right)^{-1}$.

We use the property of exponents that states for any non-zero rational number $\frac{a}{b}$ and any integer $n$, $\left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^n$.

Applying this property to $\left( \frac{2}{5} \right)^{-1}$, where $a=2$, $b=5$, and $n=1$, we get:

$\left( \frac{2}{5} \right)^{-1} = \left( \frac{5}{2} \right)^1 = \frac{5}{2}$

So, the value of the expression is $\frac{5}{2}$.

Now, we need to find the reciprocal of $\frac{5}{2}$. The reciprocal of a non-zero number $x$ is $\frac{1}{x}$.

The reciprocal of $\frac{5}{2}$ is $\frac{1}{\frac{5}{2}}$.

To divide by a fraction, we multiply by its reciprocal:

$\frac{1}{\frac{5}{2}} = 1 \times \frac{2}{5} = \frac{2}{5}$

Thus, the reciprocal of $\left( \frac{2}{5} \right)^{-1}$ is $\frac{2}{5}$.

Comparing this result with the given options:

(a) $\frac{2}{5}$ - Correct

(b) $\frac{5}{2}$ - Incorrect (This is the value of the expression itself)

(c) $-\frac{5}{2}$ - Incorrect

(d) $-\frac{2}{5}$ - Incorrect

The correct answer is (a) $\frac{2}{5}$.

The multiplicative inverse of a non-zero number $a$ is a number $b$ such that $a \times b = 1$. This number $b$ is also known as the reciprocal of $a$, written as $\frac{1}{a}$ or $a^{-1}$.

We need to find the multiplicative inverse of $10^{-100}$.

Let the multiplicative inverse be $x$. Then, $10^{-100} \times x = 1$.

So, $x = \frac{1}{10^{-100}}$.

Using the property of exponents that states $\frac{1}{a^{-n}} = a^n$, we can simplify $\frac{1}{10^{-100}}$.

Here, $a=10$ and $n=100$.

So, $\frac{1}{10^{-100}} = 10^{100}$.

Thus, the multiplicative inverse of $10^{-100}$ is $10^{100}$.

Alternatively, the multiplicative inverse of $a^m$ is $a^{-m}$. In this case, $a=10$ and $m=-100$. The multiplicative inverse is $10^{-(-100)} = 10^{100}$.

Comparing this result with the given options:

(a) 10 - Incorrect

(b) 100 - Incorrect

(c) $10^{100}$ - Correct

(d) $10^{-100}$ - Incorrect (This is the number itself)

The correct answer is (c) $10^{100}$.

We are asked to find the value of the expression $(–2)^{2 \times 3 – 1}$.

First, let's simplify the exponent by following the order of operations (multiplication before subtraction):

Exponent = $2 \times 3 - 1 = 6 - 1 = 5$

So, the expression becomes $(-2)^5$.

Now, we evaluate $(-2)^5$, which means multiplying $-2$ by itself 5 times:

$(-2)^5 = (-2) \times (-2) \times (-2) \times (-2) \times (-2)$

Let's perform the multiplication step-by-step:

$(-2) \times (-2) = 4$

$4 \times (-2) = -8$

$(-8) \times (-2) = 16$

$16 \times (-2) = -32$

Alternatively, since the base is negative and the exponent is an odd number (5), the result will be negative.

$(-2)^5 = -(2^5) = -(2 \times 2 \times 2 \times 2 \times 2) = -(32) = -32$

Thus, the value of the expression is $-32$.

Comparing this result with the given options:

(a) 32 - Incorrect

(b) 64 - Incorrect

(c) – 32 - Correct

(d) -64 - Incorrect

The correct answer is (c) – 32.

We need to find the value of $ \left( -\frac{2}{3} \right)^{4} $. This means we multiply the fraction $ -\frac{2}{3} $ by itself 4 times.

When a negative number is raised to an even power, the result is positive.

$ \left( -\frac{2}{3} \right)^{4} = \left( -\frac{2}{3} \right) \times \left( -\frac{2}{3} \right) \times \left( -\frac{2}{3} \right) \times \left( -\frac{2}{3} \right) $

First, we calculate the numerator:

$ (-2) \times (-2) \times (-2) \times (-2) $

$ = (4) \times (4) $

$ = 16 $

Next, we calculate the denominator:

$ 3 \times 3 \times 3 \times 3 $

$ = (9) \times (9) $

$ = 81 $

So, the value of $ \left( -\frac{2}{3} \right)^{4} $ is $ \frac{16}{81} $.

Now, we compare this result with the given options:

(a) $ \frac{16}{18} $

(b) $ \frac{81}{16} $

(c) $ \frac{-16}{81} $

(d) $ \frac{81}{-16} $

The calculated value $ \frac{16}{81} $ does not exactly match any of the options as written. However, option (a) has the correct numerator (16) and is the most likely option to be a typo for $ \frac{16}{81} $.

Assuming there is a typographical error in the options and that option (a) was intended to be $ \frac{16}{81} $, we select option (a).

The correct answer is (a) $\frac{16}{18}$ (assuming a typo in the option).

The multiplicative inverse of a non-zero number 'a' is the number 'b' such that $ a \times b = 1 $. This number 'b' is equal to $ \frac{1}{a} $.

The given expression is $ \left( -\frac{5}{9} \right)^{-99} $.

Let $ a = \left( -\frac{5}{9} \right)^{-99} $.

The multiplicative inverse is $ \frac{1}{a} = \frac{1}{\left( -\frac{5}{9} \right)^{-99}} $.

We use the property of exponents that states $ \frac{1}{x^{-m}} = x^m $.

Here, $ x = -\frac{5}{9} $ and $ m = 99 $.

Applying this property, we get:

$ \frac{1}{\left( -\frac{5}{9} \right)^{-99}} = \left( -\frac{5}{9} \right)^{99} $

So, the multiplicative inverse of $ \left( -\frac{5}{9} \right)^{-99} $ is $ \left( -\frac{5}{9} \right)^{99} $.

Now, we compare this result with the given options:

(a) $ \left( -\frac{5}{9} \right)^{99} $

(b) $ \left( \frac{5}{9} \right)^{99} $

(c) $ \left( \frac{9}{- 5} \right)^{99} $

(d) $ \left( \frac{9}{5} \right)^{99} $

The result $ \left( -\frac{5}{9} \right)^{99} $ matches option (a).

The correct answer is (a) $\left( -\frac{5}{9} \right)^{99}$.

We are given the expression $ x^m \times x^n $, where $x$ is a non-zero integer and $m$, $n$ are negative integers.

We use the property of exponents which states that for any non-zero base $a$ and any integers $m$ and $n$, $ a^m \times a^n = a^{m+n} $.

Applying this property to the given expression, where the base is $x$, we have:

$ x^m \times x^n = x^{m+n} $

This rule holds true regardless of whether the exponents $m$ and $n$ are positive, negative, or zero, as long as the base $x$ is non-zero.

Comparing this result with the given options:

(a) $ x^m $

(b) $ x^{m + n} $

(c) $ x^n $

(d) $ x^{m – n} $

Our result $ x^{m+n} $ matches option (b).

The correct answer is (b) $x^{m + n}$.

We are asked to find the value of $ y^0 $, where $ y $ is any non-zero integer.

There is a fundamental property of exponents which states that any non-zero number raised to the power of zero is equal to 1.

Mathematically, for any non-zero number $ a $, we have:

$ a^0 = 1 $

In this question, the base is $ y $, and it is given that $ y $ is a non-zero integer. Therefore, we can apply the property directly.

$ y^0 = 1 $

Comparing this result with the given options:

(a) 1

(b) 0

(c) – 1

(d) Not defined

Our result $ 1 $ matches option (a).

The correct answer is (a) 1.

We are asked to find the value of $ x^{-1} $, where $ x $ is any non-zero integer.

We use the property of exponents which states that for any non-zero base $ a $ and any positive integer $ m $, $ a^{-m} = \frac{1}{a^m} $.

In this specific case, the exponent is $ -1 $. So, we can write $ x^{-1} $ using the property with $ a=x $ and $ m=1 $:

$ x^{-1} = \frac{1}{x^1} $

Since $ x^1 = x $, we have:

$ x^{-1} = \frac{1}{x} $

Comparing this result with the given options:

(a) $ x $

(b) $ \frac{1}{x} $

(c) $ -x $

(d) $ \frac{-1}{x} $

Our result $ \frac{1}{x} $ matches option (b).

The correct answer is (b) $\frac{1}{x}$.

We are asked to find the value of $ x^{-m} $, where $ x $ is any integer different from zero and $ m $ is any positive integer.

We use the definition of a negative exponent. For any non-zero number $ a $ and any positive integer $ m $, the negative exponent $ a^{-m} $ is defined as the reciprocal of $ a^m $.

The property is stated as:

$ a^{-m} = \frac{1}{a^m} $

In this question, the base is $ x $, and the positive integer exponent is $ m $. Applying the property, we get:

$ x^{-m} = \frac{1}{x^m} $

Comparing this result with the given options:

(a) $ x^{m} $

(b) $ -x^{m} $

(c) $ \frac{1}{x^m} $

(d) $ \frac{-1}{x^m} $

Our result $ \frac{1}{x^m} $ matches option (c).

The correct answer is (c) $\frac{1}{x^m}$.

We are given the expression $ (x^m)^n $, where $x$ is a non-zero integer and $m$, $n$ are any integers.

We use the property of exponents known as the Power of a Power rule, which states that for any non-zero base $a$ and any integers $m$ and $n$, $ (a^m)^n = a^{m \times n} = a^{mn} $.

Applying this property to the given expression, where the base is $x$, we multiply the exponents $m$ and $n$:

$ (x^m)^n = x^{m \times n} = x^{mn} $

This rule holds true for all integer values of $m$ and $n$, as long as the base $x$ is non-zero.

Comparing this result with the given options:

(a) $ x^{m + n} $

(b) $ x^{mn} $

(c) $ x^{\frac{m}{n}} $

(d) $ x^{m – n} $

Our result $ x^{mn} $ matches option (b).

The correct answer is (b) $x^{mn}$.

We are asked to find the value equal to $ \left( -\frac{3}{4} \right)^{-3} $.

We use the property of exponents which states that for any non-zero fraction $ \frac{a}{b} $ and any integer $ m $, $ \left(\frac{a}{b}\right)^{-m} = \left(\frac{b}{a}\right)^m $.

In the given expression, the base is $ -\frac{3}{4} $ and the exponent is $ -3 $. Here, $ a = -3 $ and $ b = 4 $, or we can consider the base as $ \frac{-3}{4} $.

Applying the property $ \left(\frac{a}{b}\right)^{-m} = \left(\frac{b}{a}\right)^m $ to $ \left( -\frac{3}{4} \right)^{-3} $, we get:

$ \left( -\frac{3}{4} \right)^{-3} = \left( \frac{4}{-3} \right)^{3} $

Since $ \frac{4}{-3} $ is the same as $ -\frac{4}{3} $, we can write the expression as:

$ \left( \frac{4}{-3} \right)^{3} = \left( -\frac{4}{3} \right)^{3} $

Alternatively, we can first use the property $ x^{-m} = \frac{1}{x^m} $:

$ \left( -\frac{3}{4} \right)^{-3} = \frac{1}{\left( -\frac{3}{4} \right)^{3}} $

Now, calculate the denominator $ \left( -\frac{3}{4} \right)^{3} $:

$ \left( -\frac{3}{4} \right)^{3} = \left( -\frac{3}{4} \right) \times \left( -\frac{3}{4} \right) \times \left( -\frac{3}{4} \right) = \frac{(-3) \times (-3) \times (-3)}{4 \times 4 \times 4} = \frac{-27}{64} $

Substitute this back:

$ \frac{1}{\left( -\frac{3}{4} \right)^{3}} = \frac{1}{\frac{-27}{64}} = 1 \times \frac{64}{-27} = \frac{64}{-27} = -\frac{64}{27} $

Now let's evaluate the options:

(a) $ \left( \frac{3}{4} \right)^{-3} = \left( \frac{4}{3} \right)^{3} = \frac{4^3}{3^3} = \frac{64}{27} $

(b) $ -\left( \frac{3}{4} \right)^{-3} = - \left( \left( \frac{3}{4} \right)^{-3} \right) = - \left( \left( \frac{4}{3} \right)^{3} \right) = - \frac{64}{27} $

(c) $ \left( \frac{4}{3} \right)^{3} = \frac{64}{27} $

(d) $ \left( -\frac{4}{3} \right)^{3} = \left( -\frac{4}{3} \right) \times \left( -\frac{4}{3} \right) \times \left( -\frac{4}{3} \right) = \frac{(-4)^3}{3^3} = \frac{-64}{27} = -\frac{64}{27} $

Both option (b) and option (d) are equal to $ -\frac{64}{27} $, which is the value of the original expression $ \left( -\frac{3}{4} \right)^{-3} $.

However, the direct application of the property $ \left(\frac{a}{b}\right)^{-m} = \left(\frac{b}{a}\right)^m $ to $ \left(-\frac{3}{4}\right)^{-3} $ yields $ \left(\frac{4}{-3}\right)^{3} $ which is $ \left(-\frac{4}{3}\right)^{3} $. This form matches option (d) exactly.

The correct answer is (d) $\left( -\frac{4}{3} \right)^{3}$.

We are asked to find the value equal to $ \left( -\frac{5}{7} \right)^{-5} $.

We use the property of exponents which states that for any non-zero fraction $ \frac{a}{b} $ and any integer $ m $, $ \left(\frac{a}{b}\right)^{-m} = \left(\frac{b}{a}\right)^m $.

In the given expression, the base is $ -\frac{5}{7} $ and the exponent is $ -5 $.

We can write $ -\frac{5}{7} $ as $ \frac{-5}{7} $.

Applying the property $ \left(\frac{a}{b}\right)^{-m} = \left(\frac{b}{a}\right)^m $ to $ \left( \frac{-5}{7} \right)^{-5} $, where $ a = -5 $, $ b = 7 $ and $ m = 5 $, we get:

$ \left( \frac{-5}{7} \right)^{-5} = \left( \frac{7}{-5} \right)^{5} $

Since $ \frac{7}{-5} $ is the same as $ -\frac{7}{5} $, the expression becomes:

$ \left( \frac{7}{-5} \right)^{5} = \left( -\frac{7}{5} \right)^{5} $

Comparing this result with the given options:

(a) $ \left( \frac{5}{7} \right)^{-5} = \left( \frac{7}{5} \right)^{5} $

(b) $ \left( \frac{5}{7} \right)^{5} $

(c) $ \left( \frac{7}{5} \right)^{5} $

(d) $ \left( -\frac{7}{5} \right)^{5} $

Our result $ \left( -\frac{7}{5} \right)^{5} $ matches option (d).

The correct answer is (d) $\left( -\frac{7}{5} \right)^{5}$.

We are asked to find the value of $ \left( \frac{-7}{5} \right)^{-1} $.

We use the property of exponents which states that for any non-zero number $ a $, $ a^{-1} = \frac{1}{a} $.

Alternatively, for a non-zero fraction $ \frac{a}{b} $, the property is $ \left(\frac{a}{b}\right)^{-1} = \frac{b}{a} $.

In the given expression, the base is $ \frac{-7}{5} $. Applying the property $ \left(\frac{a}{b}\right)^{-1} = \frac{b}{a} $, where $ a = -7 $ and $ b = 5 $, we get:

$ \left( \frac{-7}{5} \right)^{-1} = \frac{5}{-7} $

The fraction $ \frac{5}{-7} $ is equivalent to $ -\frac{5}{7} $.

So, $ \left( \frac{-7}{5} \right)^{-1} = -\frac{5}{7} $.

Comparing this result with the given options:

(a) $ \frac{5}{7} $

(b) $ -\frac{5}{7} $

(c) $ \frac{7}{5} $

(d) $ \frac{-7}{5} $

Our result $ -\frac{5}{7} $ matches option (b).

The correct answer is (b) $-\frac{5}{7}$.

We are asked to evaluate the expression $ (-9)^3 \div (-9)^8 $.

We use the property of exponents for division with the same base, which states that for any non-zero base $ a $ and any integers $ m $ and $ n $, $ a^m \div a^n = a^{m-n} $.

In this expression, the base is $ -9 $, the exponent in the numerator is $ m=3 $, and the exponent in the denominator is $ n=8 $.

Applying the property, we subtract the exponents:

$ (-9)^3 \div (-9)^8 = (-9)^{3 - 8} $

$ = (-9)^{-5} $

So, $ (-9)^3 \div (-9)^8 $ is equal to $ (-9)^{-5} $.

Comparing this result with the given options:

(a) $ (9)^5 $

(b) $ (9)^{-5} $

(c) $ (- 9)^5 $

(d) $ (- 9)^{-5} $

Our result $ (-9)^{-5} $ matches option (d).

The correct answer is (d) (– 9)^–5.

We are asked to evaluate the expression $ x^7 \div x^{12} $, where $x$ is a non-zero integer.

We use the property of exponents for division with the same base, which states that for any non-zero base $ a $ and any integers $ m $ and $ n $, $ a^m \div a^n = a^{m-n} $.

In this expression, the base is $ x $, the exponent in the numerator is $ m=7 $, and the exponent in the denominator is $ n=12 $.

Applying the property, we subtract the exponents:

$ x^7 \div x^{12} = x^{7 - 12} $

$ = x^{-5} $

So, $ x^7 \div x^{12} $ is equal to $ x^{-5} $.

Comparing this result with the given options:

(a) $ x^5 $

(b) $ x^{19} $

(c) $ x^{–5} $

(d) $ x^{–19} $

Our result $ x^{-5} $ matches option (c).

The correct answer is (c) $x^{–5}$.

We are asked to evaluate the expression $ (x^4)^{-3} $, where $ x $ is a non-zero integer.

We use the property of exponents known as the Power of a Power rule, which states that for any non-zero base $ a $ and any integers $ m $ and $ n $, $ (a^m)^n = a^{m \times n} = a^{mn} $.

In this expression, the base is $ x $, the inner exponent is $ m=4 $, and the outer exponent is $ n=-3 $.

Applying the property, we multiply the exponents $ 4 $ and $ -3 $:

$ (x^4)^{-3} = x^{4 \times (-3)} $

$ = x^{-12} $

So, $ (x^4)^{-3} $ is equal to $ x^{-12} $.

Comparing this result with the given options:

(a) $ x^{12} $

(b) $ x^{–12} $

(c) $ x^{64} $

(d) $ x^{–64} $

Our result $ x^{-12} $ matches option (b).

The correct answer is (b) $x^{–12}$.

We need to evaluate the expression $ (7^{-1} – 8^{-1})^{-1} – (3^{-1} – 4^{-1})^{-1} $.

First, let's evaluate the terms inside the first parenthesis: $ (7^{-1} – 8^{-1})^{-1} $.

Using the property $ a^{-1} = \frac{1}{a} $:

$ 7^{-1} = \frac{1}{7} $

$ 8^{-1} = \frac{1}{8} $

So, $ 7^{-1} – 8^{-1} = \frac{1}{7} – \frac{1}{8} $.

To subtract these fractions, we find a common denominator, which is the LCM of 7 and 8. LCM(7, 8) = $ 7 \times 8 = 56 $.

$ \frac{1}{7} – \frac{1}{8} = \frac{1 \times 8}{7 \times 8} – \frac{1 \times 7}{8 \times 7} = \frac{8}{56} – \frac{7}{56} = \frac{8-7}{56} = \frac{1}{56} $

Now, we evaluate the outer exponent: $ \left( \frac{1}{56} \right)^{-1} $.

Using the property $ \left( \frac{a}{b} \right)^{-1} = \frac{b}{a} $:

$ \left( \frac{1}{56} \right)^{-1} = \frac{56}{1} = 56 $

Next, let's evaluate the terms inside the second parenthesis: $ (3^{-1} – 4^{-1})^{-1} $.

Using the property $ a^{-1} = \frac{1}{a} $:

$ 3^{-1} = \frac{1}{3} $

$ 4^{-1} = \frac{1}{4} $

So, $ 3^{-1} – 4^{-1} = \frac{1}{3} – \frac{1}{4} $.

To subtract these fractions, we find a common denominator, which is the LCM of 3 and 4. LCM(3, 4) = $ 3 \times 4 = 12 $.

$ \frac{1}{3} – \frac{1}{4} = \frac{1 \times 4}{3 \times 4} – \frac{1 \times 3}{4 \times 3} = \frac{4}{12} – \frac{3}{12} = \frac{4-3}{12} = \frac{1}{12} $

Now, we evaluate the outer exponent: $ \left( \frac{1}{12} \right)^{-1} $.

Using the property $ \left( \frac{a}{b} \right)^{-1} = \frac{b}{a} $:

$ \left( \frac{1}{12} \right)^{-1} = \frac{12}{1} = 12 $

Finally, we subtract the second result from the first result:

$ (7^{-1} – 8^{-1})^{-1} – (3^{-1} – 4^{-1})^{-1} = 56 – 12 $

$ 56 - 12 = 44 $

Comparing this result with the given options:

(a) 44

(b) 56

(c) 68

(d) 12

Our result 44 matches option (a).

The correct answer is (a) 44.

To write a number in standard form (scientific notation), we express it as the product of a number between 1 (inclusive) and 10 (exclusive), and a power of 10.

The general form is $ a \times 10^b $, where $ 1 \le a < 10 $ and $ b $ is an integer.

The given number is 0.000064.

We need to move the decimal point to the right until it is placed after the first non-zero digit, which is 6.

0.000064

Move the decimal point past the first 0, second 0, third 0, fourth 0, fifth 0, and then after the 6.

The new position of the decimal point is between 6 and 4, giving us the number 6.4.

The number of places the decimal point was moved is 5.

Since the original number is less than 1 (a small number), the exponent of 10 will be negative.

The number of places moved is 5, so the exponent is -5.

Thus, the standard form of 0.000064 is $ 6.4 \times 10^{-5} $.

Comparing this result with the given options:

(a) $ 64 \times 10^4 $

(b) $ 64 \times 10^{-4} $

(c) $ 6.4 \times 10^5 $

(d) $ 6.4 \times 10^{-5} $

Our result $ 6.4 \times 10^{-5} $ matches option (d).

The correct answer is (d) $6.4 \times 10^{–5}$.

To write a number in standard form (scientific notation), we express it as the product of a number between 1 (inclusive) and 10 (exclusive), and a power of 10.

The general form is $ a \times 10^b $, where $ 1 \le a < 10 $ and $ b $ is an integer.

The given number is 234000000.

For a whole number, the decimal point is implicitly at the end, so we can write it as 234000000.

To get a number between 1 and 10, we need to move the decimal point to the left until it is after the first non-zero digit (which is 2).

We move the decimal point from its current position 8 places to the left:

$ 234000000. \to 2.34000000 $

The resulting number between 1 and 10 is 2.34.

Since we moved the decimal point 8 places to the left, the exponent of 10 is positive 8 (because the original number is large).

So, the standard form of 234000000 is $ 2.34 \times 10^{8} $.

Comparing this result with the given options:

(a) $ 2.34 \times 10^8 $

(b) $ 0.234 \times 10^9 $ (Not in standard form as $0.234 < 1$)

(c) $ 2.34 \times 10^{–8} $ (Incorrect exponent sign)

(d) $ 0.234 \times 10^{–9} $ (Not in standard form and incorrect exponent sign)

Our result $ 2.34 \times 10^8 $ matches option (a).

The correct answer is (a) $2.34 \times 10^{8}$.

We are given the number in standard form as $ 2.03 \times 10^{-5} $ and asked to convert it to its usual form.

The standard form of a number is $ a \times 10^b $, where $ 1 \le a < 10 $ and $ b $ is an integer. To convert from standard form to usual form, we move the decimal point of the number $ a $ by $ b $ places.

If the exponent $ b $ is positive, we move the decimal point to the right.

If the exponent $ b $ is negative, we move the decimal point to the left.

In the given number $ 2.03 \times 10^{-5} $, the number $ a = 2.03 $ and the exponent $ b = -5 $.

Since the exponent is -5, we need to move the decimal point in 2.03 five places to the left. We add leading zeros as needed.

Starting with 2.03:

Move 1 place left: 0.203

Move 2 places left: 0.0203

Move 3 places left: 0.00203

Move 4 places left: 0.000203

Move 5 places left: 0.0000203

So, the usual form of $ 2.03 \times 10^{-5} $ is 0.0000203.

Comparing this result with the given options:

(a) 0.203

(b) 0.00203

(c) 203000

(d) 0.0000203

Our result 0.0000203 matches option (d).

The correct answer is (d) 0.0000203.

We are asked to find the value of $ \left( \frac{1}{10} \right)^{0} $.

There is a fundamental property of exponents which states that any non-zero number raised to the power of zero is equal to 1.

Mathematically, for any non-zero number $ a $, we have:

$ a^0 = 1 $

In this question, the base is $ \frac{1}{10} $. Since $ \frac{1}{10} $ is a non-zero number, we can apply the property directly.

$ \left( \frac{1}{10} \right)^{0} = 1 $

Comparing this result with the given options:

(a) 0

(b) $ \frac{1}{10} $

(c) 1

(d) 10

Our result 1 matches option (c).

The correct answer is (c) 1.

We are asked to evaluate the expression $ \left( \frac{3}{4} \right)^{5} \div \left( \frac{5}{3} \right)^{5} $.

We can use the property of exponents for division of terms with the same exponent but different bases, which states that for any non-zero numbers $ a $ and $ b $, and any integer $ m $, $ a^m \div b^m = \left(\frac{a}{b}\right)^m $.

In this expression, the bases are $ a = \frac{3}{4} $ and $ b = \frac{5}{3} $, and the exponent is $ m=5 $.

Applying the property, we divide the bases and keep the exponent:

$ \left( \frac{3}{4} \right)^{5} \div \left( \frac{5}{3} \right)^{5} = \left( \frac{\frac{3}{4}}{\frac{5}{3}} \right)^{5} $

The division of fractions inside the parenthesis is calculated as:

$ \frac{\frac{3}{4}}{\frac{5}{3}} = \frac{3}{4} \div \frac{5}{3} = \frac{3}{4} \times \frac{3}{5} = \frac{3 \times 3}{4 \times 5} = \frac{9}{20} $

So, the expression becomes $ \left( \frac{9}{20} \right)^{5} $.

Now let's look at the structure of the given options. They are in the form $ \left(\frac{3}{4} \div \frac{5}{3} \right)^{\text{exponent}} $.

We know that $ \frac{3}{4} \div \frac{5}{3} = \frac{9}{20} $.

So the options are equivalent to $ \left( \frac{9}{20} \right)^{\text{exponent}} $.

Comparing our result $ \left( \frac{9}{20} \right)^{5} $ with the form $ \left( \frac{9}{20} \right)^{\text{exponent}} $, we see that the exponent must be 5.

Let's check the exponents in the options:

(a) The exponent is 5.

(b) The exponent is 1.

(c) The exponent is 0.

(d) The exponent is 10.

Only option (a) has the correct exponent 5.

The correct answer is (a) $\left(\frac{3}{4} \div \frac{5}{3} \right)^{5}$.

We are asked to evaluate the expression $ x^4 \div y^4 $, where $ x $ and $ y $ are any two non-zero rational numbers.

We use the property of exponents for division where the bases are different but the exponents are the same. This property states that for any non-zero numbers $ a $ and $ b $, and any integer $ m $, $ a^m \div b^m = \left(\frac{a}{b}\right)^m $.

In the given expression, the bases are $ x $ and $ y $, and the exponent is $ m=4 $.

Applying the property, we divide the bases and keep the exponent:

$ x^4 \div y^4 = \left(\frac{x}{y}\right)^4 $

Since $ \frac{x}{y} $ is equivalent to $ x \div y $, we can write the result as:

$ x^4 \div y^4 = (x \div y)^4 $

Comparing this result with the given options:

(a) $ (x \div y)^0 $

(b) $ (x \div y)^1 $

(c) $ (x \div y)^4 $

(d) $ (x \div y)^8 $

Our result $ (x \div y)^4 $ matches option (c).

The correct answer is (c) $(x \div y)^{4}$.

We are asked to evaluate the expression $ p^{13} \div p^{8} $, where $ p $ is a non-zero rational number.

We use the property of exponents for division with the same base, which states that for any non-zero base $ a $ and any integers $ m $ and $ n $, $ a^m \div a^n = a^{m-n} $.

In this expression, the base is $ p $, the exponent in the numerator is $ m=13 $, and the exponent in the denominator is $ n=8 $.

Applying the property, we subtract the exponents:

$ p^{13} \div p^{8} = p^{13 - 8} $

$ = p^{5} $

So, $ p^{13} \div p^{8} $ is equal to $ p^{5} $.

Comparing this result with the given options:

(a) $ p^{5} $

(b) $ p^{21} $

(c) $ p^{–5} $

(d) $ p^{–19} $

Our result $ p^{5} $ matches option (a).

The correct answer is (a) $p^{5}$.

We are asked to evaluate the expression $ (z^{-2})^{3} $, where $ z $ is a non-zero rational number.

We use the property of exponents known as the Power of a Power rule, which states that for any non-zero base $ a $ and any integers $ m $ and $ n $, $ (a^m)^n = a^{m \times n} = a^{mn} $.

In this expression, the base is $ z $, the inner exponent is $ m=-2 $, and the outer exponent is $ n=3 $.

Applying the property, we multiply the exponents $ -2 $ and $ 3 $:

$ (z^{-2})^{3} = z^{(-2) \times 3} $

$ = z^{-6} $

So, $ (z^{-2})^{3} $ is equal to $ z^{-6} $.

Comparing this result with the given options:

(a) $ z^6 $

(b) $ z^{–6} $

(c) $ z^1 $

(d) $ z^4 $

Our result $ z^{-6} $ matches option (b).

The correct answer is (b) $z^{–6}$.

We are asked to find the cube of $ -\frac{1}{2} $. The cube of a number is the number multiplied by itself three times.

So, we need to calculate $ \left( -\frac{1}{2} \right)^{3} $.

$ \left( -\frac{1}{2} \right)^{3} = \left( -\frac{1}{2} \right) \times \left( -\frac{1}{2} \right) \times \left( -\frac{1}{2} \right) $

To multiply fractions, we multiply the numerators together and the denominators together.

Numerator: $ (-1) \times (-1) \times (-1) $

$ = (1) \times (-1) $

$ = -1 $

Denominator: $ 2 \times 2 \times 2 $

$ = 4 \times 2 $

$ = 8 $

So, $ \left( -\frac{1}{2} \right)^{3} = \frac{-1}{8} = -\frac{1}{8} $.

Comparing this result with the given options:

(a) $ \frac{1}{8} $

(b) $ \frac{1}{16} $

(c) $ -\frac{1}{8} $

(d) $ -\frac{1}{16} $

Our result $ -\frac{1}{8} $ matches option (c).

The correct answer is (c) $-\frac{1}{8}$.

The reciprocal of a non-zero number $ a $ is $ \frac{1}{a} $ or $ a^{-1} $.

We need to find the reciprocal of $ \left( \frac{2}{3} \right)^{4} $.

The reciprocal is $ \frac{1}{\left( \frac{2}{3} \right)^{4}} $.

Using the property $ \frac{1}{x^m} = x^{-m} $, we can write:

$ \frac{1}{\left( \frac{2}{3} \right)^{4}} = \left( \frac{2}{3} \right)^{-4} $

Using the property $ \left(\frac{a}{b}\right)^{-m} = \left(\frac{b}{a}\right)^m $, we can rewrite $ \left( \frac{2}{3} \right)^{-4} $ as:

$ \left( \frac{2}{3} \right)^{-4} = \left( \frac{3}{2} \right)^{4} $

Using the property $ \left(\frac{a}{b}\right)^m = \frac{a^m}{b^m} $, we can rewrite $ \left( \frac{3}{2} \right)^{4} $ as:

$ \left( \frac{3}{2} \right)^{4} = \frac{3^{4}}{2^{4}} $

So, the reciprocal of $ \left( \frac{2}{3} \right)^{4} $ is equal to $ \left( \frac{2}{3} \right)^{-4} $, $ \left( \frac{3}{2} \right)^{4} $, and $ \frac{3^{4}}{2^{4}} $.

Now, let's examine the given options:

(a) $ \left( \frac{3}{2} \right)^{4} $ - This is equal to the reciprocal.

(b) $ \left( \frac{3}{2} \right)^{-4} $ - Using the property $ \left(\frac{a}{b}\right)^{-m} = \left(\frac{b}{a}\right)^m $, $ \left( \frac{3}{2} \right)^{-4} = \left( \frac{2}{3} \right)^{4} $. This is the original number, not its reciprocal.

(c) $ \left( \frac{2}{3} \right)^{-4} $ - This is equal to the reciprocal.

(d) $ \frac{3^4}{2^4} $ - This is equal to $ \left( \frac{3}{2} \right)^{4} $, which is the reciprocal.

The option that is not the reciprocal of $ \left( \frac{2}{3} \right)^{4} $ is $ \left( \frac{3}{2} \right)^{-4} $.

The correct answer is (b) $\left( \frac{3}{2} \right)^{-4}$.

The multiplicative inverse of a non-zero number $ a $ is the number $ b $ such that $ a \times b = 1 $.

This means the multiplicative inverse of $ a $ is $ \frac{1}{a} $.

The given number is $ 10^{10} $.

Its multiplicative inverse is $ \frac{1}{10^{10}} $.

Using the property of exponents $ \frac{1}{a^m} = a^{-m} $, we can write:

$ \frac{1}{10^{10}} = 10^{-10} $

So, the multiplicative inverse of $ 10^{10} $ is $ 10^{-10} $.

The multiplicative inverse of $ 10^{10} $ is $10^{-10}$.

We need to evaluate the expression $ a^3 \times a^{-10} $.

We use the property of exponents for multiplication with the same base, which states that for any non-zero base $ x $ and any integers $ m $ and $ n $, $ x^m \times x^n = x^{m+n} $.

In this expression, the base is $ a $, the first exponent is $ m=3 $, and the second exponent is $ n=-10 $.

Applying the property, we add the exponents:

$ a^3 \times a^{-10} = a^{3 + (-10)} $

$ = a^{3 - 10} $

$ = a^{-7} $

So, $ a^3 \times a^{-10} $ is equal to $ a^{-7} $.

a³ × a^–10 = $a^{-7}$.

We need to find the value of $ 5^0 $.

There is a fundamental property of exponents which states that any non-zero number raised to the power of zero is equal to 1.

Mathematically, for any non-zero number $ a $, we have:

$ a^0 = 1 $

In this case, the base is 5, which is a non-zero number. Applying the property, we get:

$ 5^0 = 1 $

5⁰ = 1.

We need to evaluate the expression $ 5^5 \times 5^{-5} $.

We use the property of exponents for multiplication with the same base, which states that for any non-zero base $ a $ and any integers $ m $ and $ n $, $ a^m \times a^n = a^{m+n} $.

In this expression, the base is 5, the first exponent is $ m=5 $, and the second exponent is $ n=-5 $.

Applying the property, we add the exponents:

$ 5^5 \times 5^{-5} = 5^{5 + (-5)} $

$ = 5^{5 - 5} $

$ = 5^0 $

Now, we use the property that any non-zero number raised to the power of zero is 1:

$ 5^0 = 1 $

So, $ 5^5 \times 5^{-5} $ is equal to 1.

5⁵ × 5^–5 = 1.

We need to evaluate the expression $ \left( \frac{1}{2^3} \right)^2 $.

First, let's evaluate the expression inside the parenthesis: $ \frac{1}{2^3} $.

$ 2^3 = 2 \times 2 \times 2 = 8 $

So, $ \frac{1}{2^3} = \frac{1}{8} $.

Now, we need to find the square of $ \frac{1}{8} $:

$ \left( \frac{1}{8} \right)^2 = \left( \frac{1}{8} \right) \times \left( \frac{1}{8} \right) $

Multiply the numerators and the denominators:

$ = \frac{1 \times 1}{8 \times 8} = \frac{1}{64} $

Alternatively, we can use the property of exponents $ \left(\frac{a}{b}\right)^m = \frac{a^m}{b^m} $ and $ (a^m)^n = a^{mn} $.

$ \left( \frac{1}{2^3} \right)^2 = \frac{1^2}{(2^3)^2} $

$ 1^2 = 1 \times 1 = 1 $

$ (2^3)^2 = 2^{3 \times 2} = 2^6 $

$ 2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64 $

So, $ \frac{1^2}{(2^3)^2} = \frac{1}{2^6} = \frac{1}{64} $.

The value of $ \left( \frac{1}{2^3} \right)^2 $ is equal to $ \frac{1}{64} $.

The value of $\left( \frac{1}{2^3} \right)^2$ is equal to $\frac{1}{64}$.

We are asked to express $ 8^{-2} $ as a power with the base 2.

First, we need to express the base 8 as a power of 2. We know that $ 8 = 2 \times 2 \times 2 = 2^3 $.

Now, substitute $ 8 $ with $ 2^3 $ in the given expression $ 8^{-2} $:

$ 8^{-2} = (2^3)^{-2} $

We use the property of exponents known as the Power of a Power rule, which states that for any non-zero base $ a $ and any integers $ m $ and $ n $, $ (a^m)^n = a^{m \times n} = a^{mn} $.

Applying this property to $ (2^3)^{-2} $, where $ a=2 $, $ m=3 $, and $ n=-2 $, we multiply the exponents:

$ (2^3)^{-2} = 2^{3 \times (-2)} = 2^{-6} $

So, the expression for $ 8^{-2} $ as a power with the base 2 is $ 2^{-6} $.

The expression for 8^–2 as a power with the base 2 is $2^{-6}$.

Standard form (scientific notation) is used to represent very large or very small numbers conveniently.

A number in standard form is written as $ a \times 10^b $, where $ 1 \le a < 10 $ and $ b $ is an integer.

When we express a very small positive number (like 0.000001) in standard form, we move the decimal point to the right to get a number between 1 and 10. The number of places the decimal point is moved to the right determines the absolute value of the exponent.

For example, let's write 0.000001 in standard form:

0.000001 = $ 1 \times 10^{-6} $

Here, the decimal point was moved 6 places to the right, and the exponent is -6.

Another example, 0.000064 = $ 6.4 \times 10^{-5} $ (from a previous question). The decimal point was moved 5 places to the right, and the exponent is -5.

In general, for very small numbers (numbers between 0 and 1), the exponent of 10 in the standard form $ a \times 10^b $ is a negative integer.

Very small numbers can be expressed in standard form by using negative exponents.

Standard form (scientific notation) is used to represent very large or very small numbers conveniently.

A number in standard form is written as $ a \times 10^b $, where $ 1 \le a < 10 $ and $ b $ is an integer.

When we express a very large number (like 1,000,000) in standard form, we move the decimal point to the left until it is after the first non-zero digit to get a number between 1 and 10. The number of places the decimal point is moved to the left determines the absolute value of the exponent.

For example, let's write 1,000,000 in standard form:

1,000,000 = 1,000,000.0

Move the decimal point 6 places to the left: 1.000000

The number between 1 and 10 is 1.0.

The number of places the decimal point was moved is 6.

Since the original number is large (greater than 10), the exponent of 10 will be positive.

$ 1,000,000 = 1.0 \times 10^{6} $

Here, the exponent is +6.

Another example, 234000000 = $ 2.34 \times 10^8 $ (from a previous question). The decimal point was moved 8 places to the left, and the exponent is +8.

In general, for very large numbers (numbers greater than 10), the exponent of 10 in the standard form $ a \times 10^b $ is a positive integer.

Very large numbers can be expressed in standard form by using positive exponents.

We need to find the result of multiplying $ (10)^5 $ by $ (10)^{-10} $.

We have the expression $ 10^5 \times 10^{-10} $.

We use the property of exponents for multiplication with the same base, which states that for any non-zero base $ a $ and any integers $ m $ and $ n $, $ a^m \times a^n = a^{m+n} $.

In this expression, the base is 10, the first exponent is $ m=5 $, and the second exponent is $ n=-10 $.

Applying the property, we add the exponents:

$ 10^5 \times 10^{-10} = 10^{5 + (-10)} $

$ = 10^{5 - 10} $

$ = 10^{-5} $

So, by multiplying $ (10)^5 $ by $ (10)^{-10} $, we get $ 10^{-5} $.

By multiplying (10)⁵ by (10)^–10 we get $10^{-5}$.

We need to simplify the expression $ \left[ \left( \frac{2}{13} \right)^{-6}\div\left( \frac{2}{13} \right)^{3} \right]^3\times\left( \frac{2}{13} \right)^{-9} $.

Let the base be $ b = \frac{2}{13} $.

The expression can be written as $ \left[ b^{-6} \div b^{3} \right]^3 \times b^{-9} $.

First, simplify the expression inside the square brackets using the property $ a^m \div a^n = a^{m-n} $:

$ b^{-6} \div b^{3} = b^{-6 - 3} = b^{-9} $

Now, the expression inside the brackets is $ b^{-9} $. The expression becomes $ \left[ b^{-9} \right]^3 \times b^{-9} $.

Next, apply the outer exponent using the property $ (a^m)^n = a^{mn} $:

$ \left[ b^{-9} \right]^3 = b^{(-9) \times 3} = b^{-27} $

The expression is now $ b^{-27} \times b^{-9} $.

Finally, multiply the terms using the property $ a^m \times a^n = a^{m+n} $:

$ b^{-27} \times b^{-9} = b^{-27 + (-9)} = b^{-27 - 9} = b^{-36} $

Substitute the base $ b = \frac{2}{13} $ back into the result:

$ b^{-36} = \left( \frac{2}{13} \right)^{-36} $

The expression $ \left[ \left( \frac{2}{13} \right)^{-6}\div\left( \frac{2}{13} \right)^{3} \right]^3\times\left( \frac{2}{13} \right)^{-9} $ is equal to $ \left( \frac{2}{13} \right)^{-36} $.

$\left[ \left( \frac{2}{13} \right)^{-6}\div\left( \frac{2}{13} \right)^{3} \right]^3\times\left( \frac{2}{13} \right)^{-9}$ = $\left( \frac{2}{13} \right)^{-36}$.

We need to find the value of the expression $ [4^{-1} + 3^{-1} + 6^{-2}]^{-1} $.

First, we evaluate the terms inside the square brackets. We use the property $ a^{-m} = \frac{1}{a^m} $.

$ 4^{-1} = \frac{1}{4^1} = \frac{1}{4} $

$ 3^{-1} = \frac{1}{3^1} = \frac{1}{3} $

$ 6^{-2} = \frac{1}{6^2} = \frac{1}{36} $

Now, we sum these values:

$ 4^{-1} + 3^{-1} + 6^{-2} = \frac{1}{4} + \frac{1}{3} + \frac{1}{36} $

To add these fractions, we find the least common multiple (LCM) of the denominators 4, 3, and 36. The LCM(4, 3, 36) is 36.

Convert each fraction to have a denominator of 36:

$ \frac{1}{4} = \frac{1 \times 9}{4 \times 9} = \frac{9}{36} $

$ \frac{1}{3} = \frac{1 \times 12}{3 \times 12} = \frac{12}{36} $

$ \frac{1}{36} $ remains $ \frac{1}{36} $

Add the fractions:

$ \frac{9}{36} + \frac{12}{36} + \frac{1}{36} = \frac{9 + 12 + 1}{36} = \frac{22}{36} $

Simplify the fraction:

$ \frac{22}{36} = \frac{\cancel{22}^{11}}{\cancel{36}_{18}} = \frac{11}{18} $

So, the expression inside the brackets is $ \frac{11}{18} $.

Now, we need to find the value of $ \left( \frac{11}{18} \right)^{-1} $.

Using the property $ \left( \frac{a}{b} \right)^{-1} = \frac{b}{a} $:

$ \left( \frac{11}{18} \right)^{-1} = \frac{18}{11} $

Thus, the value of the expression $ [4^{-1} + 3^{-1} + 6^{-2}]^{-1} $ is $ \frac{18}{11} $.

The value [4^–1 + 3^–1 + 6^–2]^–1 is $\frac{18}{11}$.

We need to evaluate the expression $ [2^{-1} + 3^{-1} + 4^{-1}]^{0} $.

We observe that the entire expression inside the square brackets is raised to the power of 0.

According to the property of exponents, any non-zero number raised to the power of 0 is equal to 1.

Mathematically, for any non-zero number $ a $, $ a^0 = 1 $.

Let's evaluate the expression inside the brackets to confirm it is non-zero:

$ 2^{-1} = \frac{1}{2} $

$ 3^{-1} = \frac{1}{3} $

$ 4^{-1} = \frac{1}{4} $

Sum inside the brackets = $ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} $.

The LCM of 2, 3, and 4 is 12.

$ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6}{12} + \frac{4}{12} + \frac{3}{12} = \frac{6+4+3}{12} = \frac{13}{12} $

Since $ \frac{13}{12} $ is a non-zero number, raising it to the power of 0 gives 1.

$ \left( \frac{13}{12} \right)^{0} = 1 $

Thus, the value of the expression $ [2^{-1} + 3^{-1} + 4^{-1}]^{0} $ is 1.

[2^–1 + 3^–1 + 4^–1]⁰ = 1.

We need to express the number $ \frac{1}{100000000} $ in standard form.

First, let's write the denominator as a power of 10. Count the number of zeros in the denominator:

100,000,000 has 8 zeros.

So, $ 100000000 = 10^8 $.

The given number is $ \frac{1}{10^8} $.

Using the property of exponents $ \frac{1}{a^m} = a^{-m} $, we can write:

$ \frac{1}{10^8} = 10^{-8} $

The standard form is $ a \times 10^b $, where $ 1 \le a < 10 $. In the expression $ 10^{-8} $, we can write it as $ 1 \times 10^{-8} $.

Here, $ a = 1 $ and $ b = -8 $. This satisfies the condition for standard form ($ 1 \le 1 < 10 $ and -8 is an integer).

Alternatively, we can write the fraction as a decimal first:

$ \frac{1}{100000000} = 0.00000001 $

To write 0.00000001 in standard form, move the decimal point to the right until it is after the first non-zero digit (which is 1).

0.00000001

Move the decimal point 8 places to the right to get 1. The original number is less than 1, so the exponent is negative 8.

$ 0.00000001 = 1 \times 10^{-8} $

The standard form of $ \left( \frac{1}{100000000} \right) $ is $ 1 \times 10^{-8} $ (or simply $ 10^{-8} $).

The standard form of $\left( \frac{1}{100000000} \right)$ is $1 \times 10^{-8}$.

To write a number in standard form (scientific notation), we express it as the product of a number between 1 (inclusive) and 10 (exclusive), and a power of 10.

The general form is $ a \times 10^b $, where $ 1 \le a < 10 $ and $ b $ is an integer.

The given number is 12340000.

For a whole number, the decimal point is implicitly at the end, so we can write it as 12340000.

To get a number between 1 and 10, we need to move the decimal point to the left until it is after the first non-zero digit (which is 1).

We move the decimal point from its current position 7 places to the left:

$ 12340000. \to 1.2340000 $

The resulting number between 1 and 10 is 1.234.

Since we moved the decimal point 7 places to the left, the exponent of 10 is positive 7 (because the original number is large).

So, the standard form of 12340000 is $ 1.234 \times 10^{7} $.

The standard form of 12340000 is $1.234 \times 10^{7}$.

We are given the number in standard form as $ 3.41 \times 10^{6} $ and asked to convert it to its usual form.

The standard form of a number is $ a \times 10^b $, where $ 1 \le a < 10 $ and $ b $ is an integer. To convert from standard form to usual form, we move the decimal point of the number $ a $ by $ b $ places.

If the exponent $ b $ is positive, we move the decimal point to the right.

If the exponent $ b $ is negative, we move the decimal point to the left.

In the given number $ 3.41 \times 10^{6} $, the number $ a = 3.41 $ and the exponent $ b = 6 $.

Since the exponent is positive 6, we need to move the decimal point in 3.41 six places to the right. We add trailing zeros as needed.

Starting with 3.41:

Move 1 place right: 34.1

Move 2 places right: 341.

Move 3 places right: 3410.

Move 4 places right: 34100.

Move 5 places right: 341000.

Move 6 places right: 3410000.

So, the usual form of $ 3.41 \times 10^{6} $ is 3,410,000.

The usual form of 3.41 × 10⁶ is 3410000.

We are given the number in standard form as $ 2.39461 \times 10^{6} $ and asked to convert it to its usual form.

To convert from standard form $ a \times 10^b $ to usual form, we move the decimal point of the number $ a $ by $ b $ places.

If the exponent $ b $ is positive, we move the decimal point to the right.

If the exponent $ b $ is negative, we move the decimal point to the left.

In the given number $ 2.39461 \times 10^{6} $, the number $ a = 2.39461 $ and the exponent $ b = 6 $.

Since the exponent is positive 6, we need to move the decimal point in 2.39461 six places to the right. We add trailing zeros as needed.

Starting with 2.39461:

Move 1 place right: 23.9461

Move 2 places right: 239.461

Move 3 places right: 2394.61

Move 4 places right: 23946.1

Move 5 places right: 239461.

Move 6 places right: 2394610.

So, the usual form of $ 2.39461 \times 10^{6} $ is 2,394,610.

The usual form of 2.39461 × 10⁶ is 2394610.

We are given that $ 36 = 6^2 $ and asked to express $ \frac{1}{36} $ as a power with the base 6.

We have the expression $ \frac{1}{36} $.

Substitute 36 with $ 6^2 $:

$ \frac{1}{36} = \frac{1}{6^2} $

We use the property of exponents which states that for any non-zero base $ a $ and any positive integer $ m $, $ \frac{1}{a^m} = a^{-m} $.

Applying this property to $ \frac{1}{6^2} $, where $ a=6 $ and $ m=2 $, we get:

$ \frac{1}{6^2} = 6^{-2} $

So, $ \frac{1}{36} $ expressed as a power with the base 6 is $ 6^{-2} $.

If 36 = 6 × 6 = 6², then $\frac{1}{36}$ expressed as a power with the base 6 is $6^{-2}$.

Let the unknown number be $ N $.

The problem states that when $ \left( \frac{5}{3} \right)^{4} $ is multiplied by $ N $, the result is $ 5^4 $.

So, we have the equation: $ \left( \frac{5}{3} \right)^{4} \times N = 5^4 $.

We need to solve for $ N $. To isolate $ N $, we can divide both sides of the equation by $ \left( \frac{5}{3} \right)^{4} $.

$ N = \frac{5^4}{\left( \frac{5}{3} \right)^{4}} $

We use the property of exponents $ \left(\frac{a}{b}\right)^m = \frac{a^m}{b^m} $. So, $ \left( \frac{5}{3} \right)^{4} = \frac{5^4}{3^4} $.

Substitute this into the equation for $ N $:

$ N = \frac{5^4}{\frac{5^4}{3^4}} $

Dividing by a fraction is the same as multiplying by its reciprocal:

$ N = 5^4 \times \frac{3^4}{5^4} $

We can cancel out the common term $ 5^4 $ in the numerator and the denominator:

$ N = \cancel{5^4} \times \frac{3^4}{\cancel{5^4}} $

$ N = 3^4 $

Alternatively, we can use the property $ \frac{a^m}{b^m} = \left(\frac{a}{b}\right)^m $.

$ N = \frac{5^4}{\left( \frac{5}{3} \right)^{4}} $

$ N = \frac{5^4}{\frac{5^4}{3^4}} $

$ N = 5^4 \times \frac{3^4}{5^4} = \frac{5^4 \times 3^4}{5^4} $

Using $ a^m \times b^m = (ab)^m $, we have $ 5^4 \times 3^4 = (5 \times 3)^4 = 15^4 $. This is not helpful here.

Let's stick to cancellation:

$ N = \frac{5^4 \times 3^4}{5^4} = \frac{\cancel{5^4} \times 3^4}{\cancel{5^4}} = 3^4 $

Another approach: $ \left( \frac{5}{3} \right)^{4} \times N = 5^4 $

$ \frac{5^4}{3^4} \times N = 5^4 $

Multiply both sides by $ 3^4 $:

$ \frac{5^4}{3^4} \times N \times 3^4 = 5^4 \times 3^4 $

$ 5^4 \times N = 5^4 \times 3^4 $

Divide both sides by $ 5^4 $:

$ \frac{5^4 \times N}{5^4} = \frac{5^4 \times 3^4}{5^4} $

$ N = 3^4 $

So, the number we need to multiply by is $ 3^4 $.

By multiplying $\left( \frac{5}{3} \right)^{4}$ by $3^4$ we get 5⁴.

We need to simplify the expression $ 3^5 \div 3^{-6} $.

We use the property of exponents for division with the same base, which states that for any non-zero base $ a $ and any integers $ m $ and $ n $, $ a^m \div a^n = a^{m-n} $.

In this expression, the base is 3, the exponent in the numerator is $ m=5 $, and the exponent in the denominator is $ n=-6 $.

Applying the property, we subtract the exponents:

$ 3^5 \div 3^{-6} = 3^{5 - (-6)} $

$ = 3^{5 + 6} $

$ = 3^{11} $

So, $ 3^5 \div 3^{-6} $ can be simplified as $ 3^{11} $.

3⁵ ÷ 3^–6 can be simplified as $3^{11}$.

We are given the number in standard form as $ 3 \times 10^{-7} $ and asked to find its value in usual form.

To convert from standard form $ a \times 10^b $ to usual form, we move the decimal point of the number $ a $ by $ b $ places.

If the exponent $ b $ is positive, we move the decimal point to the right.

If the exponent $ b $ is negative, we move the decimal point to the left.

In the given number $ 3 \times 10^{-7} $, the number $ a = 3 $. We can write 3 as 3.0. The exponent is $ b = -7 $.

Since the exponent is negative 7, we need to move the decimal point in 3.0 seven places to the left. We add leading zeros as needed.

Starting with 3.0:

Move 1 place left: 0.3

Move 2 places left: 0.03

Move 3 places left: 0.003

Move 4 places left: 0.0003

Move 5 places left: 0.00003

Move 6 places left: 0.000003

Move 7 places left: 0.0000003

So, the value of $ 3 \times 10^{-7} $ in usual form is 0.0000003.

The value of 3 × 10^-7 is equal to 0.0000003.

When adding numbers written in standard form, such as $ a \times 10^m $ and $ c \times 10^n $, it is easiest to convert them so that they have the same power of 10 (i.e., the same exponent).

For example, to add $ 3 \times 10^5 $ and $ 2.5 \times 10^4 $, we can rewrite $ 2.5 \times 10^4 $ with an exponent of 5:

$ 2.5 \times 10^4 = 0.25 \times 10^5 $

Then we can add the numbers:

$ 3 \times 10^5 + 0.25 \times 10^5 = (3 + 0.25) \times 10^5 = 3.25 \times 10^5 $

Alternatively, we could rewrite $ 3 \times 10^5 $ with an exponent of 4:

$ 3 \times 10^5 = 30 \times 10^4 $

Then add:

$ 30 \times 10^4 + 2.5 \times 10^4 = (30 + 2.5) \times 10^4 = 32.5 \times 10^4 $

Note that $ 32.5 \times 10^4 $ is the same value as $ 3.25 \times 10^5 $ in standard form.

In both cases, the key step for addition is to make the exponents of 10 the same.

To add the numbers given in standard form, we first convert them into numbers with same exponents.

We need to write the number 32,50,00,00,000 in standard form (scientific notation).

The standard form is $ a \times 10^b $, where $ 1 \le a < 10 $ and $ b $ is an integer.

The given number is 32,500,000,000 (removing the commas for calculation).

For a whole number, the decimal point is implicitly at the end: 32500000000.

To get a number between 1 and 10, we need to move the decimal point to the left until it is after the first non-zero digit (which is 3).

Starting with 32500000000.:

We move the decimal point 10 places to the left to get 3.25.

The resulting number between 1 and 10 is 3.25.

Since we moved the decimal point 10 places to the left, the exponent of 10 is positive 10 (because the original number is large).

So, the standard form of 32,50,00,00,000 is $ 3.25 \times 10^{10} $.

The standard form for 32,50,00,00,000 is $3.25 \times 10^{10}$.

We need to write the number 0.000000008 in standard form (scientific notation).

The standard form is $ a \times 10^b $, where $ 1 \le a < 10 $ and $ b $ is an integer.

The given number is 0.000000008.

To get a number between 1 and 10, we need to move the decimal point to the right until it is after the first non-zero digit (which is 8).

Starting with 0.000000008:

We move the decimal point 9 places to the right to get 8.

The resulting number between 1 and 10 is 8.

Since we moved the decimal point 9 places to the right, and the original number is less than 1, the exponent of 10 is negative 9.

So, the standard form of 0.000000008 is $ 8 \times 10^{-9} $.

The standard form for 0.000000008 is $8 \times 10^{-9}$.

We are given the number in standard form as $ 2.3 \times 10^{-10} $ and asked to convert it to its usual form.

To convert from standard form $ a \times 10^b $ to usual form, we move the decimal point of the number $ a $ by $ b $ places.

If the exponent $ b $ is positive, we move the decimal point to the right.

If the exponent $ b $ is negative, we move the decimal point to the left.

In the given number $ 2.3 \times 10^{-10} $, the number $ a = 2.3 $ and the exponent $ b = -10 $.

Since the exponent is negative 10, we need to move the decimal point in 2.3 ten places to the left. We add leading zeros as needed.

Starting with 2.3:

Move 1 place left: 0.23

To move 10 places, we need 9 more places to the left of the current leading zero. So we add 9 zeros between the decimal point and the digit 2.

0. (9 zeros) 23

This gives: 0.00000000023

So, the usual form of $ 2.3 \times 10^{-10} $ is 0.00000000023.

The usual form for 2.3 × 10^-10 is 0.00000000023.

Let the unknown number be $ N $.

The problem states that when $ 8^5 $ is divided by $ N $, the result is $ 8 $.

So, we have the equation: $ 8^5 \div N = 8 $.

We can write this as: $ \frac{8^5}{N} = 8 $.

We need to solve for $ N $.

Multiply both sides by $ N $:

$ \frac{8^5}{N} \times N = 8 \times N $

$ 8^5 = 8 \times N $

Divide both sides by 8:

$ \frac{8^5}{8} = \frac{8 \times N}{8} $

$ \frac{8^5}{8^1} = N $

We use the property of exponents for division with the same base, which states that for any non-zero base $ a $ and any integers $ m $ and $ n $, $ a^m \div a^n = a^{m-n} $.

In this expression, the base is 8, the exponent in the numerator is $ m=5 $, and the exponent in the denominator is $ n=1 $.

Applying the property, we subtract the exponents:

$ N = 8^{5 - 1} $

$ N = 8^4 $

So, on dividing $ 8^5 $ by $ 8^4 $, we get 8.

On dividing 8⁵ by $8^4$ we get 8.

Let the unknown number be $ N $.

The problem states that when $ N $ is multiplied by $ 2^{-5} $, the result is $ 2^5 $.

So, we have the equation: $ N \times 2^{-5} = 2^5 $.

We need to solve for $ N $. To isolate $ N $, we can divide both sides of the equation by $ 2^{-5} $.

$ N = \frac{2^5}{2^{-5}} $

We use the property of exponents for division with the same base, which states that for any non-zero base $ a $ and any integers $ m $ and $ n $, $ a^m \div a^n = a^{m-n} $.

In this expression, the base is 2, the exponent in the numerator is $ m=5 $, and the exponent in the denominator is $ n=-5 $.

Applying the property, we subtract the exponents:

$ N = 2^{5 - (-5)} $

$ = 2^{5 + 5} $

$ = 2^{10} $

So, on multiplying $ 2^{10} $ by $ 2^{-5} $, we get $ 2^5 $.

Check: $ 2^{10} \times 2^{-5} = 2^{10 + (-5)} = 2^{10 - 5} = 2^5 $.

On multiplying $2^{10}$ by 2^–5 we get 2⁵.

We need to evaluate the expression $ [3^{-1} \times 4^{-1}]^{2} $.

First, let's evaluate the terms inside the square brackets: $ 3^{-1} \times 4^{-1} $.

Using the property $ a^{-1} = \frac{1}{a} $:

$ 3^{-1} = \frac{1}{3} $

$ 4^{-1} = \frac{1}{4} $

Now, multiply these values:

$ 3^{-1} \times 4^{-1} = \frac{1}{3} \times \frac{1}{4} = \frac{1 \times 1}{3 \times 4} = \frac{1}{12} $

Alternatively, we can use the property of exponents $ a^m \times b^m = (a \times b)^m $. However, the exponents are different here (-1 and -1), but the property works if exponents are the same.

Using the property $ a^m \times b^m = (ab)^m $ with $ m=-1 $:

$ 3^{-1} \times 4^{-1} = (3 \times 4)^{-1} = 12^{-1} = \frac{1}{12} $

So, the expression inside the brackets is $ \frac{1}{12} $.

Now, we need to find the square of $ \frac{1}{12} $:

$ \left( \frac{1}{12} \right)^{2} = \left( \frac{1}{12} \right) \times \left( \frac{1}{12} \right) $

Multiply the numerators and the denominators:

$ = \frac{1 \times 1}{12 \times 12} = \frac{1}{144} $

Alternatively, using the property $ \left(\frac{a}{b}\right)^m = \frac{a^m}{b^m} $:

$ \left( \frac{1}{12} \right)^{2} = \frac{1^2}{12^2} = \frac{1}{144} $

The value of $ [3^{-1} \times 4^{-1}]^{2} $ is $ \frac{1}{144} $.

The value of [3^–1 × 4^–1]² is $\frac{1}{144}$.

We need to evaluate the expression $ [2^{-1} \times 3^{-1}]^{-1} $.

First, let's evaluate the expression inside the square brackets: $ 2^{-1} \times 3^{-1} $.

Using the property $ a^{-1} = \frac{1}{a} $:

$ 2^{-1} = \frac{1}{2} $

$ 3^{-1} = \frac{1}{3} $

Now, multiply these values:

$ 2^{-1} \times 3^{-1} = \frac{1}{2} \times \frac{1}{3} = \frac{1 \times 1}{2 \times 3} = \frac{1}{6} $

Alternatively, we can use the property of exponents $ a^m \times b^m = (a \times b)^m $. Here, the exponents are the same (-1).

$ 2^{-1} \times 3^{-1} = (2 \times 3)^{-1} = 6^{-1} = \frac{1}{6} $

So, the expression inside the brackets is $ \frac{1}{6} $.

Now, we need to find the value of $ \left( \frac{1}{6} \right)^{-1} $.

Using the property $ \left( \frac{a}{b} \right)^{-1} = \frac{b}{a} $:

$ \left( \frac{1}{6} \right)^{-1} = \frac{6}{1} = 6 $

Alternatively, using the property $ x^{-1} = \frac{1}{x} $:

$ \left( \frac{1}{6} \right)^{-1} = \frac{1}{\frac{1}{6}} = 1 \times \frac{6}{1} = 6 $

Thus, the value of the expression $ [2^{-1} \times 3^{-1}]^{-1} $ is 6.

The value of [2^–1 × 3^–1]^–1 is 6.

We need to evaluate the expression $ (6^{0} – 7^{0}) \times (6^{0} + 7^{0}) $.

We use the property of exponents which states that any non-zero number raised to the power of zero is equal to 1.

So, $ 6^0 = 1 $ (since 6 is non-zero) and $ 7^0 = 1 $ (since 7 is non-zero).

Substitute these values into the expression:

$ (6^{0} – 7^{0}) \times (6^{0} + 7^{0}) = (1 – 1) \times (1 + 1) $

Now, perform the operations inside the parentheses:

$ 1 – 1 = 0 $

$ 1 + 1 = 2 $

Substitute these results back into the expression and multiply:

$ 0 \times 2 = 0 $

Thus, the value of the expression $ (6^{0} – 7^{0}) \times (6^{0} + 7^{0}) $ is 0.

By solving (6⁰ – 7⁰) × (6⁰ + 7⁰) we get 0.

We are asked to express $ 3^5 $ using a negative exponent.

We use the property of exponents that relates positive and negative exponents: For any non-zero base $ a $ and any integer $ m $, $ a^{-m} = \frac{1}{a^m} $.

This property can be rearranged as $ a^m = \frac{1}{a^{-m}} $.

In the expression $ 3^5 $, the base is 3 and the positive exponent is 5.

Using the property $ a^m = \frac{1}{a^{-m}} $ with $ a=3 $ and $ m=5 $, we get:

$ 3^5 = \frac{1}{3^{-5}} $

However, the question asks for the expression with a negative exponent. While $ \frac{1}{3^{-5}} $ contains a negative exponent in the denominator, the entire expression is not necessarily written as a single term with a negative exponent in the form $ a^b $ where $ b $ is negative.

Let's reconsider the core relationship: $ a^{-m} = \frac{1}{a^m} $.

We want to express $ 3^5 $ using a negative exponent.

Consider the reciprocal of $ 3^5 $, which is $ \frac{1}{3^5} $.

Using the property $ \frac{1}{a^m} = a^{-m} $ with $ a=3 $ and $ m=5 $, we have $ \frac{1}{3^5} = 3^{-5} $.

The question asks for the expression for $ 3^5 $ *with* a negative exponent. This implies finding an equivalent form where the exponent is negative.

From $ 3^5 = \frac{1}{3^{-5}} $, the term with the negative exponent is in the denominator.

From $ 3^{-5} = \frac{1}{3^5} $, the term with the negative exponent is $ 3^{-5} $. This expression is equal to the reciprocal of $ 3^5 $.

Let's re-read the question carefully: "The expression for 3⁵ with a negative exponent is _________". This likely means an expression *equal* to $ 3^5 $ that *uses* a negative exponent in its structure. The most common way to express a number with a positive exponent using a negative exponent is by taking the reciprocal of the base raised to the negative of the exponent.

$ 3^5 = \frac{1}{3^{-5}} $

Let's check if there's another interpretation. Could it be asking for a representation like $ (3^{-1})^{-5} $? Yes, using the power of a power rule $ (a^m)^n = a^{mn} $, we have $ (3^{-1})^{-5} = 3^{(-1) \times (-5)} = 3^{5} $.

The expression $ (3^{-1})^{-5} $ is equal to $ 3^5 $ and uses negative exponents.

Let's consider $ 3^5 = \left( \frac{1}{3} \right)^{-5} $. Using the property $ \left(\frac{a}{b}\right)^{-m} = \left(\frac{b}{a}\right)^m $, $ \left( \frac{1}{3} \right)^{-5} = \left( \frac{3}{1} \right)^{5} = 3^5 $.

The expression $ \left( \frac{1}{3} \right)^{-5} $ is equal to $ 3^5 $ and uses a negative exponent.

Without options, there are multiple valid answers that express $ 3^5 $ using negative exponents, such as $ \frac{1}{3^{-5}} $, $ (3^{-1})^{-5} $, or $ \left( \frac{1}{3} \right)^{-5} $.

However, the blank suggests a single, relatively simple expression. Among the possibilities, $ \left( \frac{1}{3} \right)^{-5} $ is a direct application of changing the base to its reciprocal and negating the exponent.

Let's assume the intended answer involves changing the base to its reciprocal and using the negative of the original exponent.

$ 3^5 = \left( \frac{1}{3} \right)^{-5} $

The expression for 3⁵ with a negative exponent is $\left( \frac{1}{3} \right)^{-5}$.

We need to evaluate the expression $ (-7)^6 \div 7^6 $.

We have the expression $ \frac{(-7)^6}{7^6} $.

Consider the numerator $ (-7)^6 $. When a negative number is raised to an even power, the result is positive.

$ (-7)^6 = (-7) \times (-7) \times (-7) \times (-7) \times (-7) \times (-7) = 7 \times 7 \times 7 \times 7 \times 7 \times 7 = 7^6 $

Substitute $ (-7)^6 $ with $ 7^6 $ in the expression:

$ \frac{(-7)^6}{7^6} = \frac{7^6}{7^6} $

Any non-zero number divided by itself is equal to 1.

$ \frac{7^6}{7^6} = 1 $

Alternatively, we can use the property of exponents for division where the bases are different but the exponents are the same: $ a^m \div b^m = \left(\frac{a}{b}\right)^m $.

$ (-7)^6 \div 7^6 = \left( \frac{-7}{7} \right)^{6} $

Simplify the fraction inside the parenthesis:

$ \frac{-7}{7} = -1 $

Now, raise -1 to the power of 6:

$ (-1)^6 $

When -1 is raised to an even power, the result is 1.

$ (-1)^6 = (-1) \times (-1) \times (-1) \times (-1) \times (-1) \times (-1) = 1 $

Thus, the value of $ (–7)^{6} \div 7^{6} $ is 1.

The value for (–7)⁶ ÷ 7⁶ is 1.

We need to evaluate the expression $ [1^{-2} + 2^{-2} + 3^{-2}] \times 6^{2} $.

First, let's evaluate the terms with negative exponents inside the square brackets using the property $ a^{-m} = \frac{1}{a^m} $.

$ 1^{-2} = \frac{1}{1^2} = \frac{1}{1} = 1 $

$ 2^{-2} = \frac{1}{2^2} = \frac{1}{4} $

$ 3^{-2} = \frac{1}{3^2} = \frac{1}{9} $

Now, sum these values inside the brackets:

$ 1^{-2} + 2^{-2} + 3^{-2} = 1 + \frac{1}{4} + \frac{1}{9} $

To add these fractions, find the least common multiple (LCM) of the denominators 1, 4, and 9. The LCM(1, 4, 9) is 36.

Convert each term to have a denominator of 36:

$ 1 = \frac{1}{1} = \frac{1 \times 36}{1 \times 36} = \frac{36}{36} $

$ \frac{1}{4} = \frac{1 \times 9}{4 \times 9} = \frac{9}{36} $

$ \frac{1}{9} = \frac{1 \times 4}{9 \times 4} = \frac{4}{36} $

Add the fractions:

$ 1 + \frac{1}{4} + \frac{1}{9} = \frac{36}{36} + \frac{9}{36} + \frac{4}{36} = \frac{36 + 9 + 4}{36} = \frac{49}{36} $

So, the expression inside the brackets is $ \frac{49}{36} $.

Next, evaluate the term outside the brackets: $ 6^2 $.

$ 6^2 = 6 \times 6 = 36 $

Finally, multiply the result from the brackets by $ 6^2 $:

$ \left[ \frac{49}{36} \right] \times 36 $

$ = \frac{49}{\cancel{36}} \times \cancel{36} $

$ = 49 $

Thus, the value of the expression $ [1^{-2} + 2^{-2} + 3^{-2}] \times 6^{2} $ is 49.

The value of [1^–2 + 2^–2 + 3^–2] × 6² is 49.

The multiplicative inverse of a non-zero number $a$ is $1/a$ or $a^{-1}$. Their product is 1, i.e., $a \times (1/a) = 1$.

We are asked to find the multiplicative inverse of $(-4)^{-2}$.

Using the property $a^{-n} = \frac{1}{a^n}$, we have $(-4)^{-2} = \frac{1}{(-4)^2}$.

So, the multiplicative inverse of $(-4)^{-2}$ is $\frac{1}{(-4)^{-2}}$.

This simplifies to $(-4)^2$, using the property $\frac{1}{a^{-n}} = a^n$.

Let's calculate the value of $(-4)^2$:

So, the multiplicative inverse of $(-4)^{-2}$ is $16$.

Now, let's look at the given statement's proposed inverse, $(4)^{-2}$.

Using the property $a^{-n} = \frac{1}{a^n}$, we have $(4)^{-2} = \frac{1}{4^2}$.

Let's calculate the value of $4^2$:

So, $(4)^{-2} = \frac{1}{16}$.

Comparing the two values: The multiplicative inverse of $(-4)^{-2}$ is $16$, and $(4)^{-2}$ is $\frac{1}{16}$.

Since $16 \neq \frac{1}{16}$, the given statement is false.

The statement is False.

Let's find the multiplicative inverse of $\left( \frac{3}{2} \right)^{2}$.

The multiplicative inverse of a non-zero number $x$ is $\frac{1}{x}$.

So, the multiplicative inverse of $\left( \frac{3}{2} \right)^{2}$ is $\frac{1}{\left( \frac{3}{2} \right)^{2}}$.

Using the property $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$, we have $\left( \frac{3}{2} \right)^{2} = \frac{3^2}{2^2} = \frac{9}{4}$.

The multiplicative inverse is $\frac{1}{\frac{9}{4}} = \frac{4}{9}$.

Now, let's evaluate $\left( \frac{2}{3} \right)^{-2}$.

Using the property $\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^{n}$, we have:

Evaluating the right side:

So, $\left( \frac{2}{3} \right)^{-2} = \frac{9}{4}$.

Comparing the multiplicative inverse of $\left( \frac{3}{2} \right)^{2}$ and $\left( \frac{2}{3} \right)^{-2}$:

Multiplicative inverse of $\left( \frac{3}{2} \right)^{2}$ is $\frac{4}{9}$.

$\left( \frac{2}{3} \right)^{-2}$ is $\frac{9}{4}$.

Since $\frac{4}{9} \neq \frac{9}{4}$, the multiplicative inverse of $\left( \frac{3}{2} \right)^{2}$ is indeed not equal to $\left( \frac{2}{3} \right)^{-2}$.

Therefore, the given statement is true.

The statement is True.

We need to check if the given statement $10^{-2} = \frac{1}{100}$ is true or false.

Let's evaluate the left side of the equation, which is $10^{-2}$.

Using the property of exponents $a^{-n} = \frac{1}{a^n}$, where $a \neq 0$, we can write:

Now, we calculate $10^2$:

Substituting this value back, we get:

The right side of the given statement is $\frac{1}{100}$.

Comparing the left side ($10^{-2} = \frac{1}{100}$) and the right side ($\frac{1}{100}$), we see that they are equal.

Thus, the statement $10^{-2} = \frac{1}{100}$ is true.

The statement is True.

We need to check if the given statement $24.58 = 2 \times 10 + 4 \times 1 + 5 \times 10 + 8 \times 100$ is true or false.

Let's evaluate the right side of the equation:

Calculate the products:

Add the numbers:

The left side of the given statement is $24.58$.

Comparing the left side ($24.58$) and the right side ($874$), we see that they are not equal.

Therefore, the given statement is false.

The correct expanded form of 24.58 in terms of powers of 10 is:

The statement is False.

We need to check if the given statement is true or false.

The statement is $329.25 = 3 \times 10^2 + 2 \times 10^1 + 9 \times 10^0 + 2 \times 10^{–1} + 5 \times 10^{–2}$.

Let's evaluate the right-hand side (RHS) of the equation:

We evaluate each term:

Now, we sum these values:

The left-hand side (LHS) of the equation is $329.25$.

Comparing LHS and RHS, we have $329.25 = 329.25$.

The equality holds true.

The statement is True.

We need to check if the given statement $(-5)^{-2} \times (-5)^{-3} = (-5)^{-6}$ is true or false.

Let's evaluate the left-hand side (LHS) of the equation.

Using the property of exponents $a^m \times a^n = a^{m+n}$, we have:

The right-hand side (RHS) of the equation is given as:

Now, we compare the LHS and the RHS.

LHS is $(-5)^{-5}$ and RHS is $(-5)^{-6}$.

Since the exponents $-5$ and $-6$ are different, and the base $-5$ is not 0 or 1 or -1, the values $(-5)^{-5}$ and $(-5)^{-6}$ are different.

Therefore, the given statement is false.

The statement is False.

We need to check if the given statement $(-4)^{-4} \times (4)^{-1} = (4)^{5}$ is true or false.

Let's evaluate the left-hand side (LHS) of the equation:

We use the property of exponents that for an even exponent $n$, $(-a)^{-n} = (a)^{-n}$. Here, $a=4$ and $n=4$ (which is even).

Substitute this back into the LHS:

Now, using the property $a^m \times a^n = a^{m+n}$, we add the exponents:

The right-hand side (RHS) of the equation is given as:

Now, we compare the LHS and the RHS.

LHS is $(4)^{-5}$ and RHS is $(4)^{5}$.

Using the property $a^{-n} = \frac{1}{a^n}$, we have $(4)^{-5} = \frac{1}{4^5}$.

So, we are comparing $\frac{1}{4^5}$ and $4^5$.

Since $\frac{1}{4^5} \neq 4^5$, the LHS is not equal to the RHS.

Therefore, the given statement is false.

The statement is False.

We need to check if the given statement $\left( \frac{2}{3} \right)^{-2} \times \left( \frac{2}{3} \right)^{-5} = \left( \frac{2}{3} \right)^{10}$ is true or false.

Let's evaluate the left-hand side (LHS) of the equation.

Using the property of exponents $a^m \times a^n = a^{m+n}$, where $a = \frac{2}{3}$, $m = -2$, and $n = -5$, we have:

The right-hand side (RHS) of the equation is given as:

Now, we compare the LHS and the RHS.

LHS is $\left( \frac{2}{3} \right)^{-7}$ and RHS is $\left( \frac{2}{3} \right)^{10}$.

Since the base $\frac{2}{3}$ is the same, the expressions are equal if and only if their exponents are equal.

Comparing the exponents, we have $-7$ and $10$.

Therefore, the given statement is false.

The statement is False.

We need to check if the given statement $5^0 = 5$ is true or false.

Let's evaluate the left-hand side (LHS) of the equation, which is $5^0$.

According to the properties of exponents, any non-zero number raised to the power of zero is equal to 1.

Mathematically, for any $a \neq 0$, $a^0 = 1$.

In this case, the base is 5, which is a non-zero number.

The right-hand side (RHS) of the given statement is $5$.

Comparing the LHS and the RHS, we have $1$ and $5$.

Since $1 \neq 5$, the LHS is not equal to the RHS.

Therefore, the given statement is false.

The statement is False.

We need to check if the given statement $(-2)^0 = 2$ is true or false.

Let's evaluate the left-hand side (LHS) of the equation, which is $(-2)^0$.

According to the properties of exponents, any non-zero number raised to the power of zero is equal to 1.

Mathematically, for any $a \neq 0$, $a^0 = 1$.

In this case, the base is $-2$, which is a non-zero number.

The right-hand side (RHS) of the given statement is $2$.

Comparing the LHS and the RHS, we have $1$ and $2$.

Since $1 \neq 2$, the LHS is not equal to the RHS.

Therefore, the given statement is false.

The statement is False.

We need to check if the given statement $\left( -\frac{8}{2} \right)^{0} = 0$ is true or false.

Let's evaluate the left-hand side (LHS) of the equation:

First, we simplify the expression inside the parentheses:

So, the LHS becomes:

According to the properties of exponents, any non-zero number raised to the power of zero is equal to 1.

Mathematically, for any $a \neq 0$, $a^0 = 1$.

In this case, the base is $-4$, which is a non-zero number.

The right-hand side (RHS) of the given statement is $0$.

Comparing the LHS and the RHS, we have $1$ and $0$.

Since $1 \neq 0$, the LHS is not equal to the RHS.

Therefore, the given statement is false.

The statement is False.

We need to check if the given statement $(-6)^0 = -1$ is true or false.

Let's evaluate the left-hand side (LHS) of the equation, which is $(-6)^0$.

According to the properties of exponents, any non-zero number raised to the power of zero is equal to 1.

Mathematically, for any $a \neq 0$, $a^0 = 1$.

In this case, the base is $-6$, which is a non-zero number.

The right-hand side (RHS) of the given statement is $-1$.

Comparing the LHS and the RHS, we have $1$ and $-1$.

Since $1 \neq -1$, the LHS is not equal to the RHS.

Therefore, the given statement is false.

The statement is False.

We need to check if the given statement $(-7)^{-4} \times (-7)^{2} = (-7)^{-2}$ is true or false.

Let's evaluate the left-hand side (LHS) of the equation:

We use the property of exponents for multiplication with the same base: $a^m \times a^n = a^{m+n}$.

Here, the base is $a = -7$, the first exponent is $m = -4$, and the second exponent is $n = 2$.

Applying the rule:

The right-hand side (RHS) of the equation is given as:

Now, we compare the LHS and the RHS.

LHS is $(-7)^{-2}$ and RHS is $(-7)^{-2}$.

Since the LHS is equal to the RHS, the given statement is true.

The statement is True.

We need to determine if the value of $\frac{1}{4^{-2}}$ is equal to 16.

Let's evaluate the expression $\frac{1}{4^{-2}}$.

We use the property of exponents that states $\frac{1}{a^{-n}} = a^n$ for any non-zero base $a$ and any integer $n$.

In this case, $a = 4$ and $n = 2$.

Applying the property:

Now, we calculate the value of $4^2$:

So, the value of $\frac{1}{4^{-2}}$ is $16$.

The statement claims that the value of $\frac{1}{4^{-2}}$ is equal to 16.

Our calculation shows that the value is indeed 16.

Therefore, the given statement is true.

The statement is True.

We need to check if the expression for $4^{-3}$ as a power with base 2 is equal to $2^6$.

Let's express $4^{-3}$ as a power with base 2.

We know that $4$ can be written as a power of 2:

Substitute this into the expression $4^{-3}$:

Using the property of exponents $(a^m)^n = a^{m \times n}$, where $a=2$, $m=2$, and $n=-3$, we multiply the exponents:

So, the expression for $4^{-3}$ as a power with the base 2 is $2^{-6}$.

The given statement claims that the expression for $4^{-3}$ as a power with base 2 is $2^6$.

Comparing our result ($2^{-6}$) with the claimed result ($2^6$), we see that they are not equal.

Therefore, the given statement is false.

The statement is False.

We need to check if the given statement $a^p \times b^q = (ab)^{pq}$ is true or false.

Let's analyze the left-hand side (LHS) and the right-hand side (RHS).

The LHS is $a^p \times b^q$. This is a product of two terms with different bases ($a$ and $b$) and different exponents ($p$ and $q$). There is no general exponent rule to simplify this expression further when the bases and exponents are different.

The RHS is $(ab)^{pq}$. Using the property of exponents $(xy)^n = x^n y^n$, where $x=a$, $y=b$, and $n=pq$, we can expand the RHS:

Now, we compare the LHS and the RHS:

For the statement to be true, LHS must equal RHS, i.e., $a^p \times b^q = a^{pq} \times b^{pq}$.

This equality does not hold for all values of $a, b, p, q$.

For example, let $a=2$, $b=3$, $p=2$, $q=1$.

LHS $= 2^2 \times 3^1 = 4 \times 3 = 12$.

RHS $= (2 \times 3)^{2 \times 1} = 6^2 = 36$.

Since $12 \neq 36$, the statement is false in this case.

The statement would only be true under specific conditions, such as when $p=pq$ and $q=pq$ (which implies $p=1$ and $q=1$, or $p=0, q=0$, etc.), or if $a=0$ or $b=0$ (with some restrictions on exponents). However, the statement is presented as a general rule, which is incorrect.

The correct rule when bases are different and exponents are the same is $a^n \times b^n = (ab)^n$.

The correct rule when bases are the same and exponents are different is $a^m \times a^n = a^{m+n}$.

The given statement $a^p \times b^q = (ab)^{pq}$ mixes these rules incorrectly.

Therefore, the given statement is false.

The statement is False.

We need to check if the given statement $\frac{x^m}{y^m} = \left( \frac{y}{x} \right)^{-m}$ is true or false.

Let's evaluate the left-hand side (LHS) of the equation:

Using the property of exponents $\frac{a^n}{b^n} = \left(\frac{a}{b}\right)^n$, we can combine the terms with the same exponent:

Now, let's evaluate the right-hand side (RHS) of the equation:

Using the property of exponents $\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^{n}$, we can rewrite the expression:

Now, we compare the LHS and the RHS.

LHS is $\left(\frac{x}{y}\right)^m$ and RHS is $\left(\frac{x}{y}\right)^m$.

Since the LHS is equal to the RHS, the given statement is true.

The statement is True.

We need to check if the given statement $a^m = \frac{1}{a^{-m}}$ is true or false. This statement assumes that $a \neq 0$, as $a^{-m}$ appears in the denominator when rewritten with a positive exponent if $m > 0$, or the original expression $a^{-m}$ would be $1/a^m$ if $m<0$, requiring $a \neq 0$.

Let's consider the right-hand side (RHS) of the equation:

We use the property of exponents that states $a^{-n} = \frac{1}{a^n}$ for any non-zero base $a$ and any integer $n$. We can apply this property in reverse, i.e., $\frac{1}{a^{-n}} = a^n$.

Let $n = -m$. Then, the expression in the denominator is $a^{-m}$.

Applying the property $\frac{1}{a^{-n}} = a^n$ with $n=-m$:

Simplifying the exponent $-(-m)$, we get $m$.

So, the RHS simplifies to $a^m$.

The left-hand side (LHS) of the given statement is $a^m$.

Comparing the LHS and the RHS, we have $a^m$ and $a^m$.

Since the LHS is equal to the RHS, the given statement is true (provided $a \neq 0$).

The statement is True.

We need to check if the exponential form for $(-2)^4 \times \left( \frac{5}{2} \right)^{4}$ is $5^4$.

Let's evaluate the expression $(-2)^4 \times \left( \frac{5}{2} \right)^{4}$.

We have a product of two terms with different bases ($-2$ and $\frac{5}{2}$) but the same exponent (4).

Using the property of exponents $a^n \times b^n = (a \times b)^n$, we can combine the bases:

Now, simplify the expression inside the parentheses:

Substitute this back into the expression:

Since the exponent is an even number (4), the result of $(-5)^4$ is the same as $5^4$:

So, the exponential form of the given expression is $(-5)^4$, which is equal to $5^4$.

The given statement claims that the exponential form for $(-2)^4 \times \left( \frac{5}{2} \right)^{4}$ is $5^4$.

Our result is $5^4$.

Since our result matches the claimed result, the given statement is true.

The statement is True.

We need to check if the standard form of $0.000037$ is $3.7 \times 10^{-5}$.

The standard form (or scientific notation) of a number is written in the form $a \times 10^n$, where $a$ is a number greater than or equal to 1 and less than 10 ($1 \leq |a| < 10$), and $n$ is an integer.

To convert the number $0.000037$ into standard form, we need to move the decimal point so that there is only one non-zero digit to the left of the decimal point.

The number is $0.000037$. The first non-zero digit is 3.

We move the decimal point to the right, past the digit 3, to get $3.7$.

The original position of the decimal point is before the first 0.

The new position of the decimal point is between 3 and 7.

Let's count the number of places we moved the decimal point:

From $0.000037$, we move 1 place to get $0.00037$

Move 2 places to get $0.0037$

Move 3 places to get $0.037$

Move 4 places to get $0.37$

Move 5 places to get $3.7$

We moved the decimal point 5 places to the right.

When we move the decimal point to the right for a number less than 1, the exponent of 10 is negative, and its absolute value is equal to the number of places moved.

So, the power of 10 is $10^{-5}$.

The number in standard form is the number with the decimal moved, multiplied by the calculated power of 10.

The given statement is that the standard form for $0.000037$ is $3.7 \times 10^{-5}$.

Our calculation shows that the standard form is indeed $3.7 \times 10^{-5}$.

Therefore, the given statement is true.

The statement is True.

We need to check if the standard form of $203000$ is $2.03 \times 10^{5}$.

The standard form (or scientific notation) of a number is written in the form $a \times 10^n$, where $a$ is a number such that $1 \leq |a| < 10$, and $n$ is an integer.

To convert the number $203000$ into standard form, we need to move the decimal point so that there is only one non-zero digit to the left of the decimal point.

The number $203000$ is an integer, so its decimal point is implicitly at the end: $203000.$

The first non-zero digit from the left is 2. We need to place the decimal point after the digit 2 to get a number between 1 and 10.

We move the decimal point from its original position (after the last 0) to the position after the digit 2.

Let's count the number of places the decimal point is moved to the left:

$203000.$ -> $20300.0$ (1 place)

$20300.0$ -> $2030.00$ (2 places)

$2030.00$ -> $203.000$ (3 places)

$203.000$ -> $20.3000$ (4 places)

$20.3000$ -> $2.03000$ (5 places)

The number with the decimal point moved is $2.03$.

We moved the decimal point 5 places to the left. When we move the decimal point to the left for a number greater than 10, the exponent of 10 is positive and equal to the number of places moved.

So, the power of 10 is $10^{5}$.

The number in standard form is the number with the decimal moved, multiplied by the calculated power of 10.

The given statement is that the standard form for $203000$ is $2.03 \times 10^{5}$.

Our calculation shows that the standard form is indeed $2.03 \times 10^{5}$.

Therefore, the given statement is true.

The statement is True.

We need to check if the given statement "The usual form for $2 \times 10^{-2}$ is not equal to $0.02$" is true or false.

This is equivalent to checking if $2 \times 10^{-2} \neq 0.02$.

Let's convert the number from scientific notation to its usual form.

We have $2 \times 10^{-2}$.

The exponent of 10 is $-2$. A negative exponent means we need to move the decimal point to the left.

The absolute value of the exponent is 2, so we move the decimal point 2 places to the left.

Start with the number 2. The decimal point is implicitly after the 2, i.e., $2.$

Move the decimal point 1 place to the left: $0.2$

Move the decimal point 2 places to the left: $0.02$

So, the usual form of $2 \times 10^{-2}$ is $0.02$.

The given statement says that the usual form for $2 \times 10^{-2}$ is not equal to $0.02$.

Our calculation shows that the usual form is equal to $0.02$.

Since $0.02 = 0.02$, the statement that they are not equal is false.

The statement is False.

We need to check if the given statement "The value of $5^{-2}$ is equal to $25$" is true or false.

Let's evaluate the value of $5^{-2}$.

We use the property of exponents that states $a^{-n} = \frac{1}{a^n}$ for any non-zero base $a$ and any integer $n$.

In this case, the base is $a = 5$ and the exponent is $n = 2$.

Applying the property:

Now, we calculate the value of $5^2$:

Substitute this value back:

The given statement claims that the value of $5^{-2}$ is equal to $25$.

Our calculation shows that the value is $\frac{1}{25}$.

Comparing our result ($\frac{1}{25}$) with the claimed value ($25$), we see that they are not equal.

Therefore, the given statement is false.

The statement is False.

We need to determine if the statement "Large numbers can be expressed in the standard form by using positive exponents" is true or false.

The standard form of a number is $a \times 10^n$, where $1 \leq |a| < 10$.

For a large number (absolute value $\geq 10$), such as 540000, the decimal point is moved to the left to get $5.4$. The number of places moved left determines the positive exponent of 10.

Since the exponent $5$ is positive, large numbers require positive exponents in standard form.

The statement is True.

We need to check if the given statement $a^m \times b^m = (ab)^m$ is true or false.

This statement represents a fundamental property of exponents.

The property states that when multiplying two or more terms with the same exponent but different bases, we can multiply the bases first and then raise the product to that common exponent.

Mathematically, this property is written as:

where $a$ and $b$ are any real numbers and $m$ is an integer.

Let's verify with an example. Let $a=2$, $b=3$, and $m=4$.

Left-hand side (LHS):

Right-hand side (RHS):

Since $1296 = 1296$, LHS = RHS for this example.

The property $a^m \times b^m = (ab)^m$ is a valid rule of exponents that holds true for all allowed values of $a, b,$ and $m$.

The statement is True.

(i)

Given expression is $100^{-10}$.

Using the property of exponents $a^{-n} = \frac{1}{a^n}$, we have:

$100^{-10} = \frac{1}{100^{10}}$

The solution is: $\frac{1}{100^{10}}$

(ii)

Given expression is $2^{-2} \times 2^{-3}$.

Using the property of exponents $a^m \times a^n = a^{m+n}$, we have:

$2^{-2} \times 2^{-3} = 2^{-2 + (-3)} = 2^{-5}$

Now, using the property $a^{-n} = \frac{1}{a^n}$:

$2^{-5} = \frac{1}{2^5}$

Calculating the value of $2^5$:

$2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32$

So, $2^{-5} = \frac{1}{32}$.

The solution is: $\frac{1}{32}$

(iii)

Given expression is $\left( \frac{1}{2} \right)^{-2}$ ÷ $\left( \frac{1}{2} \right)^{-3}$.

Using the property of exponents $a^m \div a^n = a^{m-n}$, where $a = \frac{1}{2}$, $m = -2$, and $n = -3$, we have:

$\left( \frac{1}{2} \right)^{-2}$ ÷ $\left( \frac{1}{2} \right)^{-3} = \left( \frac{1}{2} \right)^{-2 - (-3)}$

$\left( \frac{1}{2} \right)^{-2 - (-3)} = \left( \frac{1}{2} \right)^{-2 + 3} = \left( \frac{1}{2} \right)^1$

Using the property $a^1 = a$:

$\left( \frac{1}{2} \right)^1 = \frac{1}{2}$

The solution is: $\frac{1}{2}$

Given expression is $3^{-5} \times 3^{-4}$.

Using the property of exponents $a^m \times a^n = a^{m+n}$, where $a=3$, $m=-5$, and $n=-4$, we have:

$3^{-5} \times 3^{-4} = 3^{-5 + (-4)}$

$3^{-5 + (-4)} = 3^{-9}$

The expression simplified is $3^{-9}$. The question asks to express this as a power of 3 with a positive exponent.

The exponent $-9$ is negative. To express this in a form involving a positive exponent, we use the property $a^{-n} = \frac{1}{a^n}$.

$3^{-9} = \frac{1}{3^9}$

The expression $\frac{1}{3^9}$ contains $3^9$, which is a power of 3 with a positive exponent ($9$). While the entire expression is not in the form $3^{\text{positive number}}$, this is the standard way to represent the value using a positive exponent of the base 3.

The final answer expressed as a power of 3 with positive exponent is $\frac{1}{3^9}$.

Given expression is $16^{-2}$.

We need to express this as a power with the base 2.

First, express the base 16 as a power of 2.

$16 = 2 \times 2 \times 2 \times 2 = 2^4$

Now substitute $2^4$ for 16 in the given expression:

$16^{-2} = (2^4)^{-2}$

Using the property of exponents $(a^m)^n = a^{m \times n}$, we have:

$(2^4)^{-2} = 2^{4 \times (-2)}$

$2^{4 \times (-2)} = 2^{-8}$

The expression $16^{-2}$ expressed as a power with base 2 is $2^{-8}$.

Part 1: Express $\frac{27}{64}$ as a power of a rational number.

We find the prime factorization of the numerator and the denominator:

$27 = 3 \times 3 \times 3 = 3^3$

$64 = 4 \times 4 \times 4 = 4^3$

So, $\frac{27}{64} = \frac{3^3}{4^3}$.

Using the property $\frac{a^n}{b^n} = \left(\frac{a}{b}\right)^n$, we can write:

$\frac{3^3}{4^3} = \left(\frac{3}{4}\right)^3$

Thus, $\frac{27}{64}$ expressed as a power of a rational number is $\left(\frac{3}{4}\right)^3$.

Part 2: Express $\frac{-27}{64}$ as a power of a rational number.

We know that $27 = 3^3$ and $64 = 4^3$. So, $\frac{27}{64} = \left(\frac{3}{4}\right)^3$.

The expression is $\frac{-27}{64}$. This can be written as $-\left(\frac{27}{64}\right) = -\left(\frac{3}{4}\right)^3$.

Since the exponent is odd (3), a negative base raised to this power will result in a negative value.

Consider the rational number $-\frac{3}{4}$. Let's raise it to the power of 3:

$\left(-\frac{3}{4}\right)^3 = \frac{(-3)^3}{4^3} = \frac{-27}{64}$

Thus, $\frac{-27}{64}$ expressed as a power of a rational number is $\left(-\frac{3}{4}\right)^3$.

Part 1: Express $\frac{16}{81}$ as a power of a rational number.

First, we find the prime factorization of the numerator and the denominator:

$16 = 2 \times 2 \times 2 \times 2 = 2^4$

$81 = 3 \times 3 \times 3 \times 3 = 3^4$

So, $\frac{16}{81} = \frac{2^4}{3^4}$.

Using the property of exponents $\frac{a^n}{b^n} = \left(\frac{a}{b}\right)^n$, we have:

$\frac{16}{81} = \left(\frac{2}{3}\right)^4$

The solution for $\frac{16}{81}$ is: $\left(\frac{2}{3}\right)^4$

Part 2: Express $\frac{-16}{81}$ as a power of a rational number.

We need to express $\frac{-16}{81}$ in the form $(\frac{a}{b})^n$ where $\frac{a}{b}$ is a rational number and $n$ is an integer.

We know that $\frac{16}{81} = \left(\frac{2}{3}\right)^4$.

The expression $\frac{-16}{81}$ is a negative number.

If an even integer power $n$ is used, $(\frac{a}{b})^n$ would be positive (or zero if $\frac{a}{b}=0$, which is not the case here).

If an odd integer power $n > 1$ is used, $(\frac{a}{b})^n = \frac{-16}{81}$ would require $a^n = -16$ and $b^n = 81$ (or scaled versions), which does not have rational solutions for $a$ and $b$ when $n > 1$ is odd (e.g., for $n=3$, $\sqrt[3]{-16}$ and $\sqrt[3]{81}$ are not rational, so the base $\frac{a}{b}$ would not be rational).

The only integer power $n$ for which $\frac{-16}{81} = (\frac{a}{b})^n$ is possible with a rational base $\frac{a}{b}$ is $n=1$, where the base is $\frac{-16}{81}$ itself.

$\frac{-16}{81} = \left(\frac{-16}{81}\right)^1$

The solution for $\frac{-16}{81}$ is: $\left(\frac{-16}{81}\right)^1$

(a)

Given expression is $\left( \left( \frac{-3}{2} \right)^{-2} \right)^{-3}$.

We use the property of exponents $(a^m)^n = a^{mn}$.

Here, $a = \frac{-3}{2}$, $m = -2$, and $n = -3$.

$\left( \left( \frac{-3}{2} \right)^{-2} \right)^{-3} = \left( \frac{-3}{2} \right)^{(-2) \times (-3)}$

$ = \left( \frac{-3}{2} \right)^6$

We need to express this as a power of a rational number with a negative exponent.

We use the property $\left(\frac{a}{b}\right)^n = \left(\frac{b}{a}\right)^{-n}$.

So, $\left( \frac{-3}{2} \right)^6 = \left( \frac{2}{-3} \right)^{-6}$.

The rational number $\frac{2}{-3}$ is the same as $-\frac{2}{3}$.

Thus, the expression is $\left( -\frac{2}{3} \right)^{-6}$.

(b)

Given expression is $(2^5 \div 2^8) \times 2^{-7}$.

First, simplify the expression inside the parentheses using the property $a^m \div a^n = a^{m-n}$.

$2^5 \div 2^8 = 2^{5-8} = 2^{-3}$

Now, multiply the result by $2^{-7}$ using the property $a^m \times a^n = a^{m+n}$.

$(2^5 \div 2^8) \times 2^{-7} = 2^{-3} \times 2^{-7}$

$ = 2^{-3 + (-7)}$

$ = 2^{-10}$

The result is $2^{-10}$. The base is 2, which is a rational number ($\frac{2}{1}$), and the exponent is $-10$, which is a negative exponent.

The expression is already in the required form.

The expression as a power of a rational number with negative exponent is $2^{-10}$.

We need to find the product of the cube of $(-2)$ and the square of $(+4)$.

First, find the cube of $(-2)$.

Cube of $(-2) = (-2)^3$

$(-2)^3 = (-2) \times (-2) \times (-2)$

$(-2)^3 = 4 \times (-2)$

$(-2)^3 = -8$

Next, find the square of $(+4)$.

Square of $(+4) = (+4)^2$

$(+4)^2 = 4 \times 4$

$(+4)^2 = 16$

Now, find the product of the cube of $(-2)$ and the square of $(+4)$, which is the product of $-8$ and $16$.

Product = $(-8) \times 16$

Product = $-128$

The product is $-128$.

(i)

We are asked to simplify the expression $\left( \frac{1}{4} \right)^{-2} + \left( \frac{1}{2} \right)^{-2} + \left( \frac{1}{3} \right)^{-2}$.

Using the property $\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$, we can rewrite each term:

$\left( \frac{1}{4} \right)^{-2} = \left( \frac{4}{1} \right)^2 = 4^2 = 16$

$\left( \frac{1}{2} \right)^{-2} = \left( \frac{2}{1} \right)^2 = 2^2 = 4$

$\left( \frac{1}{3} \right)^{-2} = \left( \frac{3}{1} \right)^2 = 3^2 = 9$

Now, add the simplified terms:

$16 + 4 + 9 = 20 + 9 = 29$

The simplified value is $29$.

(ii)

We are asked to simplify $\left( \left( \frac{-2}{3} \right)^{-2} \right)^{3}$ × $\left( \frac{1}{3} \right)^{-4}$ × 3^-1 × $\frac{1}{6}$.

Simplify the first term using $(a^m)^n = a^{mn}$ and $\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$:

$\left( \left( \frac{-2}{3} \right)^{-2} \right)^{3} = \left( \frac{-2}{3} \right)^{(-2) \times 3} = \left( \frac{-2}{3} \right)^{-6}$

$ = \left( \frac{3}{-2} \right)^6 = \left( -\frac{3}{2} \right)^6 = \frac{(-3)^6}{2^6} = \frac{3^6}{2^6}$

Simplify the second term using $\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$:

$\left( \frac{1}{3} \right)^{-4} = \left( \frac{3}{1} \right)^4 = 3^4$

Rewrite $3^{-1}$ as $\frac{1}{3^1} = \frac{1}{3}$.

Rewrite $\frac{1}{6}$ as $\frac{1}{2 \times 3}$.

Now substitute these back into the expression:

$\frac{3^6}{2^6} \times 3^4 \times \frac{1}{3} \times \frac{1}{2 \times 3}$

Combine terms with the same base:

$ = \frac{3^6 \times 3^4}{2^6 \times 3^1 \times 2^1 \times 3^1}$

Using $a^m \times a^n = a^{m+n}$ in the numerator and denominator:

$ = \frac{3^{6+4}}{2^{6+1} \times 3^{1+1}}$

$ = \frac{3^{10}}{2^7 \times 3^2}$

Using $\frac{a^m}{a^n} = a^{m-n}$:

$ = \frac{3^{10-2}}{2^7}$

$ = \frac{3^8}{2^7}$

Calculate the values:

$3^8 = 6561$

$2^7 = 128$

The simplified value is $\frac{6561}{128}$.

(iii)

We are asked to simplify $\frac{49 \;×\; z^{−3}}{7^{−3} \;×\; 10 \;×\; z^{−5}}$ (z ≠ 0).

Rewrite $49$ as $7^2$. The expression becomes:

$\frac{7^2 \;×\; z^{−3}}{7^{−3} \;×\; 10 \;×\; z^{−5}}$

Group terms with the same base and the constant term:

$ \left( \frac{7^2}{7^{-3}} \right) \times \left( \frac{z^{-3}}{z^{-5}} \right) \times \frac{1}{10} $

Using the property $\frac{a^m}{a^n} = a^{m-n}$:

$\frac{7^2}{7^{-3}} = 7^{2 - (-3)} = 7^{2+3} = 7^5$

$\frac{z^{-3}}{z^{-5}} = z^{-3 - (-5)} = z^{-3+5} = z^2$

Substitute these back:

$ 7^5 \times z^2 \times \frac{1}{10} $

Calculate $7^5$:

$7^5 = 7 \times 7 \times 7 \times 7 \times 7 = 49 \times 49 \times 7 = 2401 \times 7 = 16807$

The simplified expression is $\frac{16807 z^2}{10}$.

(iv)

We are asked to simplify $(2^5 \div 2^8) \times 2^{-7}$.

Simplify the expression inside the parentheses using the property $a^m \div a^n = a^{m-n}$:

$2^5 \div 2^8 = 2^{5-8} = 2^{-3}$

Now, multiply the result by $2^{-7}$ using the property $a^m \times a^n = a^{m+n}$:

$2^{-3} \times 2^{-7} = 2^{-3 + (-7)} = 2^{-10}$

The simplified value is $2^{-10}$.

(i)

Given equation is $\left( \frac{5}{3} \right)^{-2} \times \left( \frac{5}{3} \right)^{-14} = \left( \frac{5}{3} \right)^{8x}$.

Using the property $a^m \times a^n = a^{m+n}$ on the left side:

$\left( \frac{5}{3} \right)^{-2 + (-14)} = \left( \frac{5}{3} \right)^{8x}$

$\left( \frac{5}{3} \right)^{-16} = \left( \frac{5}{3} \right)^{8x}$

Since the bases are equal ($\frac{5}{3}$) and not equal to 0, -1, or 1, the exponents must be equal.

$-16 = 8x$

To find the value of $x$, divide both sides by 8:

$x = \frac{-16}{8}$

$x = -2$

The value of $x$ is $-2$.

(ii)

Given equation is $(–2)^3 \times (–2)^{–6} = (–2)^{2x – 1}$.

Using the property $a^m \times a^n = a^{m+n}$ on the left side:

$(–2)^{3 + (-6)} = (–2)^{2x – 1}$

$(–2)^{-3} = (–2)^{2x – 1}$

Since the bases are equal (–2) and not equal to 0 or 1, the exponents must be equal.

$-3 = 2x - 1$

Add 1 to both sides:

$-3 + 1 = 2x$

$-2 = 2x$

To find the value of $x$, divide both sides by 2:

$x = \frac{-2}{2}$

$x = -1$

The value of $x$ is $-1$.

(iii)

Given equation is $(2^{–1} + 4^{–1} + 6^{–1} + 8^{–1})^x = 1$.

First, simplify the expression inside the parentheses using the property $a^{-n} = \frac{1}{a^n}$:

$2^{-1} = \frac{1}{2}$

$4^{-1} = \frac{1}{4}$

$6^{-1} = \frac{1}{6}$

$8^{-1} = \frac{1}{8}$

The sum inside the parentheses is $\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8}$.

Find the least common multiple (LCM) of the denominators 2, 4, 6, and 8.

LCM(2, 4, 6, 8) = 24.

Rewrite each fraction with the denominator 24:

$\frac{1}{2} = \frac{1 \times 12}{2 \times 12} = \frac{12}{24}$

$\frac{1}{4} = \frac{1 \times 6}{4 \times 6} = \frac{6}{24}$

$\frac{1}{6} = \frac{1 \times 4}{6 \times 4} = \frac{4}{24}$

$\frac{1}{8} = \frac{1 \times 3}{8 \times 3} = \frac{3}{24}$

Now, add the fractions:

$\frac{12}{24} + \frac{6}{24} + \frac{4}{24} + \frac{3}{24} = \frac{12 + 6 + 4 + 3}{24} = \frac{25}{24}$

Substitute this sum back into the original equation:

$\left(\frac{25}{24}\right)^x = 1$

We know that for any non-zero base $a$, $a^0 = 1$. So, $1 = \left(\frac{25}{24}\right)^0$.

The equation becomes:

$\left(\frac{25}{24}\right)^x = \left(\frac{25}{24}\right)^0$

Since the bases are equal ($\frac{25}{24}$) and not equal to 0 or 1, the exponents must be equal.

$x = 0$

The value of $x$ is $0$.

We need to divide 293 by 10,00,000 and express the result in standard form.

The division is $\frac{293}{10,00,000}$.

The number 10,00,000 can be written in scientific notation as $10^6$.

So, the expression becomes $\frac{293}{10^6}$.

Using the property $\frac{1}{a^n} = a^{-n}$, we can write $\frac{1}{10^6}$ as $10^{-6}$.

Thus, $\frac{293}{10^6} = 293 \times 10^{-6}$.

Now, we need to express $293 \times 10^{-6}$ in standard form.

Standard form is $a \times 10^n$, where $1 \leq |a| < 10$ and $n$ is an integer.

Express the number 293 in standard form:

$293 = 2.93 \times 100 = 2.93 \times 10^2$

Substitute this back into the expression:

$ (2.93 \times 10^2) \times 10^{-6} $

Using the property $a^m \times a^n = a^{m+n}$ for the powers of 10:

$2.93 \times 10^{2 + (-6)}$

$2.93 \times 10^{2 - 6}$

$2.93 \times 10^{-4}$

This result is in standard form, as $1 \leq 2.93 < 10$ and the exponent $-4$ is an integer.

The result in standard form is $2.93 \times 10^{-4}$.

Given the expression for $x$:

$x = (100)^{1 – 4} ÷ (100)^{0}$

First, we simplify the exponent in the first term:

$1 - 4 = -3$

So, $(100)^{1 - 4} = (100)^{-3}$.

Next, we simplify the second term using the property $a^0 = 1$ (for $a \neq 0$):

$(100)^0 = 1$

Now, substitute these simplified terms back into the expression for $x$:

$x = (100)^{-3} ÷ 1$

Dividing any number by 1 results in the same number:

$x = (100)^{-3}$

We are asked to find the value of $x^{-3}$.

Substitute the value of $x$ we found:

$x^{-3} = ((100)^{-3})^{-3}$

Using the property of exponents $(a^m)^n = a^{m \times n}$:

$x^{-3} = (100)^{(-3) \times (-3)}$

Calculate the exponent:

$(-3) \times (-3) = 9$

So,

$x^{-3} = (100)^9$

The value of $x^{-3}$ is $(100)^9$.

We are asked to find a number, let's call it $y$, such that when we multiply $(–29)^0$ by $y$, the result is $(+29)^0$.

The given condition can be written as an equation:

We use the property of exponents which states that any non-zero number raised to the power of 0 is equal to 1.

So, $(–29)^0 = 1$ since $-29 \neq 0$.

Also, $(+29)^0 = 1$ since $+29 \neq 0$.

Substitute these values into equation (i):

Solving equation (ii) for $y$:

$y = \frac{1}{1}$

$y = 1$

Thus, we should multiply by 1.

The number is $1$.

Let the number by which $(–15)^{–1}$ should be divided be $y$.

According to the question, the division of $(–15)^{–1}$ by $y$ results in the quotient $(–15)^{–1}$.

We can write this as an equation:

We know that for any non-zero number $a$, $a^{-1} = \frac{1}{a}$.

So, $(–15)^{–1} = \frac{1}{–15}$.

Substitute this value into equation (i):

Rewrite the division as multiplication by the reciprocal of $y$ (which is $\frac{1}{y}$):

$\frac{1}{-15} \times \frac{1}{y} = \frac{1}{-15}$

$\frac{1}{-15y} = \frac{1}{-15}$

To solve for $y$, we can take the reciprocal of both sides (since both sides are non-zero):

$-15y = -15$

Divide both sides by -15:

$y = \frac{-15}{-15}$

$y = 1$

Alternatively, from equation (i), if we have $A ÷ y = A$ where $A \neq 0$, dividing both sides by $A$ gives $1 ÷ y = 1$. Multiplying both sides by $y$ gives $1 = y$.

Thus, the number by which $(–15)^{–1}$ should be divided is 1.

The number is $1$.

We need to find the multiplicative inverse of the expression $(–7)^{–2} ÷ (90)^{–1}$.

First, let's evaluate the given expression.

Using the property of exponents $a^{-n} = \frac{1}{a^n}$:

$(–7)^{–2} = \frac{1}{(-7)^2} = \frac{1}{49}$

$(90)^{–1} = \frac{1}{90^1} = \frac{1}{90}$

Now, perform the division:

$(–7)^{–2} ÷ (90)^{–1} = \frac{1}{49} ÷ \frac{1}{90}$

To divide by a fraction, we multiply by its reciprocal:

$ \frac{1}{49} ÷ \frac{1}{90} = \frac{1}{49} \times \frac{90}{1} $

$ = \frac{90}{49} $

So, the value of the expression $(–7)^{–2} ÷ (90)^{–1}$ is $\frac{90}{49}$.

We need to find the multiplicative inverse of $\frac{90}{49}$.

The multiplicative inverse of a non-zero rational number $\frac{a}{b}$ is $\frac{b}{a}$.

The multiplicative inverse of $\frac{90}{49}$ is $\frac{49}{90}$.

The multiplicative inverse is $\frac{49}{90}$.

Given equation is $5^{3x – 1} ÷ 25 = 125$.

We can express 25 and 125 as powers of 5:

$25 = 5 \times 5 = 5^2$

$125 = 5 \times 5 \times 5 = 5^3$

Substitute these into the given equation:

$5^{3x – 1} ÷ 5^2 = 5^3$

Using the property of exponents $a^m \div a^n = a^{m-n}$ on the left side:

$5^{(3x – 1) - 2} = 5^3$

Simplify the exponent on the left side:

$3x - 1 - 2 = 3x - 3$

So the equation becomes:

$5^{3x - 3} = 5^3$

Since the bases are equal (5) and not equal to 0, -1, or 1, the exponents must be equal:

Solve equation (i) for $x$. Add 3 to both sides:

$3x = 3 + 3$

$3x = 6$

Divide both sides by 3:

$x = \frac{6}{3}$

$x = 2$

The value of $x$ is $2$.

We need to write the number 39,00,00,000 in standard form.

Standard form is represented as $a \times 10^n$, where $1 \leq |a| < 10$ and $n$ is an integer.

The given number is 39,00,00,000.

To express this in standard form, we need to move the decimal point so that there is only one non-zero digit to the left of the decimal point.

The implied decimal point in 39,00,00,000 is at the end (390000000.).

We want to place the decimal point after the digit 3, so the number becomes 3.9.

To get from 390000000 to 3.9, we need to move the decimal point to the left.

Let's count the number of places the decimal point is moved:

390000000.

Moving the decimal 1 place left gives 39000000.

Moving the decimal 2 places left gives 3900000.

... and so on, until we get 3.9.

The number of places the decimal point is moved to the left is 8 (past the 9 and the seven 0s).

Since the decimal point is moved to the left, the exponent of 10 is positive.

The number of places moved is 8, so the exponent is 8.

Therefore, 39,00,00,000 in standard form is $3.9 \times 10^8$.

The number in standard form is $3.9 \times 10^8$.

We need to write the number 0.000005678 in standard form.

Standard form is represented as $a \times 10^n$, where $1 \leq |a| < 10$ and $n$ is an integer.

The given number is 0.000005678.

To express this in standard form, we need to move the decimal point so that there is only one non-zero digit to the left of the decimal point.

The first non-zero digit from the left is 5. We need to move the decimal point to the right so that it is placed after the digit 5.

Let's count the number of places the decimal point is moved:

0.000005678

Moving the decimal point to the right past the first 0, second 0, third 0, fourth 0, fifth 0, and sixth 0, we get 5.678.

The number of places the decimal point is moved to the right is 6.

Since the decimal point is moved to the right, the exponent of 10 is negative.

The number of places moved is 6, so the exponent is -6.

Therefore, 0.000005678 in standard form is $5.678 \times 10^{-6}$.

The number in standard form is $5.678 \times 10^{-6}$.

We need to find the product of $3.2 \times 10^6$ and $4.1 \times 10^{–1}$.

Product = $(3.2 \times 10^6) \times (4.1 \times 10^{–1})$

We can rearrange the terms and group the decimal parts and the powers of 10:

Product = $(3.2 \times 4.1) \times (10^6 \times 10^{–1})$

First, multiply the decimal parts:

$3.2 \times 4.1$

$\begin{array}{cc}& & 3 & . & 2 \\ \times & & 4 & . & 1 \\ \hline && 3 & 2 \\ 1 & 2 & 8 & \times \\ \hline 1 & 3 & . & 1 & 2 \\ \hline \end{array}$

$3.2 \times 4.1 = 13.12$

Next, multiply the powers of 10 using the property $a^m \times a^n = a^{m+n}$:

$10^6 \times 10^{–1} = 10^{6 + (–1)}$

$ = 10^{6 - 1}$

$ = 10^5$

Combine the results:

Product = $13.12 \times 10^5$

We need to express this result in standard form, which is $a \times 10^n$ where $1 \leq |a| < 10$ and $n$ is an integer.

The current decimal part, $13.12$, is not between 1 and 10. We need to move the decimal point one place to the left to get $1.312$.

Moving the decimal point one place to the left is equivalent to dividing by 10, which can be compensated by multiplying by $10^1$.

So, $13.12 = 1.312 \times 10^1$.

Substitute this back into the product expression:

Product = $(1.312 \times 10^1) \times 10^5$

Product = $1.312 \times (10^1 \times 10^5)$

Using the property $a^m \times a^n = a^{m+n}$ again:

$10^1 \times 10^5 = 10^{1+5} = 10^6$

The product in standard form is:

Product = $1.312 \times 10^6$

This is in standard form because $1 \leq 1.312 < 10$ and 6 is an integer.

The product in standard form is $1.312 \times 10^6$.

We need to express the value of the expression $\frac{1.5 \;×\; 10^{6}}{2.5 \;×\; 10^{−4}}$ in standard form.

We can rewrite the expression by separating the decimal parts and the powers of 10:

$\frac{1.5 \;×\; 10^{6}}{2.5 \;×\; 10^{−4}} = \left(\frac{1.5}{2.5}\right) \times \left(\frac{10^{6}}{10^{−4}}\right)$

First, calculate the division of the decimal parts:

$\frac{1.5}{2.5} = \frac{1.5 \times 10}{2.5 \times 10} = \frac{15}{25}$

Simplify the fraction by dividing the numerator and the denominator by 5:

$\frac{15}{25} = \frac{\cancel{15}^{3}}{\cancel{25}_{5}} = \frac{3}{5}$

Convert the fraction to a decimal:

$\frac{3}{5} = 0.6$

Next, calculate the division of the powers of 10 using the property $\frac{a^m}{a^n} = a^{m-n}$:

$\frac{10^{6}}{10^{−4}} = 10^{6 - (−4)}$

$ = 10^{6 + 4}$

$ = 10^{10}$

Now, combine the results from the two parts:

$\left(\frac{1.5}{2.5}\right) \times \left(\frac{10^{6}}{10^{−4}}\right) = 0.6 \times 10^{10}$

We need to express this result in standard form, which is $a \times 10^n$, where $1 \leq |a| < 10$ and $n$ is an integer.

The current decimal part, $0.6$, is not in the required range ($1 \leq 0.6 < 10$ is false).

To make the decimal part $6.0$ (which is between 1 and 10), we move the decimal point one place to the right. Moving the decimal one place right means we are effectively multiplying by 10. To keep the value the same, we must compensate by dividing the power of 10 by 10 (i.e., multiplying by $10^{-1}$).

$0.6 = 6.0 \times 10^{-1}$

Substitute this back into the combined expression:

$0.6 \times 10^{10} = (6.0 \times 10^{-1}) \times 10^{10}$

Using the property $a^m \times a^n = a^{m+n}$ for the powers of 10:

$ = 6.0 \times 10^{−1 + 10}$

$ = 6.0 \times 10^{9}$

This result is in standard form, as $1 \leq 6.0 < 10$ and 9 is an integer.

The expression in standard form is $6.0 \times 10^{9}$.

The distance traveled by the migratory birds is given as 15,000 km.

We need to express this distance in metres using scientific notation.

We know that 1 kilometer (km) is equal to 1000 metres (m).

$1 \text{ km} = 1000 \text{ m}$

To convert kilometers to metres, we multiply the distance in kilometers by 1000.

Distance in metres = $15,000 \text{ km} \times 1000 \frac{\text{m}}{\text{km}}$

Distance in metres = $15,000 \times 1000 \text{ m}$

Distance in metres = $15,000,000 \text{ m}$

Now, we need to express the number 15,000,000 in scientific notation.

Scientific notation (also known as standard form) is written in the format $a \times 10^n$, where $1 \leq |a| < 10$ and $n$ is an integer.

The number is 15,000,000. The decimal point is currently at the end (15000000.).

We need to move the decimal point to the left until only one non-zero digit is before the decimal point. The first non-zero digit is 1.

Moving the decimal point from its original position to after the digit 1 requires moving it 7 places to the left (past the last six 0s and the digit 5).

Since the decimal point is moved to the left, the exponent of 10 will be positive, and its value will be the number of places the decimal point was moved, which is 7.

So, 15,000,000 can be written as $1.5 \times 10^7$.

The value $a = 1.5$ satisfies $1 \leq 1.5 < 10$, and the exponent $n = 7$ is an integer.

Therefore, the distance in metres in scientific notation is $1.5 \times 10^7$ metres.

The distance of Pluto from the sun is given as 59,1,30,00,000 m.

We need to express this distance in standard form.

The number written without commas is 5,913,000,000.

Standard form (also known as scientific notation) is expressed as $a \times 10^n$, where $1 \leq |a| < 10$ and $n$ is an integer.

The given number is 5,913,000,000.

The decimal point is implicitly at the end of the number (5913000000.).

To express this in standard form, we need to move the decimal point to the left so that only one non-zero digit remains to the left of the decimal point.

The first non-zero digit from the left is 5. We need to move the decimal point to place it after the digit 5.

The number of places the decimal point is moved to the left is 9 (for the trailing zeros) + 3 (for the digits 3, 1, 9) = 12 places.

Since the decimal point is moved to the left, the exponent of 10 is positive, and its value is equal to the number of places moved.

The number of places moved is 12, so the exponent is 12.

The number $a$ is obtained by placing the decimal point after the first digit: 5.913.

The number $a = 5.913$ satisfies the condition $1 \leq 5.913 < 10$.

Therefore, 5,913,000,000 in standard form is $5.913 \times 10^{12}$.

The distance in standard form is $5.913 \times 10^{12}$ metres.

We need to express the number 0.00000001 in standard form.

Standard form (also known as scientific notation) is written as $a \times 10^n$, where $1 \leq |a| < 10$ and $n$ is an integer.

The given number is 0.00000001.

To express this in standard form, we need to move the decimal point to the right so that the first non-zero digit is immediately to the left of the decimal point.

The first non-zero digit is 1. We need to move the decimal point to place it after the digit 1.

Let's count the number of places the decimal point is moved to the right:

0.00000001

Moving the decimal point 1 place to the right gives 0.0000001

Moving the decimal point 2 places to the right gives 0.000001

...and so on, until moving it 8 places to the right gives 1.0.

The number of places the decimal point is moved to the right is 8.

Since the decimal point is moved to the right, the exponent of 10 is negative.

The number of places moved is 8, so the exponent is -8.

The number $a$ is obtained by placing the decimal point after the first non-zero digit: $1.0$.

The value $a = 1.0$ satisfies the condition $1 \leq 1.0 < 10$.

Therefore, 0.00000001 in standard form is $1.0 \times 10^{-8}$ or simply $1 \times 10^{-8}$.

The number in standard form is $1 \times 10^{-8}$ gram.

The annual sales of the sugar factory are given as 3 billion 720 million kilograms.

First, let's write the number numerically:

3 billion = 3,000,000,000

720 million = 720,000,000

Total sales = 3,000,000,000 + 720,000,000 = 3,720,000,000 kilograms.

We need to express the number 3,720,000,000 in standard form.

Standard form (also known as scientific notation) is written as $a \times 10^n$, where $1 \leq |a| < 10$ and $n$ is an integer.

The given number is 3,720,000,000.

The decimal point is implicitly at the end of the number (3720000000.).

To express this in standard form, we move the decimal point to the left until only one non-zero digit remains to the left of the decimal point.

The first non-zero digit from the left is 3. We need to place the decimal point after the digit 3 to get the number 3.72.

Let's count the number of places the decimal point is moved to the left:

Starting from the end (3720000000.), we move past the six 0s, the 0, the 2, and the 7.

The total number of places the decimal point is moved to the left is 6 (from the trailing zeros) + 3 (from the digits 0, 2, 7) = 9 places.

Since the decimal point is moved to the left, the exponent of 10 is positive, and its value is the number of places moved.

The number of places moved is 9, so the exponent is 9.

The number $a$ is obtained by placing the decimal point after the first digit: 3.72.

The value $a = 3.72$ satisfies the condition $1 \leq 3.72 < 10$.

Therefore, 3,720,000,000 in standard form is $3.72 \times 10^9$.

The number in standard form is $3.72 \times 10^9$ kilograms.

Given Information:

Number of red blood cells per cubic millimetre = 5.5 million

Average body blood volume = 5 litres

Conversion factor: 1 litre = 1,00,000 mm$^3$

First, convert the total blood volume from litres to cubic millimetres (mm$^3$).

1 litre = 1,00,000 mm$^3 = 10^5$ mm$^3$

Total volume in mm$^3$ = $5 \text{ litres} \times 10^5 \text{ mm}^3/\text{litre}$

Total volume in mm$^3$ = $5 \times 10^5 \text{ mm}^3$

Next, express the number of red blood cells per cubic millimetre in scientific notation.

5.5 million = $5.5 \times 1,000,000$

$1,000,000 = 10^6$

Number of red blood cells per mm$^3$ = $5.5 \times 10^6$ cells/mm$^3$

Now, calculate the total number of red blood cells in the body by multiplying the number of cells per mm$^3$ by the total volume in mm$^3$.

Total number of red cells = (Number of cells per mm$^3$) $\times$ (Total volume in mm$^3$)

Total number of red cells = $(5.5 \times 10^6) \times (5 \times 10^5)$

Group the decimal parts and the powers of 10:

Total number of red cells = $(5.5 \times 5) \times (10^6 \times 10^5)$

Perform the multiplication of the decimal parts:

$5.5 \times 5 = 27.5$

Perform the multiplication of the powers of 10 using the property $a^m \times a^n = a^{m+n}$:

$10^6 \times 10^5 = 10^{6+5} = 10^{11}$

Combine the results:

Total number of red cells = $27.5 \times 10^{11}$

Finally, express this number in standard form ($a \times 10^n$, where $1 \leq |a| < 10$).

The current coefficient $27.5$ is not in the range $1 \leq a < 10$. We need to move the decimal point one place to the left to get $2.75$.

$27.5 = 2.75 \times 10^1$

Substitute this back into the expression:

Total number of red cells = $(2.75 \times 10^1) \times 10^{11}$

Total number of red cells = $2.75 \times (10^1 \times 10^{11})$

Using the property $a^m \times a^n = a^{m+n}$ again:

$10^1 \times 10^{11} = 10^{1+11} = 10^{12}$

Total number of red cells = $2.75 \times 10^{12}$

This is in standard form.

The total number of red blood cells in the body in standard form is $2.75 \times 10^{12}$.

(a)

The mass of a proton is given as $\frac{1673}{1000000000000000000000000000}$ gram.

The denominator is 1 followed by 27 zeros, which can be written as $10^{27}$.

So, the mass is $\frac{1673}{10^{27}} \text{ g}$.

Using the property $\frac{1}{a^n} = a^{-n}$, we get $1673 \times 10^{-27} \text{ g}$.

To express this in standard form ($a \times 10^n$, where $1 \leq |a| < 10$), we need to write 1673 in standard form.

$1673 = 1.673 \times 10^3$

Substitute this back into the expression:

Mass = $(1.673 \times 10^3) \times 10^{-27} \text{ g}$

Using the property $a^m \times a^n = a^{m+n}$:

Mass = $1.673 \times 10^{3 + (-27)} \text{ g}$

Mass = $1.673 \times 10^{-24} \text{ g}$

The standard form is $1.673 \times 10^{-24}$ g.

(b)

The diameter of a Helium atom is 0.000000022 cm.

To express this in standard form ($a \times 10^n$, where $1 \leq |a| < 10$), we need to move the decimal point to the right until the first non-zero digit (2) is before the decimal point.

0.000000022

Moving the decimal point 8 places to the right gives 2.2.

Since the decimal point is moved to the right, the exponent of 10 is negative, and the value of the exponent is the number of places moved, which is 8.

So, 0.000000022 cm = $2.2 \times 10^{-8}$ cm.

The standard form is $2.2 \times 10^{-8}$ cm.

(c)

The mass of a molecule of hydrogen gas is about 0.00000000000000000000334 tons.

To express this in standard form ($a \times 10^n$, where $1 \leq |a| < 10$), we need to move the decimal point to the right until the first non-zero digit (3) is before the decimal point.

0.00000000000000000000334

Count the number of zeros between the decimal point and the first non-zero digit (3). There are 21 zeros.

Moving the decimal point 22 places to the right gives 3.34.

Since the decimal point is moved to the right, the exponent of 10 is negative, and the value of the exponent is the number of places moved, which is 22.

So, 0.00000000000000000000334 tons = $3.34 \times 10^{-22}$ tons.

The standard form is $3.34 \times 10^{-22}$ tons.

(d)

The human body has 1 trillion cells.

Assuming 1 trillion equals $10^{12}$ (short scale), the number of cells is 1,000,000,000,000.

To express this in standard form ($a \times 10^n$, where $1 \leq |a| < 10$), we need to write 1,000,000,000,000 in standard form.

The number is $10^{12}$. This can be written as $1 \times 10^{12}$.

The value $a = 1$ satisfies $1 \leq 1 < 10$, and the exponent $n = 12$ is an integer.

The standard form is $1 \times 10^{12}$ cells.

(e)

Express 56 km in m.

We know that 1 km = 1000 m.

$1000 = 10^3$

So, 1 km = $10^3$ m.

To convert 56 km to metres, we multiply by $10^3$:

$56 \text{ km} = 56 \times 10^3 \text{ m}$

To express this in standard form ($a \times 10^n$, where $1 \leq |a| < 10$), we need to write 56 in standard form.

$56 = 5.6 \times 10^1$

Substitute this back into the expression:

$56 \times 10^3 \text{ m} = (5.6 \times 10^1) \times 10^3 \text{ m}$

Using the property $a^m \times a^n = a^{m+n}$:

$ = 5.6 \times 10^{1 + 3} \text{ m}$

$ = 5.6 \times 10^4 \text{ m}$

The standard form is $5.6 \times 10^4$ m.

(f)

Express 5 tons in g.

Assuming "ton" refers to the metric ton (tonne), we know that 1 metric ton = 1000 kg.

Also, 1 kg = 1000 g.

So, 1 metric ton = $1000 \text{ kg} = 1000 \times (1000 \text{ g}) = 1,000,000 \text{ g}$.

In terms of powers of 10:

$1000 = 10^3$

1 metric ton = $10^3 \text{ kg} = 10^3 \times 10^3 \text{ g} = 10^{3+3} \text{ g} = 10^6 \text{ g}$.

To convert 5 tons to grams, we multiply by $10^6$:

$5 \text{ tons} = 5 \times 10^6 \text{ g}$

This is already in standard form ($a \times 10^n$) where $a=5$ and $n=6$. The value $a=5$ satisfies $1 \leq 5 < 10$, and $n=6$ is an integer.

The standard form is $5 \times 10^6$ g.

(g)

Express 2 years in seconds.

We need to calculate the number of seconds in 2 years. We will use the approximation that 1 year has 365 days.

1 year = 365 days

1 day = 24 hours

1 hour = 60 minutes

1 minute = 60 seconds

Number of seconds in 1 year = $365 \times 24 \times 60 \times 60$ seconds

$365 \times 24 = 8760$

$60 \times 60 = 3600$

Number of seconds in 1 year = $8760 \times 3600$ seconds

$8760 \times 3600 = 31,536,000$ seconds.

Number of seconds in 2 years = $2 \times (31,536,000 \text{ seconds})$

$2 \times 31,536,000 = 63,072,000$ seconds.

Now, express 63,072,000 in standard form ($a \times 10^n$, where $1 \leq |a| < 10$).

The number is 63,072,000. The decimal point is at the end (63072000.).

Move the decimal point to the left until the first non-zero digit (6) is before the decimal point. We want to get 6.3072.

Moving the decimal point from 63072000. to 6.3072 requires moving it 7 places to the left (past the three 0s, the 2, the 7, the 0, and the 3).

Since the decimal point is moved to the left, the exponent of 10 is positive, and its value is the number of places moved, which is 7.

So, 63,072,000 seconds = $6.3072 \times 10^7$ seconds.

The standard form is $6.3072 \times 10^7$ seconds.

(h)

Express 5 hectares in cm$^2$.

Given that 1 hectare = 10000 m$^2$.

We need to convert m$^2$ to cm$^2$.

We know that 1 m = 100 cm.

$1 \text{ m}^2 = (1 \text{ m}) \times (1 \text{ m}) = (100 \text{ cm}) \times (100 \text{ cm}) = 10000 \text{ cm}^2$.

In terms of powers of 10:

$10000 = 10^4$

$1 \text{ m}^2 = 10^4 \text{ cm}^2$.

Now, substitute this into the hectare conversion:

1 hectare = 10000 m$^2 = 10^4 \text{ m}^2$

1 hectare = $10^4 \times (10^4 \text{ cm}^2) = 10^{4+4} \text{ cm}^2 = 10^8 \text{ cm}^2$.

To convert 5 hectares to cm$^2$, we multiply by $10^8$:

$5 \text{ hectares} = 5 \times 10^8 \text{ cm}^2$

This is already in standard form ($a \times 10^n$) where $a=5$ and $n=8$. The value $a=5$ satisfies $1 \leq 5 < 10$, and $n=8$ is an integer.

The standard form is $5 \times 10^8$ cm$^2$.

The given equation is:

$\left( \frac{2}{9} \right)^{3} \times \left( \frac{2}{9} \right)^{-6}$ = $\left( \frac{2}{9} \right)^{2x-1}$

We use the exponent rule for multiplication of terms with the same base: $a^m \times a^n = a^{m+n}$.

Applying this rule to the left side of the equation:

$\left( \frac{2}{9} \right)^{3} \times \left( \frac{2}{9} \right)^{-6} = \left( \frac{2}{9} \right)^{3 + (-6)}$

$= \left( \frac{2}{9} \right)^{3 - 6}$

$= \left( \frac{2}{9} \right)^{-3}$

Substituting this back into the original equation, we get:

$\left( \frac{2}{9} \right)^{-3} = \left( \frac{2}{9} \right)^{2x-1}$

Since the bases on both sides of the equation are equal ($\frac{2}{9}$), the exponents must also be equal.

Therefore,

$-3 = 2x - 1$

Now, we solve the linear equation for $x$:

Add 1 to both sides:

$-3 + 1 = 2x - 1 + 1$

$-2 = 2x$

Divide both sides by 2:

$\frac{-2}{2} = \frac{2x}{2}$

$-1 = x$

So, the value of $x$ is $\textbf{-1}$.

Given:

The first number is $\left( \frac{-3}{2} \right)^{-3}$. The required quotient is $\left( \frac{4}{27} \right)^{-2}$.

To Find:

The number by which $\left( \frac{-3}{2} \right)^{-3}$ should be divided to get $\left( \frac{4}{27} \right)^{-2}$.

Solution:

Let the required number be $x$.

According to the problem, we can write the equation:

$\left( \frac{-3}{2} \right)^{-3} \div x = \left( \frac{4}{27} \right)^{-2}$

This can also be written as:

$\frac{\left( \frac{-3}{2} \right)^{-3}}{x} = \left( \frac{4}{27} \right)^{-2}$

First, let's simplify the terms with negative exponents using the rule $a^{-n} = \left(\frac{1}{a}\right)^n$.

Simplify the left side:

$\left( \frac{-3}{2} \right)^{-3} = \left( \frac{2}{-3} \right)^3 = \left( \frac{-2}{3} \right)^3 = \frac{(-2)^3}{3^3} = \frac{-8}{27}$

Simplify the right side:

$\left( \frac{4}{27} \right)^{-2} = \left( \frac{27}{4} \right)^2 = \frac{27^2}{4^2} = \frac{729}{16}$

Now, substitute these simplified values back into the equation:

$\frac{-8}{27} \div x = \frac{729}{16}$

To solve for $x$, we can rewrite the equation as:

$\frac{-8}{27} = x \times \frac{729}{16}$

Now, isolate $x$ by dividing both sides by $\frac{729}{16}$ (which is the same as multiplying by its reciprocal $\frac{16}{729}$):

$x = \frac{-8}{27} \div \frac{729}{16}$

$x = \frac{-8}{27} \times \frac{16}{729}$

Multiply the numerators and the denominators:

$x = \frac{-8 \times 16}{27 \times 729}$

$x = \frac{-128}{19683}$

There are no common factors between $128 = 2^7$ and $19683 = 3^9$, so the fraction is in its simplest form.

Thus, the required number is $\frac{-128}{19683}$.

Given:

The equation is $\frac{6^{n}}{6^{-2}} = 6^{3}$.

To Find:

The value of $n$.

Solution:

We are given the equation:

$\frac{6^{n}}{6^{-2}} = 6^{3}$

Using the law of exponents for division with the same base, which states $\frac{a^m}{a^n} = a^{m-n}$, we can simplify the left side of the equation.

$6^{n - (-2)} = 6^{3}$

Simplify the exponent:

$6^{n+2} = 6^{3}$

Since the bases on both sides of the equation are the same (both are 6), the exponents must be equal.

Therefore, we can equate the exponents:

$n+2 = 3$

To find $n$, subtract 2 from both sides:

$n = 3 - 2$

$n = 1$

Thus, the value of $n$ is 1.

Given:

The equation is $\frac{2^{n}\;\times\;2^{6}}{2^{-3}} = 2^{18}$.

To Find:

The value of $n$.

Solution:

We are given the equation:

$\frac{2^{n}\;\times\;2^{6}}{2^{-3}} = 2^{18}$

First, simplify the numerator using the law of exponents for multiplication with the same base, which states $a^m \times a^n = a^{m+n}$.

$2^{n+6}$

Now the equation becomes:

$\frac{2^{n+6}}{2^{-3}} = 2^{18}$

Next, simplify the left side using the law of exponents for division with the same base, which states $\frac{a^m}{a^n} = a^{m-n}$.

$2^{(n+6) - (-3)} = 2^{18}$

Simplify the exponent on the left side:

$2^{n+6+3} = 2^{18}$

$2^{n+9} = 2^{18}$

Since the bases on both sides of the equation are the same (both are 2), the exponents must be equal.

Therefore, we can equate the exponents:

$n+9 = 18$

To find $n$, subtract 9 from both sides:

$n = 18 - 9$

$n = 9$

Thus, the value of $n$ is 9.

Given:

The expression to be simplified is $\frac{125\;\times\;x^{-3}}{5^{-3}\;\times\;25\;\times\;x^{-6}}$.

To Simplify:

Simplify the given expression.

Solution:

We have the expression:

$\frac{125\;\times\;x^{-3}}{5^{-3}\;\times\;25\;\times\;x^{-6}}$

First, express the constant terms 125 and 25 as powers of 5:

$125 = 5 \times 5 \times 5 = 5^3$

$25 = 5 \times 5 = 5^2$

Substitute these into the expression:

$\frac{5^{3}\;\times\;x^{-3}}{5^{-3}\;\times\;5^{2}\;\times\;x^{-6}}$

Now, simplify the denominator by combining the powers of 5 using the rule $a^m \times a^n = a^{m+n}$:

$5^{-3} \times 5^{2} = 5^{-3+2} = 5^{-1}$

The expression becomes:

$\frac{5^{3}\;\times\;x^{-3}}{5^{-1}\;\times\;x^{-6}}$

Now, use the rule for dividing powers with the same base, $\frac{a^m}{a^n} = a^{m-n}$. Apply this separately to the powers of 5 and the powers of $x$.

For the powers of 5: $\frac{5^3}{5^{-1}} = 5^{3 - (-1)} = 5^{3+1} = 5^4$

For the powers of $x$: $\frac{x^{-3}}{x^{-6}} = x^{-3 - (-6)} = x^{-3+6} = x^3$

Combine these results to get the simplified expression:

$5^4 \times x^3$

Calculate $5^4$:

$5^4 = 5 \times 5 \times 5 \times 5 = 25 \times 25 = 625$

So, the simplified expression is:

$625 x^3$

Given:

The expression to be simplified is $\frac{16\;\times\;10^{2}\;\times\;64}{2^{4}\;\times\;4^{2}}$.

To Simplify:

Simplify the given expression.

Solution:

We have the expression:

$\frac{16\;\times\;10^{2}\;\times\;64}{2^{4}\;\times\;4^{2}}$

Express all the numbers in the numerator and the denominator as powers of their prime factors, preferably 2 and 5 where possible.

$16 = 2 \times 2 \times 2 \times 2 = 2^4$

$10 = 2 \times 5$, so $10^2 = (2 \times 5)^2 = 2^2 \times 5^2$

$64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6$

$4 = 2 \times 2 = 2^2$, so $4^2 = (2^2)^2 = 2^{2 \times 2} = 2^4$

Substitute these into the expression:

$\frac{2^{4}\;\times\;(2^{2}\;\times\;5^{2})\;\times\;2^{6}}{2^{4}\;\times\;2^{4}}$

Now, use the law of exponents $a^m \times a^n = a^{m+n}$ to combine the terms with the same base in the numerator and the denominator.

Numerator: $2^4 \times 2^2 \times 5^2 \times 2^6 = 2^{4+2+6} \times 5^2 = 2^{12} \times 5^2$

Denominator: $2^4 \times 2^4 = 2^{4+4} = 2^8$

The expression becomes:

$\frac{2^{12}\;\times\;5^{2}}{2^{8}}$

Now, use the law of exponents $\frac{a^m}{a^n} = a^{m-n}$ to simplify the terms with base 2:

$2^{12-8} \times 5^2 = 2^4 \times 5^2$

Calculate the values:

$2^4 = 2 \times 2 \times 2 \times 2 = 16$

$5^2 = 5 \times 5 = 25$

Multiply the results:

$16 \times 25$

$16 \times 25 = 400$

Thus, the simplified value of the expression is 400.

Given:

The equation is $\frac{5^m\;\times\;5^{3}\;\times\;5^{-2}}{5^{-5}} = 5^{12}$.

To Find:

The value of $m$.

Solution:

We are given the equation:

$\frac{5^m\;\times\;5^{3}\;\times\;5^{-2}}{5^{-5}} = 5^{12}$

First, simplify the numerator using the law of exponents for multiplication with the same base, which states $a^p \times a^q \times a^r = a^{p+q+r}$.

$5^m \times 5^3 \times 5^{-2} = 5^{m+3+(-2)} = 5^{m+3-2} = 5^{m+1}$

Now the equation becomes:

$\frac{5^{m+1}}{5^{-5}} = 5^{12}$

Next, simplify the left side using the law of exponents for division with the same base, which states $\frac{a^p}{a^q} = a^{p-q}$.

$5^{(m+1) - (-5)} = 5^{12}$

Simplify the exponent on the left side:

$5^{m+1+5} = 5^{12}$

$5^{m+6} = 5^{12}$

Since the bases on both sides of the equation are the same (both are 5), the exponents must be equal.

Therefore, we can equate the exponents:

$m+6 = 12$

To find $m$, subtract 6 from both sides:

$m = 12 - 6$

$m = 6$

Thus, the value of $m$ is 6.

Given:

Weight of a new born bear = 4 kg.

The weight increases by the power of 2 in 5 years.

To Find:

The weight of the five year old bear.

Solution:

The problem states that the weight of the bear increases by the power of 2 in 5 years. This implies that the initial weight is multiplied by $2^5$ over the period of 5 years.

Initial weight = 4 kg

Increase factor over 5 years = $2^5$

Calculate the increase factor:

$2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32$

The weight after 5 years will be the initial weight multiplied by the increase factor.

Weight after 5 years = Initial weight $\times$ Increase factor

Weight after 5 years = $4 \text{ kg} \times 32$

$4 \times 32 = 128$

So, the weight of the five year old bear might be 128 kg.

Weight of five year old bear = 128 kg.

Given:

Initial number of bacterial cells = 1.

Doubling time = 30 minutes.

To Find:

The number of cells after:

(a) 12 hours

(b) 24 hours

Solution:

The number of cells doubles every 30 minutes. This means the number of cells after $t$ minutes, starting with 1 cell, can be calculated by finding how many 30-minute periods are in $t$ minutes and then raising 2 to that power.

Number of doubling periods = $\frac{\text{Total time}}{\text{Doubling time}}$

If the initial number of cells is $N_0$ and the doubling time is $T_d$, then the number of cells $N(t)$ after time $t$ is given by $N(t) = N_0 \times 2^{t/T_d}$. Here, $N_0 = 1$ and $T_d = 30$ minutes.

So, $N(t) = 1 \times 2^{t/30} = 2^{t/30}$, where $t$ is in minutes.

(a) After 12 hours:

Convert 12 hours to minutes:

$12 \text{ hours} = 12 \times 60 \text{ minutes} = 720 \text{ minutes}$

Number of doubling periods in 12 hours = $\frac{720 \text{ minutes}}{30 \text{ minutes/period}} = 24 \text{ periods}$.

Number of cells after 12 hours = $2^{24}$

Calculate $2^{24}$:

$2^{10} = 1024$

$2^{20} = (2^{10})^2 = 1024^2 = 1,048,576$

$2^{24} = 2^{20} \times 2^4 = 1,048,576 \times 16$

$1,048,576 \times 16 = 16,777,216$

So, after 12 hours, there will be $2^{24} = 16,777,216$ cells.

(b) After 24 hours:

Convert 24 hours to minutes:

$24 \text{ hours} = 24 \times 60 \text{ minutes} = 1440 \text{ minutes}$

Number of doubling periods in 24 hours = $\frac{1440 \text{ minutes}}{30 \text{ minutes/period}} = 48 \text{ periods}$.

Number of cells after 24 hours = $2^{48}$

Calculate $2^{48}$:

$2^{48} = (2^{24})^2 = (16,777,216)^2$

Calculating $(16,777,216)^2$ gives a very large number.

Alternatively, $2^{48} = (2^{10})^4 \times 2^8 = (1024)^4 \times 256$.

Using logarithms or a calculator, $2^{48} \approx 2.81 \times 10^{14}$.

The exact value is $281,474,976,710,656$.

So, after 24 hours, there will be $2^{48} = 281,474,976,710,656$ cells.

Summary:

(a) After 12 hours, there will be $2^{24} = 16,777,216$ cells.

(b) After 24 hours, there will be $2^{48} = 281,474,976,710,656$ cells.

Given:

Distance of Planet A from Earth = $9.35 \times 10^{6}$ km.

Distance of Planet B from Earth = $6.27 \times 10^{7}$ km.

To Find:

Which planet is nearer to Earth.

Solution:

To determine which planet is nearer, we need to compare their distances from Earth. The smaller distance corresponds to the nearer planet.

The distances are given in scientific notation:

Distance of Planet A = $9.35 \times 10^{6}$ km

Distance of Planet B = $6.27 \times 10^{7}$ km

To compare these numbers easily, it is helpful to express them with the same power of 10.

Let's express the distance of Planet A with the exponent $10^7$. To change the exponent from 6 to 7 (an increase of 1), we move the decimal point one place to the left:

$9.35 \times 10^{6} = 0.935 \times 10^{7}$ km

Now we compare $0.935 \times 10^{7}$ km (Distance of Planet A) with $6.27 \times 10^{7}$ km (Distance of Planet B).

Since the powers of 10 are the same ($10^7$), we compare the decimal parts:

Compare $0.935$ and $6.27$.

Clearly, $0.935 < 6.27$.

Therefore, $0.935 \times 10^{7} < 6.27 \times 10^{7}$.

Substituting back the original value for Planet A's distance:

$9.35 \times 10^{6} \text{ km} < 6.27 \times 10^{7} \text{ km}$.

Since the distance of Planet A from Earth is less than the distance of Planet B from Earth, Planet A is nearer to Earth.

Conclusion: Planet A is nearer to Earth.

Given:

Initial number of bacterial cells = 1.

Doubling time = 1 hour.

Total time elapsed = 8 hours.

To Find:

The number of cells after 8 hours, expressed in powers.

Solution:

The number of cells doubles every hour. This means that after each hour, the number of cells is multiplied by 2.

Initial number of cells = 1

After 1 hour, the number of cells = $1 \times 2 = 2^1$

After 2 hours, the number of cells = $2 \times 2 = 2^2$

After 3 hours, the number of cells = $2^2 \times 2 = 2^3$

In general, after $t$ hours, the number of cells, starting with 1 cell and doubling every hour, will be $2^t$.

In this case, the total time is 8 hours ($t=8$).

Number of cells after 8 hours = $2^{8}$

We are asked to express the answer in powers, so $2^8$ is the required format.

If we were to calculate the value:

$2^8 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

$2^8 = 16 \times 16 = 256$

However, the question asks for the answer in powers, so the answer is $2^8$.

Thus, after 8 hours, there will be $2^8$ cells.

Answer in powers:

Number of cells after 8 hours = $2^8$.

Given:

An insect starts at the 0 point on a number line and hops towards 1.

With each hop, the insect covers half the distance from its current location to 1.

To Find:

(a) A table showing the insect's location for the first 10 hops.

(b) The insect's location after $n$ hops.

(c) Whether the insect will ever get to 1, with an explanation.

Solution:

Let $P_k$ be the position of the insect after $k$ hops. The insect starts at $P_0 = 0$. The target is 1.

The distance from the current location $P_k$ to 1 is $1 - P_k$.

With each hop, the insect covers half of this distance, which is $\frac{1}{2}(1 - P_k)$.

The new position after the $(k+1)$-th hop is $P_{k+1} = P_k + \frac{1}{2}(1 - P_k)$.

Let's find the first few positions:

$P_0 = 0$

$P_1 = P_0 + \frac{1}{2}(1 - P_0) = 0 + \frac{1}{2}(1 - 0) = \frac{1}{2}$

$P_2 = P_1 + \frac{1}{2}(1 - P_1) = \frac{1}{2} + \frac{1}{2}(1 - \frac{1}{2}) = \frac{1}{2} + \frac{1}{2}(\frac{1}{2}) = \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}$

$P_3 = P_2 + \frac{1}{2}(1 - P_2) = \frac{3}{4} + \frac{1}{2}(1 - \frac{3}{4}) = \frac{3}{4} + \frac{1}{2}(\frac{1}{4}) = \frac{3}{4} + \frac{1}{8} = \frac{6}{8} + \frac{1}{8} = \frac{7}{8}$

From these calculations, we can observe a pattern. The position after $k$ hops seems to be $P_k = 1 - (\frac{1}{2})^k$.

Let's verify this general formula. If $P_k = 1 - (\frac{1}{2})^k$, then $1 - P_k = 1 - (1 - (\frac{1}{2})^k) = (\frac{1}{2})^k$.

The next position is $P_{k+1} = P_k + \frac{1}{2}(1 - P_k) = (1 - (\frac{1}{2})^k) + \frac{1}{2}(\frac{1}{2})^k = 1 - (\frac{1}{2})^k + (\frac{1}{2})^{k+1} = 1 - \frac{1}{2} \times (\frac{1}{2})^k = 1 - (\frac{1}{2})^{k+1}$.

This confirms the formula $P_n = 1 - (\frac{1}{2})^n$ for the position after $n$ hops.

(a) Table showing the insect’s location for the first 10 hops:

We use the formula $P_n = 1 - (\frac{1}{2})^n$ for $n=1, 2, \dots, 10$.

(b) Where will the insect be after n hops?

Based on our derivation, the insect's location after $n$ hops will be $P_n = 1 - (\frac{1}{2})^n$.

(c) Will the insect ever get to 1? Explain.

The location of the insect after $n$ hops is given by $P_n = 1 - (\frac{1}{2})^n$.

For the insect to reach 1, its position must be exactly 1. This means we need to find if there is a finite number of hops $n$ such that $P_n = 1$.

$1 - (\frac{1}{2})^n = 1$

Subtracting 1 from both sides, we get:

$-(\frac{1}{2})^n = 0$

This is equivalent to:

$(\frac{1}{2})^n = 0$

The expression $(\frac{1}{2})^n$ is equal to $\frac{1}{2^n}$. For any finite, non-negative integer value of $n$, $2^n$ will be a positive finite number. Therefore, $\frac{1}{2^n}$ will always be a positive number greater than 0.

Since $\frac{1}{2^n}$ is never equal to 0 for any finite number of hops $n$, the equation $(\frac{1}{2})^n = 0$ has no solution for finite $n$.

This means that the position $P_n = 1 - (\frac{1}{2})^n$ will always be slightly less than 1 for any finite number of hops.

As the number of hops $n$ increases, $(\frac{1}{2})^n$ gets smaller and smaller, approaching 0. This means $P_n$ gets closer and closer to 1. Mathematically, we say that the limit of $P_n$ as $n$ approaches infinity is 1 ($\lim\limits_{n \to \infty} (1 - (\frac{1}{2})^n) = 1$).

However, in a finite number of steps, the insect will always cover only half of the remaining distance, leaving the other half uncovered. Thus, it will always be some distance away from 1.

Therefore, the insect will never actually get to 1 in a finite number of hops.

Below is the completed table showing the powers and the pattern of their ones digits.

(a) Describe patterns you see in the ones digits of the powers.

We observe the following patterns in the ones digits of the powers:

For bases ending in 1: The ones digit is always 1.

For bases ending in 0: The ones digit is always 0.

For bases ending in 5: The ones digit is always 5.

For bases ending in 6: The ones digit is always 6.

For bases ending in 4: The ones digits cycle through 4, 6 (cycle length 2). The ones digit is 4 if the exponent is odd, and 6 if the exponent is even.

For bases ending in 9: The ones digits cycle through 9, 1 (cycle length 2). The ones digit is 9 if the exponent is odd, and 1 if the exponent is even.

For bases ending in 2: The ones digits cycle through 2, 4, 8, 6 (cycle length 4). The ones digit depends on the remainder when the exponent is divided by 4. Remainder 1 gives 2, Remainder 2 gives 4, Remainder 3 gives 8, Remainder 0 (or 4) gives 6.

For bases ending in 3: The ones digits cycle through 3, 9, 7, 1 (cycle length 4). The ones digit depends on the remainder when the exponent is divided by 4. Remainder 1 gives 3, Remainder 2 gives 9, Remainder 3 gives 7, Remainder 0 (or 4) gives 1.

For bases ending in 7: The ones digits cycle through 7, 9, 3, 1 (cycle length 4). The ones digit depends on the remainder when the exponent is divided by 4. Remainder 1 gives 7, Remainder 2 gives 9, Remainder 3 gives 3, Remainder 0 (or 4) gives 1.

For bases ending in 8: The ones digits cycle through 8, 4, 2, 6 (cycle length 4). The ones digit depends on the remainder when the exponent is divided by 4. Remainder 1 gives 8, Remainder 2 gives 4, Remainder 3 gives 2, Remainder 0 (or 4) gives 6.

In general, the ones digit of a power depends only on the ones digit of the base and the exponent.

(b) Predict the ones digit in the following:

1. $4^{12}$

The base ends in 4. The pattern of ones digits for powers of 4 is 4, 6 (cycle length 2).

The exponent is 12, which is an even number. The second digit in the cycle is 6.

Alternatively, $12 \pmod 2 = 0$. The ones digit is the last digit in the cycle (or the second digit in the cycle of length 2).

The ones digit of $4^{12}$ is 6.

2. $9^{20}$

The base ends in 9. The pattern of ones digits for powers of 9 is 9, 1 (cycle length 2).

The exponent is 20, which is an even number. The second digit in the cycle is 1.

Alternatively, $20 \pmod 2 = 0$. The ones digit is the last digit in the cycle (or the second digit in the cycle of length 2).

The ones digit of $9^{20}$ is 1.

3. $3^{17}$

The base ends in 3. The pattern of ones digits for powers of 3 is 3, 9, 7, 1 (cycle length 4).

The exponent is 17. We find the remainder when 17 is divided by 4: $17 \div 4 = 4$ with a remainder of 1, or $17 \pmod 4 = 1$.

The remainder 1 corresponds to the first digit in the cycle, which is 3.

The ones digit of $3^{17}$ is 3.

4. $5^{100}$

The base ends in 5. The ones digit for any positive integer power of a number ending in 5 is always 5.

The ones digit of $5^{100}$ is 5.

5. $10^{500}$

The base ends in 0. The ones digit for any positive integer power of a number ending in 0 is always 0.

The ones digit of $10^{500}$ is 0.

(c) Predict the ones digit in the following:

To predict the ones digit of a power of a multi-digit number, we only need to consider the ones digit of the base.

1. $31^{10}$

The base is 31, which ends in 1. The ones digit of $31^{10}$ is the same as the ones digit of $1^{10}$.

The ones digit of $1^{10}$ is always 1.

The ones digit of $31^{10}$ is 1.

2. $12^{10}$

The base is 12, which ends in 2. The ones digit of $12^{10}$ is the same as the ones digit of $2^{10}$.

The pattern of ones digits for powers of 2 is 2, 4, 8, 6 (cycle length 4).

The exponent is 10. We find the remainder when 10 is divided by 4: $10 \div 4 = 2$ with a remainder of 2, or $10 \pmod 4 = 2$.

The remainder 2 corresponds to the second digit in the cycle, which is 4.

The ones digit of $12^{10}$ is 4.

3. $17^{21}$

The base is 17, which ends in 7. The ones digit of $17^{21}$ is the same as the ones digit of $7^{21}$.

The pattern of ones digits for powers of 7 is 7, 9, 3, 1 (cycle length 4).

The exponent is 21. We find the remainder when 21 is divided by 4: $21 \div 4 = 5$ with a remainder of 1, or $21 \pmod 4 = 1$.

The remainder 1 corresponds to the first digit in the cycle, which is 7.

The ones digit of $17^{21}$ is 7.

4. $29^{10}$

The base is 29, which ends in 9. The ones digit of $29^{10}$ is the same as the ones digit of $9^{10}$.

The pattern of ones digits for powers of 9 is 9, 1 (cycle length 2).

The exponent is 10, which is an even number. The second digit in the cycle is 1.

Alternatively, $10 \pmod 2 = 0$. The ones digit is the last digit in the cycle (or the second digit in the cycle of length 2).

The ones digit of $29^{10}$ is 1.

Below is the completed table showing the mass of each celestial body in standard notation.

(a) The mass of each planet and the Moon in scientific notation:

Mercury: $3.30 \times 10^{23}$ kg

Venus: $4.87 \times 10^{24}$ kg

Earth: $5.97 \times 10^{24}$ kg

Mars: $6.42 \times 10^{23}$ kg (Using the standard value)

Jupiter: $1.90 \times 10^{27}$ kg

Saturn: $5.68 \times 10^{26}$ kg

Uranus: $8.68 \times 10^{25}$ kg

Neptune: $1.02 \times 10^{26}$ kg

Pluto: $1.27 \times 10^{22}$ kg

Moon: $7.35 \times 10^{22}$ kg

(b) Ordering the planets and the moon by mass, from least to greatest:

To order by mass, we compare the exponents first, then the coefficient (the number before $\times 10$).

The masses (in kg) are approximately:

Pluto: $1.27 \times 10^{22}$

Moon: $7.35 \times 10^{22}$

Mercury: $3.30 \times 10^{23}$

Mars: $6.42 \times 10^{23}$

Venus: $4.87 \times 10^{24}$

Earth: $5.97 \times 10^{24}$

Uranus: $8.68 \times 10^{25}$

Neptune: $1.02 \times 10^{26}$

Saturn: $5.68 \times 10^{26}$

Jupiter: $1.90 \times 10^{27}$

Ordering from least to greatest mass:

Pluto, Moon, Mercury, Mars, Venus, Earth, Uranus, Neptune, Saturn, Jupiter.

(c) Which planet has about the same mass as earth?

Earth's mass is $5.97 \times 10^{24}$ kg.

Comparing this to the other planetary masses:

Venus has a mass of $4.87 \times 10^{24}$ kg. This is the closest mass among the planets to Earth's mass, having the same exponent ($10^{24}$) and a coefficient ($4.87$) that is relatively close to Earth's coefficient ($5.97$).

The planet with about the same mass as Earth is Venus.

Below is the completed table showing the average distance from each planet to the Sun in standard notation.

(a) The distance from each planet to the Sun in scientific notation:

Earth: $1.496 \times 10^{8}$ km

Jupiter: $7.783 \times 10^{8}$ km

Mars: $2.279 \times 10^{8}$ km

Mercury: $5.79 \times 10^{7}$ km

Neptune: $4.497 \times 10^{9}$ km

Pluto: $5.90 \times 10^{9}$ km

Saturn: $1.427 \times 10^{9}$ km

Uranus: $2.870 \times 10^{9}$ km

Venus: $1.082 \times 10^{8}$ km

(b) Ordering the planets from closest to the sun to farthest from the sun:

We compare the exponents first. The smallest exponent means the closest distance. If exponents are the same, we compare the coefficient.

The exponents are 7, 8, and 9.

$10^7$: Mercury ($5.79$)

$10^8$: Venus ($1.082$), Earth ($1.496$), Mars ($2.279$), Jupiter ($7.783$)

$10^9$: Saturn ($1.427$), Uranus ($2.870$), Neptune ($4.497$), Pluto ($5.90$)

Ordering the planets from closest to farthest:

Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, Pluto.

To compare the masses easily, we convert all masses to the same power of 10. Let's use $10^{-26}$ kg.

Mass of Titanium = $7.95 \times 10^{-26}$ kg

Mass of Lead = $3.44 \times 10^{-25}$ kg = $3.44 \times 10^{1} \times 10^{-26}$ kg = $34.4 \times 10^{-26}$ kg

Mass of Silver = $1.44 \times 10^{-25}$ kg = $1.44 \times 10^{1} \times 10^{-26}$ kg = $14.4 \times 10^{-26}$ kg

Mass of Lithium = $1.15 \times 10^{-26}$ kg

Mass of Hydrogen = $1.674 \times 10^{-27}$ kg = $1.674 \times 10^{-1} \times 10^{-26}$ kg = $0.1674 \times 10^{-26}$ kg

Now, we compare the coefficients of $10^{-26}$:

Hydrogen: $0.1674$

Lithium: $1.15$

Titanium: $7.95$

Silver: $14.4$

Lead: $34.4$

Comparing these values, we can answer the questions.

(a) The heaviest element is the one with the largest mass. The largest coefficient is $34.4$, which corresponds to Lead.

The heaviest element is Lead.

(b) We compare the mass of Silver and Titanium.

Mass of Silver = $14.4 \times 10^{-26}$ kg

Mass of Titanium = $7.95 \times 10^{-26}$ kg

Since $7.95 < 14.4$, Titanium is lighter than Silver.

The lighter element is Titanium.

(c) To list the elements from lightest to heaviest, we arrange them in increasing order of their masses (based on the coefficients of $10^{-26}$).

$0.1674 < 1.15 < 7.95 < 14.4 < 34.4$

This order corresponds to:

Hydrogen ($0.1674 \times 10^{-26}$ kg)

Lithium ($1.15 \times 10^{-26}$ kg)

Titanium ($7.95 \times 10^{-26}$ kg)

Silver ($14.4 \times 10^{-26}$ kg)

Lead ($34.4 \times 10^{-26}$ kg)

The order from lightest to heaviest is:

Hydrogen, Lithium, Titanium, Silver, Lead.

Given:

Distance of Uranus from the Sun = $2,896,819,200,000$ metres.

To Find:

The distance in standard form.

Solution:

Standard form (or scientific notation) is a way of writing very large or very small numbers concisely.

A number in standard form is written as $a \times 10^n$, where $1 \le |a| < 10$ and $n$ is an integer.

We are given the distance as $2,896,819,200,000$ metres.

To convert this number to standard form, we need to move the decimal point (which is currently at the end of the number) to the left so that only one non-zero digit is before the decimal point.

The number is $2,896,819,200,000$.

Moving the decimal point after the first digit (2):

$2\text{.}\underbrace{896819200000}_{12 \text{ digits}}$

We moved the decimal point 12 places to the left.

The value of $a$ is the number formed by the digits with the decimal point placed after the first digit, which is $2.896819200000$. We can drop the trailing zeros after the last non-zero digit in the decimal part, so $a = 2.8968192$.

The exponent $n$ is the number of places the decimal point was moved. Since we moved it 12 places to the left, the exponent is positive, $n=12$.

Therefore, the standard form is $2.8968192 \times 10^{12}$.

The distance of Uranus from the Sun in standard form is $\mathbf{2.8968192 \times 10^{12}}$ metres.

Given:

Distance equivalent to one inch = $0.02543$ metres.

To Find:

The distance in standard form.

Solution:

Standard form (or scientific notation) is a way of writing very large or very small numbers concisely.

A number in standard form is written as $a \times 10^n$, where $1 \le |a| < 10$ and $n$ is an integer.

We are given the distance as $0.02543$ metres.

To convert this number to standard form, we need to move the decimal point to the right so that only one non-zero digit is before the decimal point.

The number is $0.02543$.

Moving the decimal point after the first non-zero digit (2):

$0\text{.}\underbrace{02}_{2 \text{ places}}543 \implies 2.543$

We moved the decimal point 2 places to the right.

The value of $a$ is the number formed by the digits with the decimal point placed after the first non-zero digit, which is $2.543$. So, $a = 2.543$.

The exponent $n$ is the number of places the decimal point was moved. Since we moved it 2 places to the right, the exponent is negative, $n=-2$.

Therefore, the standard form is $2.543 \times 10^{-2}$.

The distance equivalent to one inch in standard form is $\mathbf{2.543 \times 10^{-2}}$ metres.

Given:

The ratio of the volume of the Earth to the volume of the Sun is approximately $7.67 \times 10^{-7}$.

To Find:

The given figure in usual form.

Solution:

The given figure is in standard form (scientific notation): $7.67 \times 10^{-7}$.

A number in standard form is written as $a \times 10^n$, where $1 \le |a| < 10$ and $n$ is an integer.

In this case, $a = 7.67$ and $n = -7$.

To convert a number from standard form to usual form, we look at the exponent $n$ of $10$.

If $n$ is positive, we move the decimal point $|n|$ places to the right.

If $n$ is negative, we move the decimal point $|n|$ places to the left.

Here, the exponent is $n = -7$, which is negative. So, we need to move the decimal point 7 places to the left from its current position in $7.67$.

The number is $7.67$. The decimal point is between 7 and 6.

Moving the decimal point 7 places to the left:

Starting number: $7.67$

Moving 1 place left: $0.767$

Moving 2 places left: $0.0767$

Moving 3 places left: $0.00767$

Moving 4 places left: $0.000767$

Moving 5 places left: $0.0000767$

Moving 6 places left: $0.00000767$

Moving 7 places left: $0.000000767$

The usual form of $7.67 \times 10^{-7}$ is $0.000000767$.

The figure in usual form is $\mathbf{0.000000767}$.

Given:

Mass of an electron = $9.1093826 \times 10^{-31}$ kilograms.

To Find:

The mass of the electron in grams.

Solution:

We are given the mass of the electron in kilograms (kg) and need to convert it to grams (g).

The relationship between kilograms and grams is:

We can write 1000 as a power of 10:

So, the conversion factor is $1 \text{ kg} = 10^3 \text{ g}$.

To convert the mass from kilograms to grams, we multiply the mass in kilograms by the conversion factor ($10^3$).

Mass in grams = (Mass in kilograms) $\times$ (Conversion factor)

Using the rule for multiplying powers with the same base ($a^m \times a^n = a^{m+n}$):

The resulting value $9.1093826 \times 10^{-28}$ is already in standard form, as $9.1093826$ is between 1 and 10, and $-28$ is an integer exponent.

The mass of an electron in grams is $\mathbf{9.1093826 \times 10^{-28}}$ grams.

Given:

World population at the end of the 20th century = $6.1 \times 10^9$ people.

To Find:

1. The population figure in usual form.

2. How to say this number in words.

Solution:

The given population figure is in standard form: $6.1 \times 10^9$.

This form is $a \times 10^n$, where $a = 6.1$ and $n = 9$.

Part 1: Expressing in Usual Form

To convert from standard form to usual form when the exponent $n$ is positive, we move the decimal point $|n|$ places to the right.

Here, $n=9$, so we need to move the decimal point 9 places to the right from its position in $6.1$.

The number is $6.1$. We move the decimal point 9 places to the right:

$6.1 \times 10^9 = 6\underbrace{1000000000}_{9 \text{ places}}$

This number is $6,100,000,000$.

The population in usual form is $\mathbf{6,100,000,000}$ people.

Part 2: Saying the Number in Words

The usual form of the number is $6,100,000,000$.

To read this number, we group the digits into periods of three from the right:

$6 \text{,}$ $100 \text{,}$ $000 \text{,}$ $000$

The periods are: Billions, Millions, Thousands, Ones.

We have 6 in the billions period, 100 in the millions period, and 0 in the thousands and ones periods.

So, the number is read as "Six billion, one hundred million".

In words, the number is Six billion, one hundred million.

Solution:

Shikha is counting ancestors in previous generations. Going back one generation, she has 2 parents. Going back two generations, she has 4 grandparents. Going back three generations, she has 8 great-grandparents. This pattern shows that the number of ancestors doubles with each generation going back in time. This is an example of exponential growth, specifically a geometric progression with a common ratio of 2.

(a) Table and Graph showing the number of ancestors in each of the 12 generations.

Let $n$ represent the number of generations back from the present (where $n=1$ is the generation of parents, $n=2$ is the generation of grandparents, and so on). The number of ancestors in generation $n$ is given by $2^n$. We can create a table for the first 12 generations.

The total number of ancestors over all 12 generations would be the sum of ancestors in each generation: $2^1 + 2^2 + \dots + 2^{12}$. However, the question asks for the number of ancestors *in* each generation.

For the graph, we would plot the generation number ($n$) on the horizontal axis and the number of ancestors ($2^n$) on the vertical axis. The points would be $(1, 2), (2, 4), (3, 8), \dots, (12, 4096)$. The graph would show a steep upward curve, characteristic of exponential growth.

(b) Equation for the number of ancestors in a given generation n.

Let $A$ be the number of ancestors in generation $n$, where $n$ is the number of generations back from the present.

From the pattern observed in part (a):

Generation 1: $A = 2 = 2^1$

Generation 2: $A = 4 = 2^2$

Generation 3: $A = 8 = 2^3$

...and so on.

The number of ancestors in generation $n$ is given by the formula:

$A = 2^n$

where $n$ is the number of generations back from the present ($n \geq 1$).

Given:

Amount of water flow per day = 230 billion litres.

To Find:

1. Amount of water flow in a week.

2. Amount of water flow in a year.

Write the answers in standard notation.

Solution:

We are given that the water flow per day is 230 billion litres.

One billion is equal to $10^9$.

So, the water flow per day is $230 \times 10^9$ litres.

First, let's express 230 in scientific notation:

$230 = 2.3 \times 10^2$

Water flow per day $= (2.3 \times 10^2) \times 10^9$ litres

$= 2.3 \times 10^{2+9}$ litres

$= 2.3 \times 10^{11}$ litres

Water flow in a week:

There are 7 days in a week.

Water flow in a week = Water flow per day $\times$ Number of days in a week

$= (2.3 \times 10^{11}) \times 7$ litres

$= (2.3 \times 7) \times 10^{11}$ litres

$= 16.1 \times 10^{11}$ litres

To write this in standard notation (scientific notation), the number must be between 1 and 10.

$16.1 = 1.61 \times 10^1$

Water flow in a week $= (1.61 \times 10^1) \times 10^{11}$ litres

$= 1.61 \times 10^{1+11}$ litres

$= 1.61 \times 10^{12}$ litres

So, $1.61 \times 10^{12}$ litres of water flows through the river in a week.

Water flow in a year:

There are approximately 365 days in a year (assuming a non-leap year).

Water flow in a year = Water flow per day $\times$ Number of days in a year

$= (2.3 \times 10^{11}) \times 365$ litres

$= (2.3 \times 365) \times 10^{11}$ litres

Let's calculate $2.3 \times 365$:

$2.3 \times 365 = 839.5$

Water flow in a year $= 839.5 \times 10^{11}$ litres

To write this in standard notation, the number must be between 1 and 10.

$839.5 = 8.395 \times 10^2$

Water flow in a year $= (8.395 \times 10^2) \times 10^{11}$ litres

$= 8.395 \times 10^{2+11}$ litres

$= 8.395 \times 10^{13}$ litres

So, $8.395 \times 10^{13}$ litres of water flows through the river in a year.

Given:

Initial quantity of radioactive substance = 300 grams.

Amount remaining after 3 half-lives = $300 \times 2^{-3}$ grams.

To Find:

1. Evaluate $300 \times 2^{-3}$.

2. Explain why the expression $300 \times 2^{-n}$ can be used for the amount remaining after $n$ half-lives.

Solution:

First, we need to evaluate the expression $300 \times 2^{-3}$.

Using the property of negative exponents, $a^{-m} = \frac{1}{a^m}$, we have:

$2^{-3} = \frac{1}{2^3}$

$= \frac{1}{2 \times 2 \times 2}$

$= \frac{1}{8}$

Now, substitute this value back into the given expression:

$300 \times 2^{-3} = 300 \times \frac{1}{8}$

$= \frac{300}{8}$

We can simplify this fraction:

$= \frac{\cancel{300}^{75}}{\cancel{8}_{2}}$

$= \frac{75}{2}$

$= 37.5$

So, 37.5 grams of the substance are left after 3 half-lives.

Now, we need to explain why the expression $300 \times 2^{-n}$ can be used to find the amount of the substance that remains after $n$ half-lives.

Let $Q_0$ be the initial quantity of the substance. In this case, $Q_0 = 300$ grams.

After 1 half-life, the quantity of the substance is reduced by half. The amount remaining is $Q_0 \times \frac{1}{2}$.

After 2 half-lives, the quantity is again reduced by half of the amount present after the first half-life. The amount remaining is $(Q_0 \times \frac{1}{2}) \times \frac{1}{2} = Q_0 \times (\frac{1}{2})^2$.

After 3 half-lives, the amount remaining is $(Q_0 \times (\frac{1}{2})^2) \times \frac{1}{2} = Q_0 \times (\frac{1}{2})^3$.

We can see a pattern here. After $n$ half-lives, the quantity remaining is the initial quantity multiplied by $\frac{1}{2}$ taken to the power of $n$.

The amount remaining after $n$ half-lives, denoted by $Q_n$, is given by:

$Q_n = Q_0 \times \left(\frac{1}{2}\right)^n$

Since $\frac{1}{2}$ can be written as $2^{-1}$, we have:

$Q_n = Q_0 \times (2^{-1})^n$

Using the property of exponents $(a^m)^n = a^{mn}$, we get:

$Q_n = Q_0 \times 2^{-1 \times n} = Q_0 \times 2^{-n}$

Given that the initial quantity $Q_0 = 300$ grams, the expression for the amount remaining after $n$ half-lives is:

$Q_n = 300 \times 2^{-n}$

This expression correctly models the exponential decay process where the substance halves its quantity for every period of one half-life.

Given:

The fraction of a radioactive substance that remains after $t$ half-lives is given by the expression $3^{-t}$.

To Find:

(a) The fraction of substance remaining after 7 half-lives.

(b) The number of half-lives after which the fraction will be $\frac{1}{243}$ of the original.

Solution:

(a) Fraction of substance remaining after 7 half-lives:

The given expression for the fraction remaining after $t$ half-lives is $3^{-t}$.

We need to find the fraction remaining after 7 half-lives, so we substitute $t=7$ into the expression:

Fraction remaining $= 3^{-7}$

Using the property of negative exponents, $a^{-m} = \frac{1}{a^m}$, we have:

$3^{-7} = \frac{1}{3^7}$

Now, we calculate the value of $3^7$:

$3^7 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 2187$

Therefore, the fraction of the substance remaining after 7 half-lives is:

Fraction remaining $= \frac{1}{2187}$

So, $\frac{1}{2187}$ of the substance remains after 7 half-lives.

(b) After how many half-lives will the fraction be $\frac{1}{243}$ of the original:

We need to find the value of $t$ such that the fraction remaining, $3^{-t}$, is equal to $\frac{1}{243}$.

Set up the equation:

$3^{-t} = \frac{1}{243}$

To solve for $t$, we need to express the right side of the equation as a power of 3.

Find the prime factorization of 243:

$\begin{array}{c|cc} 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $243 = 3 \times 3 \times 3 \times 3 \times 3 = 3^5$.

Substitute this back into the equation:

$3^{-t} = \frac{1}{3^5}$

Using the property $\frac{1}{a^m} = a^{-m}$, we can write $\frac{1}{3^5}$ as $3^{-5}$.

$3^{-t} = 3^{-5}$

Since the bases are equal, the exponents must also be equal:

$-t = -5$

Multiply both sides by -1:

$t = 5$

So, the fraction will be $\frac{1}{243}$ of the original after 5 half-lives.

Given:

1 Fermi = $10^{-15}$ metre

Radius of a proton = 1.3 Fermis

To Find:

The radius of a proton in metres in standard form.

Solution:

We are given the conversion factor between Fermis and metres: 1 Fermi is equal to $10^{-15}$ metre.

The radius of the proton is given as 1.3 Fermis.

To convert the radius from Fermis to metres, we multiply the value in Fermis by the conversion factor.

Radius of proton in metres = Radius in Fermis $\times$ (Metres per Fermi)

Radius of proton in metres = $1.3 \text{ Fermis} \times 10^{-15} \text{ metre/Fermi}$

$= 1.3 \times 10^{-15}$ metres

Standard form (scientific notation) requires a number in the form $a \times 10^b$, where $1 \leq a < 10$ and $b$ is an integer.

In our result, $1.3 \times 10^{-15}$ metres, the number $a$ is 1.3, which satisfies the condition $1 \leq 1.3 < 10$. The exponent $b$ is -15, which is an integer.

Therefore, the radius of the proton in metres in standard form is already $1.3 \times 10^{-15}$ metres.

The radius of a proton in metres in standard form is $\mathbf{1.3 \times 10^{-15}}$ metres.

Given:

An image showing a paper clip next to a ruler.

To Find:

The length of the paper clip in standard form.

Solution:

From the image, we can see that one end of the paper clip is aligned with the 0 cm mark on the ruler, and the other end is aligned with the 2.8 cm mark.

Therefore, the length of the paper clip is 2.8 cm.

Standard form (or scientific notation) is a way of writing numbers as a product of a number between 1 and 10 (inclusive of 1, exclusive of 10) and a power of 10.

Let's express the length in the base SI unit, which is the metre (m).

We know that 1 cm = $\frac{1}{100}$ m = $10^{-2}$ m.

So, to convert 2.8 cm to metres, we multiply by the conversion factor $10^{-2}$:

Length in metres = $2.8 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}}$

$= 2.8 \times 10^{-2}$ metres

The number $2.8 \times 10^{-2}$ is already in standard form because:

1. The first part of the number, 2.8, is between 1 and 10 ($1 \leq 2.8 < 10$).

2. The second part is a power of 10 ($10^{-2}$), where the exponent -2 is an integer.

Thus, the length of the paper clip in metres in standard form is $2.8 \times 10^{-2}$ m.

If the question intended the standard form in centimetres, then the length is 2.8 cm.

To write 2.8 in standard form:

$2.8 = 2.8 \times 1$

Since $1 = 10^0$, we can write:

$2.8 = 2.8 \times 10^0$

In standard form, this is $2.8 \times 10^0$ cm.

Assuming the question implies standard form in the standard SI unit (metres), the answer is:

The length of the paper clip in standard form is $\mathbf{2.8 \times 10^{-2}}$ metres.

Verification:

We will verify each statement using the properties of exponents:

1. $a^{-m} = \frac{1}{a^m}$ (Negative exponent property)

2. $a^m \times a^n = a^{m+n}$ (Product of powers property)

(a) Verify that $\frac{1}{4} (2^n) = 2^{n - 2}$ is true.

Consider the left side of the statement: $\frac{1}{4} (2^n)$.

We know that $4 = 2^2$. So, we can rewrite $\frac{1}{4}$ as $\frac{1}{2^2}$.

Using the negative exponent property, $\frac{1}{2^2} = 2^{-2}$.

Substitute this back into the left side:

Left Side $= 2^{-2} \times 2^n$

Now, using the product of powers property, $a^m \times a^n = a^{m+n}$, where $a=2$, $m=-2$, and $n=n$:

Left Side $= 2^{(-2) + n} = 2^{n - 2}$

This is equal to the right side of the statement.

Thus, the statement $\frac{1}{4} (2^n) = 2^{n - 2}$ is verified as True.

(b) Verify that $4^{n - 1} = \frac{1}{4} (4)^n$ is true.

Consider the right side of the statement: $\frac{1}{4} (4)^n$.

We know that $\frac{1}{4}$ can be written as $4^{-1}$ using the negative exponent property.

Substitute this back into the right side:

Right Side $= 4^{-1} \times 4^n$

Now, using the product of powers property, $a^m \times a^n = a^{m+n}$, where $a=4$, $m=-1$, and $n=n$:

Right Side $= 4^{(-1) + n} = 4^{n - 1}$

This is equal to the left side of the statement.

Thus, the statement $4^{n - 1} = \frac{1}{4} (4)^n$ is verified as True.

(c) Verify that $25(5^{n – 2}) = 5^n$ is true.

Consider the left side of the statement: $25(5^{n – 2})$.

We know that $25 = 5 \times 5 = 5^2$.

Substitute this back into the left side:

Left Side $= 5^2 \times 5^{n - 2}$

Now, using the product of powers property, $a^m \times a^n = a^{m+n}$, where $a=5$, $m=2$, and $n=n-2$:

Left Side $= 5^{2 + (n - 2)} = 5^{2 + n - 2} = 5^{n + (2 - 2)} = 5^{n + 0} = 5^n$

This is equal to the right side of the statement.

Thus, the statement $25(5^{n – 2}) = 5^n$ is verified as True.

Solution:

We need to fill in the blanks with the correct properties of exponents.

1. $\frac{a^m}{a^n} = a^{m-n}$ (Quotient of powers property)

2. $(a^m)^n = a^{mn}$ (Power of a power property)

3. $a^m \times a^n = a^{m+n}$ (Product of powers property)

4. $a^{-m} = \frac{1}{a^m}$ (Negative exponent property, where $a \neq 0$)

5. $a^0 = 1$ (Zero exponent property, where $a \neq 0$)

6. $(ab)^m = a^m b^m$ (Power of a product property)

7. $(\frac{a}{b})^m = \frac{a^m}{b^m}$ (Power of a quotient property, where $b \neq 0$)

Filling in the blanks according to the provided image, we get the following completed properties:

$\frac{a^m}{a^n} = a^{m-n}$

$(a^m)^n = a^{mn}$

$a^m \times a^n = a^{m+n}$

$a^{-m} = \frac{1}{a^m}$

$a^0 = 1$

$(ab)^m = a^m b^m$

$(\frac{a}{b})^m = \frac{a^m}{b^m}$

Given:

Number of seconds in a day = 86,400 seconds.

To Find:

How many days long is a second, expressed in scientific notation.

Solution:

We are given that:

86,400 seconds = 1 day

To find out how many days long a second is, we can divide both sides of the equation by 86,400:

$\frac{86,400}{86,400}$ seconds = $\frac{1}{86,400}$ day

1 second = $\frac{1}{86,400}$ day

Now, we need to express the value $\frac{1}{86,400}$ in scientific notation.

First, let's write 86,400 in scientific notation.

$86,400 = 8.64 \times 10^4$

So, $\frac{1}{86,400} = \frac{1}{8.64 \times 10^4}$

Using the property $\frac{1}{a^m} = a^{-m}$, we can write $\frac{1}{10^4}$ as $10^{-4}$.

$\frac{1}{8.64 \times 10^4} = \frac{1}{8.64} \times \frac{1}{10^4} = \frac{1}{8.64} \times 10^{-4}$

Now we need to calculate the value of $\frac{1}{8.64}$.

$\frac{1}{8.64} \approx 0.11574$

So, the value is approximately $0.11574 \times 10^{-4}$ days.

To express this in scientific notation, the number before the power of 10 must be between 1 and 10. We need to move the decimal point one place to the right, which means we multiply by 10 and decrease the exponent by 1.

$0.11574 = 1.1574 \times 10^{-1}$

Substitute this back into the expression:

$(1.1574 \times 10^{-1}) \times 10^{-4}$ days

Using the product of powers property, $a^m \times a^n = a^{m+n}$:

$1.1574 \times 10^{-1 + (-4)}$ days

$1.1574 \times 10^{-5}$ days

Rounding to a reasonable number of significant figures (based on the input 86,400 which has 3 significant figures), we can write the number as $1.16 \times 10^{-5}$.

Thus, 1 second is approximately $\mathbf{1.16 \times 10^{-5}}$ days long in scientific notation.

Solution:

(a) For which crop(s) did the production decrease?

We need to look at the 'Increase/Decrease (Hectare) in 2009' column of the table. A decrease is indicated by a negative value.

• Bajra: -100 (Decrease)
• Jowar: -440,000 (Decrease)
• Rice: -100 (Decrease)
• Wheat: +190,000 (Increase)

The crops for which the production decreased in 2009 compared to 2008 are Bajra, Jowar, and Rice.

(b) Write the production of all the crops in 2009 in their standard form.

The production in 2009 for each crop is the production in 2008 plus the increase or decrease in 2009.

• Bajra:

2008 Production = $1.4 \times 10^3 = 1400$ Hectares

Increase/Decrease in 2009 = -100 Hectares

2009 Production = $1400 - 100 = 1300$ Hectares

In standard form, $1300 = 1.3 \times 10^3$ Hectares.

• Jowar:

2008 Production = $1.7 \times 10^6 = 1,700,000$ Hectares

Increase/Decrease in 2009 = -440,000 Hectares

2009 Production = $1,700,000 - 440,000 = 1,260,000$ Hectares

In standard form, $1,260,000 = 1.26 \times 10^6$ Hectares.

• Rice:

2008 Production = $3.7 \times 10^3 = 3700$ Hectares

Increase/Decrease in 2009 = -100 Hectares

2009 Production = $3700 - 100 = 3600$ Hectares

In standard form, $3600 = 3.6 \times 10^3$ Hectares.

• Wheat:

2008 Production = $5.1 \times 10^5 = 510,000$ Hectares

Increase/Decrease in 2009 = +190,000 Hectares

2009 Production = $510,000 + 190,000 = 700,000$ Hectares

In standard form, $700,000 = 7 \times 10^5$ Hectares.

Summary of 2009 production in standard form:

• Bajra: $\mathbf{1.3 \times 10^3}$ Hectares
• Jowar: $\mathbf{1.26 \times 10^6}$ Hectares
• Rice: $\mathbf{3.6 \times 10^3}$ Hectares
• Wheat: $\mathbf{7 \times 10^5}$ Hectares

(c) Assuming the same decrease in rice production each year as in 2009, how many acres will be harvested in 2015? Write in standard form.

The decrease in rice production in 2009 was 100 Hectares.

We assume this same decrease occurs each year after 2009.

The number of years from the end of 2009 to the end of 2015 is $2015 - 2009 = 6$ years.

The total decrease in production from the 2009 level to the 2015 level will be the annual decrease multiplied by the number of years.

Total decrease = Decrease per year $\times$ Number of years

Total decrease = $100 \text{ Hectares/year} \times 6 \text{ years} = 600$ Hectares

The rice production in 2009 was 3600 Hectares (as calculated in part b).

Rice production in 2015 = Rice production in 2009 - Total decrease from 2009 to 2015

Rice production in 2015 = $3600 \text{ Hectares} - 600 \text{ Hectares} = 3000$ Hectares.

The question asks for the answer in "acres", while the table uses "Hectare". Assuming this is a typo and the unit should remain Hectares, we write the answer in standard form.

Rice production in 2015 = 3000 Hectares

To write 3000 in standard form, we express it as a number between 1 and 10 multiplied by a power of 10.

$3000 = 3 \times 1000 = 3 \times 10^3$

Assuming the unit remains Hectares, the production in 2015 is $\mathbf{3 \times 10^3}$ Hectares in standard form.

(Note: If a conversion to acres were required, the conversion factor would be needed, which is not provided. We proceed with the Hectare unit based on the table context).

Given:

Initial length of the carrot = 10 cm.

Stretching machine factor = $\times 4$.

To Find:

The length of the carrot after it comes out of the stretching machine.

Solution:

The stretching machine multiplies the length of the object by the given factor.

In this case, the initial length of the carrot is 10 cm, and the stretching machine factor is $\times 4$.

Length of the carrot after stretching = Initial length $\times$ Stretching factor

$= 10 \text{ cm} \times 4$

$= 40 \text{ cm}$

The length of the carrot when it comes out of the machine will be 40 cm.

The length of the carrot after stretching is 40 cm.

Given:

Various stretching machines with different multiplication factors (powers of 2, 10, and 5).

Hooking machines together means the output of the first is the input for the second.

To Find:

(a) Which two machines provide a total multiplication factor of $10^2$. Whether the arrangement order matters.

(b) Which single machine is equivalent to two (× 2) machines hooked together.

Solution:

When two stretching machines are hooked together, the total stretching factor is the product of the individual stretching factors of the two machines.

For example, if a machine has a factor of $M_1$ and another has a factor of $M_2$, hooking them together results in a total factor of $M_1 \times M_2$.

(a) Which two machines hooked together do the same work a (× 10²) machine does? Is there more than one arrangement of two machines that will work?

We are looking for two machines with factors $M_1$ and $M_2$ from the available set such that $M_1 \times M_2 = 10^2$.

The target factor is $10^2 = 100$.

Let's look at the available factors from the images:

• Powers of 2: $2^4, 2^3, 2^2, 2^1, 2^{-1}, 2^{-2}, 2^{-3}$ (i.e., 16, 8, 4, 2, 1/2, 1/4, 1/8)
• Powers of 10: $10^2, 10^1, 10^0, 10^{-1}, 10^{-2}, 10^{-3}$ (i.e., 100, 10, 1, 0.1, 0.01, 0.001)
• Powers of 5: $5^2, 5^1, 5^0, 5^{-1}, 5^{-2}, 5^{-3}$ (i.e., 25, 5, 1, 1/5, 1/25, 1/125)

We need to find two factors from these lists whose product is 100.

Consider the prime factorization of 100: $100 = 10 \times 10 = 10^2$. We see a (× 10²) machine is available, but we need *two* machines.

Another way to factor 100 is $100 = 4 \times 25$.

Is there a machine with factor 4? Yes, (× $2^2$) = (× 4).

Is there a machine with factor 25? Yes, (× $5^2$) = (× 25).

So, hooking up a (× $2^2$) machine and a (× $5^2$) machine will give a total factor of $2^2 \times 5^2 = (2 \times 5)^2 = 10^2$.

The two machines are the one with factor $2^2$ and the one with factor $5^2$. These are the (× 4) machine and the (× 25) machine.

Regarding the arrangement order: Since multiplication is commutative ($a \times b = b \times a$), the order in which the machines are hooked together does not matter. Hooking (× $2^2$) then (× $5^2$) results in a total factor of $2^2 \times 5^2 = 100$. Hooking (× $5^2$) then (× $2^2$) results in a total factor of $5^2 \times 2^2 = 100$.

There is more than one arrangement that will work.

The two machines that do the same work as a (× 10²) machine when hooked together are the (× $2^2$) machine and the (× $5^2$) machine. Yes, there is more than one arrangement; hooking them in either order gives the same result.

(b) Which stretching machine does the same work as two (× 2) machines hooked together?

Putting something through a (× 2) machine means multiplying its length by 2.

If we hook two (× 2) machines together, the initial length is multiplied by 2 by the first machine, and then that result is multiplied by 2 by the second machine.

Total stretching factor = Factor of first machine $\times$ Factor of second machine

$= 2 \times 2 = 4$

We need to find a single stretching machine that has a factor of 4.

From the list of available machines:

• Powers of 2: ... $2^2 = 4$ ...

The machine with factor $2^2$ has a factor of 4.

Two (× 2) machines hooked together do the same work as the (× $2^2$) machine.

Given:

Initial length of the strip = 4 cm.

The machine is a repeater machine with factor (× $2^3$) as shown in the image.

To Find:

The new length of the strip after being inserted into the machine.

Solution:

A repeater machine with a factor of $a^n$ means the initial length is multiplied by $a^n$.

In this case, the initial length of the strip is 4 cm, and the machine's factor is (× $2^3$).

The factor $2^3$ is calculated as:

$2^3 = 2 \times 2 \times 2 = 8$

So, the machine multiplies the length by 8.

New length of the strip = Initial length $\times$ Machine factor

$= 4 \text{ cm} \times 2^3$

$= 4 \text{ cm} \times 8$

$= 32 \text{ cm}$

The new length of the 4 cm strip inserted in the (× $2^3$) machine will be 32 cm.

The new length of the strip is 32 cm.

Solution:

In a repeater machine represented as (× $a^n$), the base machine is the one that multiplies by $a$, and this base machine is applied $n$ times. The total stretch is the value of $a^n$.

Machine 1: (× $10^4$)

In this machine, the factor is $10^4$.

The base of the power is 10, and the exponent is 4.

The base machine has a factor of ($10$).

The base machine is applied 4 times (according to the exponent).

The total stretch is the value of $10^4$.

$10^4 = 10 \times 10 \times 10 \times 10 = 10000$

The total stretch is 10000.

Machine 2: (× $5^3$)

In this machine, the factor is $5^3$.

The base of the power is 5, and the exponent is 3.

The base machine has a factor of ($5$).

The base machine is applied 3 times (according to the exponent).

The total stretch is the value of $5^3$.

$5^3 = 5 \times 5 \times 5 = 125$

The total stretch is 125.

Machine 3: (× $3^5$)

In this machine, the factor is $3^5$.

The base of the power is 3, and the exponent is 5.

The base machine has a factor of ($3$).

The base machine is applied 5 times (according to the exponent).

The total stretch is the value of $3^5$.

$3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 243$

The total stretch is 243.

To Find:

Three different repeater machines that have a total stretching factor of 64.

Solution:

A repeater machine is described as (× $a^n$), where the base machine has a factor of $a$ and is applied $n$ times, resulting in a total stretch of $a^n$. We need to find three different pairs of $(a, n)$ such that $a^n = 64$, where $a$ is the factor of the base machine and $n$ is the number of times it is applied.

We need to express 64 as a power $a^n$ in different ways.

First, let's find the prime factorization of 64:

$\begin{array}{c|cc} 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

So, $64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6$.

Using the prime factorization, we can find different ways to write 64 as $a^n$ where $a$ is an integer base and $n$ is an integer exponent ($n \geq 1$).

Possibility 1:

We can use the prime factorization directly: $64 = 2^6$.

Here, the base is $a=2$ and the exponent is $n=6$.

This corresponds to a repeater machine (× $2^6$).

Description: A base machine (× 2) applied 6 times.

Total stretch = $2^6 = 64$.

Possibility 2:

We can group the factors of 2: $64 = (2 \times 2) \times (2 \times 2) \times (2 \times 2) = 4 \times 4 \times 4 = 4^3$.

Here, the base is $a=4$ and the exponent is $n=3$.

This corresponds to a repeater machine (× $4^3$).

Description: A base machine (× 4) applied 3 times.

Total stretch = $4^3 = 64$.

Possibility 3:

We can group the factors of 2 differently: $64 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) = 8 \times 8 = 8^2$.

Here, the base is $a=8$ and the exponent is $n=2$.

This corresponds to a repeater machine (× $8^2$).

Description: A base machine (× 8) applied 2 times.

Total stretch = $8^2 = 64$.

Another possibility is $64^1$, which corresponds to a base machine (× 64) applied 1 time. This is also valid, but the question asks for three distinct machines, and $2^6$, $4^3$, and $8^2$ provide different base/exponent combinations.

Here are three repeater machines that will do the same work as a (×64) machine, described using exponents:

1. (× $2^6$) machine: A base machine (× 2) applied 6 times. Total stretch = $2^6 = 64$.

2. (× $4^3$) machine: A base machine (× 4) applied 3 times. Total stretch = $4^3 = 64$.

3. (× $8^2$) machine: A base machine (× 8) applied 2 times. Total stretch = $8^2 = 64$.

Given:

Initial length of the piece of chalk = 2 cm.

The machine is a repeater machine with the factor (× $3^{-2}$), as shown in the image.

To Find:

The new length of the piece of chalk after being inserted into the machine.

Solution:

A repeater machine with a factor of $a^n$ multiplies the initial length of an object by $a^n$.

In this case, the initial length of the chalk is 2 cm, and the machine's factor is (× $3^{-2}$).

We need to calculate the value of the machine's factor $3^{-2}$.

Using the property of negative exponents, $a^{-m} = \frac{1}{a^m}$, we have:

$3^{-2} = \frac{1}{3^2}$

Now, calculate the value of $3^2$:

$3^2 = 3 \times 3 = 9$

So, the machine's factor is $\frac{1}{9}$.

The new length of the chalk after passing through the machine is the initial length multiplied by the machine's factor:

New length = Initial length $\times$ Machine factor

$= 2 \text{ cm} \times 3^{-2}$

$= 2 \text{ cm} \times \frac{1}{9}$

$= \frac{2}{9}$ cm

The machine will multiply the length of the 2 cm piece of chalk by the factor $3^{-2} = \frac{1}{9}$. This means the chalk will become shorter.

The new length of the piece of chalk will be $\mathbf{\frac{2}{9}}$ cm.

Solution:

A repeater machine with a factor (× $a^n$) multiplies the length of an object by $a^n$. If the exponent is 0, the machine's factor is $a^0$.

(a) What do these machines do to a piece of chalk?

The image shows three repeater machines with factors $2^0$, $5^0$, and $10^0$.

Based on the definition of a repeater machine, these machines will multiply the length of the chalk by their respective factors.

The factors are $2^0$, $5^0$, and $10^0$. According to the property of exponents, any non-zero number raised to the power of 0 is equal to 1.

$2^0 = 1$

$5^0 = 1$

$10^0 = 1$

So, each of these machines has a multiplication factor of 1.

If we insert a piece of chalk with an initial length $L$ into any of these machines, the new length will be $L \times 1 = L$.

Therefore, these machines will not change the length of the piece of chalk.

(b) What do you think the value of 6⁰ is?

Based on the property of exponents demonstrated in part (a), where any non-zero base raised to the power of 0 equals 1, we can determine the value of $6^0$.

The base is 6, which is a non-zero number, and the exponent is 0.

According to the zero exponent property ($a^0 = 1$ for $a \neq 0$), the value of $6^0$ is 1.

The value of $6^0$ is 1.

Given:

Initial length of the sandwich = 9 cm.

Shrinking machine factor = (× $3^{-1}$).

To Find:

The length of the sandwich after it emerges from the machine.

Solution:

The shrinking machine multiplies the length of the object by the given factor.

In this case, the initial length of the sandwich is 9 cm, and the shrinking machine factor is $3^{-1}$.

We need to calculate the value of the machine's factor $3^{-1}$.

Using the property of negative exponents, $a^{-m} = \frac{1}{a^m}$, we have:

$3^{-1} = \frac{1}{3^1} = \frac{1}{3}$

So, the machine's factor is $\frac{1}{3}$.

The new length of the sandwich after passing through the machine is the initial length multiplied by the machine's factor:

New length = Initial length $\times$ Machine factor

$= 9 \text{ cm} \times 3^{-1}$

$= 9 \text{ cm} \times \frac{1}{3}$

$= \frac{9}{3}$ cm

$= 3$ cm

The 9 cm long sandwich will be 3 cm long when it emerges from the shrinking machine.

The new length of the sandwich is 3 cm.

Given:

Initial length of the worm = 1 cm.

Two hook-up scenarios involving machines with factors (× $2^{-1}$) and (× $2^1$).

To Find:

The final length of the worm after passing through each hook-up.

Solution:

When machines are hooked together, the total effect on the length of the object is the product of the multiplication factors of the individual machines in the order they are encountered. The machine with factor (× $2^{-1}$) multiplies the length by $2^{-1} = \frac{1}{2}$. The machine with factor (× $2^1$) multiplies the length by $2^1 = 2$.

Scenario 1: Worm goes through (× $2^{-1}$) then (× $2^1$).

Initial length = 1 cm.

Length after the first machine (× $2^{-1}$) = Initial length $\times 2^{-1}$

$= 1 \text{ cm} \times \frac{1}{2} = \frac{1}{2}$ cm.

This new length becomes the input for the second machine (× $2^1$).

Length after the second machine (× $2^1$) = Length after first machine $\times 2^1$

$= \frac{1}{2} \text{ cm} \times 2 = 1$ cm.

Scenario 2: Worm goes through (× $2^1$) then (× $2^{-1}$).

Initial length = 1 cm.

Length after the first machine (× $2^1$) = Initial length $\times 2^1$

$= 1 \text{ cm} \times 2 = 2$ cm.

This new length becomes the input for the second machine (× $2^{-1}$).

Length after the second machine (× $2^{-1}$) = Length after first machine $\times 2^{-1}$

$= 2 \text{ cm} \times \frac{1}{2} = 1$ cm.

In both hook-ups, the final length of the worm is 1 cm.

This is because the total multiplication factor for the hook-up is the product of the individual factors. For the first scenario, the total factor is $2^{-1} \times 2^1$. For the second scenario, the total factor is $2^1 \times 2^{-1}$.

Using the product of powers property ($a^m \times a^n = a^{m+n}$):

$2^{-1} \times 2^1 = 2^{-1+1} = 2^0$

$2^1 \times 2^{-1} = 2^{1+(-1)} = 2^0$

From the zero exponent property ($a^0 = 1$ for $a \neq 0$), we know that $2^0 = 1$.

So, the total multiplication factor for both hook-ups is 1. Multiplying the initial length (1 cm) by the total factor (1) gives a final length of $1 \text{ cm} \times 1 = 1$ cm.

When 1 cm worms are sent through these hook-ups, their length remains 1 cm.

Given:

Initial length of the stick of gum = 1 cm.

Machine factor = $(1 \times 3^{-2})$.

To Find:

The length of the stick when it came out of the machine.

Solution:

The machine multiplies the length of the object inserted by its factor.

The factor of the machine is given as $(1 \times 3^{-2})$.

First, let's evaluate the term with the negative exponent, $3^{-2}$.

Using the property of negative exponents, $a^{-m} = \frac{1}{a^m}$, we have:

$3^{-2} = \frac{1}{3^2}$

Calculate $3^2$:

$3^2 = 3 \times 3 = 9$

So, $3^{-2} = \frac{1}{9}$.

Now, the machine factor is $1 \times 3^{-2} = 1 \times \frac{1}{9} = \frac{1}{9}$.

The machine multiplies the initial length by $\frac{1}{9}$.

New length of the stick = Initial length $\times$ Machine factor

$= 1 \text{ cm} \times \frac{1}{9}$

$= \frac{1}{9}$ cm

The length of the stick when it came out of the machine was $\mathbf{\frac{1}{9}}$ cm.

Given:

Initial length of the piece of gum = 1 cm.

Final length of the piece of gum = $\frac{1}{100,000}$ cm.

The machine is a repeater machine with factor (× $10^?$). Let the missing value be $x$. So, the machine factor is $10^x$.

To Find:

The missing value (the exponent $x$) in the machine factor.

Solution:

When an object is put through a repeater machine with a factor of $a^n$, its initial length is multiplied by $a^n$ to get the final length.

In this case, the initial length is 1 cm, the machine factor is $10^x$, and the final length is $\frac{1}{100,000}$ cm.

We can write the equation:

Initial length $\times$ Machine factor = Final length

$1 \text{ cm} \times 10^x = \frac{1}{100,000}$ cm

Simplifying the equation, we get:

$10^x = \frac{1}{100,000}$

Now, we need to express the right side of the equation as a power of 10.

Let's write 100,000 in terms of a power of 10:

$100,000 = 10 \times 10 \times 10 \times 10 \times 10 = 10^5$

Substitute this back into the equation:

$10^x = \frac{1}{10^5}$

Using the property of negative exponents, $\frac{1}{a^m} = a^{-m}$, we can write $\frac{1}{10^5}$ as $10^{-5}$.

$10^x = 10^{-5}$

Since the bases on both sides of the equation are equal (both are 10), the exponents must also be equal.

$x = -5$

The missing value in the repeater machine factor (× $10^?$) is -5.

The missing value is $\mathbf{-5}$.

Solution:

When two machines are hooked together, the total effect is the product of their individual multiplication factors. If a machine has a factor of $M_1$ and is followed by a machine with a factor of $M_2$, the equivalent single machine has a factor of $M_1 \times M_2$. We will use the property of exponents $a^m \times a^n = a^{m+n}$ to find the equivalent single machine.

(a) a (× 2³) machine followed by (× 2^–2) machine.

The factor of the first machine is $2^3$.

The factor of the second machine is $2^{-2}$.

The factor of the equivalent single machine is the product of these two factors:

Equivalent factor $= 2^3 \times 2^{-2}$

Using the product of powers property $a^m \times a^n = a^{m+n}$:

$= 2^{3 + (-2)}$

$= 2^{3 - 2}$

$= 2^1$

So, the equivalent single machine has a factor of $2^1$.

The single machine that will do the same job is the (× $2^1$) machine.

(b) a (× 2⁴) machine followed by $\left( \times\left( \frac{1}{2} \right)^{2} \right)$ machine.

The factor of the first machine is $2^4$.

The factor of the second machine is $\left( \frac{1}{2} \right)^{2}$.

First, let's rewrite the second factor with a base of 2.

Using the property $\frac{1}{a} = a^{-1}$, we have $\frac{1}{2} = 2^{-1}$.

So, $\left( \frac{1}{2} \right)^{2} = (2^{-1})^{2}$

Using the power of a power property $(a^m)^n = a^{mn}$:

$= 2^{(-1) \times 2} = 2^{-2}$

Now, the factor of the equivalent single machine is the product of the two factors:

Equivalent factor $= 2^4 \times 2^{-2}$

Using the product of powers property $a^m \times a^n = a^{m+n}$:

$= 2^{4 + (-2)}$

$= 2^{4 - 2}$

$= 2^2$

So, the equivalent single machine has a factor of $2^2$.

The single machine that will do the same job is the (× $2^2$) machine.

(c) a (× 5⁹⁹) machine followed by a (5^–100) machine.

The factor of the first machine is $5^{99}$.

The factor of the second machine is $5^{-100}$.

The factor of the equivalent single machine is the product of these two factors:

Equivalent factor $= 5^{99} \times 5^{-100}$

Using the product of powers property $a^m \times a^n = a^{m+n}$:

$= 5^{99 + (-100)}$

$= 5^{99 - 100}$

$= 5^{-1}$

So, the equivalent single machine has a factor of $5^{-1}$.

The single machine that will do the same job is the (× $5^{-1}$) machine.

Solution:

When two or more machines are hooked together, the total multiplication factor is the product of the multiplication factors of the individual machines. A repeater machine (× $a^n$) has a multiplication factor of $a^n$. To find the equivalent single repeater machine for a hook-up, we calculate the total multiplication factor and express it in the form $b^m$, where the base $b$ is used as the base for the equivalent single repeater machine, and $m$ is the exponent.

Hook-up 1 (shown in the first image):

This hook-up consists of a (× $10^2$) machine followed by a (× $10^3$) machine.

The factor of the first machine is $10^2$.

The factor of the second machine is $10^3$.

The total multiplication factor for this hook-up is the product of these factors:

Total factor $= 10^2 \times 10^3$

Using the product of powers property, $a^m \times a^n = a^{m+n}$, we add the exponents since the bases are the same (base 10):

Total factor $= 10^{2+3} = 10^5$

This total factor $10^5$ represents a single repeater machine where the base machine is (× 10) and it is applied 5 times.

The equivalent single repeater machine for the first hook-up is (× $10^5$).

Hook-up 2 (shown in the second image):

This hook-up consists of a (× $5^2$) machine followed by a (× $5^?$) machine. The image also states that this hook-up is equal to a (× $5^4$) machine.

Let the missing exponent be $x$. The factor of the second machine is $5^x$.

The factor of the first machine is $5^2$.

The total multiplication factor for this hook-up is $5^2 \times 5^x$.

Using the product of powers property, the total factor is $5^{2+x}$.

The image states that this hook-up is equivalent to a (× $5^4$) machine, which has a factor of $5^4$.

So, we have the equation:

$5^{2+x} = 5^4$

Since the bases are equal, the exponents must be equal:

$2 + x = 4$

Solving for $x$:

$x = 4 - 2$

$x = 2$

The missing value is 2, so the second machine is (× $5^2$). The hook-up is a (× $5^2$) machine followed by a (× $5^2$) machine.

The total factor for this hook-up is indeed $5^2 \times 5^2 = 5^{2+2} = 5^4$.

The equivalent single repeater machine for the second hook-up is (× $5^4$), as indicated by the image itself.

Solution:

When machines are hooked together, the total multiplication factor is the product of the individual factors. A single repeater machine is represented with a factor in the form $a^n$. We need to find the total factor for each hook-up and check if it can be expressed in this form.

Hook-up 1: (× $10^2$) followed by (× $5^2$)

The first machine has a factor of $10^2$.

The second machine has a factor of $5^2$.

The total factor of the hook-up is the product of these factors:

Total factor $= 10^2 \times 5^2$

Using the property of exponents $(ab)^m = a^m b^m$, we can rewrite the expression:

Total factor $= (10 \times 5)^2 = 50^2$

This total factor is in the form $a^n$, where $a=50$ and $n=2$. Therefore, there is a single repeater machine that will do the same work.

The equivalent single repeater machine has the factor (× $50^2$).

Hook-up 2: (× $2^3$) followed by (× $3^2$)

The first machine has a factor of $2^3$.

The second machine has a factor of $3^2$.

The total factor of the hook-up is the product of these factors:

Total factor $= 2^3 \times 3^2$

Evaluate the powers:

$2^3 = 2 \times 2 \times 2 = 8$

$3^2 = 3 \times 3 = 9$

Total factor $= 8 \times 9 = 72$

Now, we need to check if 72 can be expressed in the form $a^n$ for integer base $a$ and integer exponent $n \geq 1$.

Prime factorization of 72 is $2^3 \times 3^2$. Since the bases are different (2 and 3) and the exponents are different (3 and 2), we cannot combine them into a single power $a^n$ where $a$ is an integer base and $n > 1$.

The only way to write 72 in the form $a^n$ with integer $a$ and $n \geq 1$ is $72^1$.

This is in the form $a^n$, where $a=72$ and $n=1$. Therefore, there is a single repeater machine that will do the same work.

Solution:

When two machines are hooked together, the total effect is found by multiplying their individual multiplication factors. In this case, a (× $2^3$) machine is followed by a (× $3^3$) machine.

The factor of the first machine is $2^3$.

The factor of the second machine is $3^3$.

The total multiplication factor for the hook-up is the product of these factors:

Total factor $= 2^3 \times 3^3$

We can use the property of exponents that states $(ab)^m = a^m b^m$. Reading this property in reverse, we have $a^m b^m = (ab)^m$.

In our case, we have $2^3 \times 3^3$. Here, the bases are different (2 and 3), but the exponents are the same (3). We can apply the property $a^m b^m = (ab)^m$ with $a=2$, $b=3$, and $m=3$:

Total factor $= (2 \times 3)^3$

$= 6^3$

The total multiplication factor is $6^3$. A single repeater machine that does the same work must have this total factor.

A repeater machine with factor $6^3$ is described as a (× $6^3$) machine, which means a base machine (× 6) is applied 3 times.

The single repeater machine that Shikha should use is the (× $6^3$) machine.

Given:

Target stretching factor = $25^2$.

The (× 25) machine is broken.

We need to find a hook-up of two repeater machines that achieves the target factor.

To Find:

A hook-up of two repeater machines equivalent to a (× $25^2$) machine.

Solution:

A repeater machine (× $a^n$) has a multiplication factor of $a^n$. When two machines are hooked together, the total multiplication factor is the product of the factors of the individual machines.

The target machine is a (× $25^2$) machine, which has a total multiplication factor of $25^2$.

We need to find two repeater machines with factors $M_1$ and $M_2$ such that $M_1 \times M_2 = 25^2$.

We know that $25^2 = 25 \times 25$.

This means we could use a machine with factor 25 followed by another machine with factor 25. However, the (× 25) machine is broken.

We need to find repeater machines that have factors which, when multiplied together, equal $25^2$.

Let's express the number 25 as a power. $25 = 5 \times 5 = 5^2$.

So, the factor 25 can be represented by a repeater machine (× $5^2$).

Now consider the target factor $25^2$. We can rewrite this using the base 5:

$25^2 = (5^2)^2$

Using the power of a power property, $(a^m)^n = a^{mn}$, we get:

$(5^2)^2 = 5^{2 \times 2} = 5^4$

So, the target total factor is $5^4$.

We need to find two repeater machines whose factors multiply to $5^4$. Since we are looking for a hook-up of two repeater machines, let their factors be $a^n$ and $b^m$. The total factor is $a^n \times b^m$. We need $a^n \times b^m = 5^4$.

The hint suggests thinking about replacing the (× 25) machine. A (× 25) machine has a factor of 25, which is $5^2$. A single repeater machine equivalent to (× 25) is (× $5^2$). This machine represents applying a base (× 5) machine 2 times.

We need the equivalent of (× $25^2$), which means applying the (× 25) effect twice (since $25^2 = 25 \times 25$). Since the (× 25) machine is broken, we can replace each instance of the (× 25) effect with its equivalent repeater machine factor, which is $5^2$.

So, instead of doing (× 25) followed by (× 25), we can do (× $5^2$) followed by (× $5^2$).

A hook-up of two repeater machines: the first machine has a factor of $5^2$, and the second machine has a factor of $5^2$. Both (× $5^2$) are repeater machines (a base 5 machine applied 2 times).

The total factor of this hook-up would be $5^2 \times 5^2 = 5^{2+2} = 5^4$. This matches the target factor $25^2 = (5^2)^2 = 5^4$.

The hook-up of two repeater machines that will do the same work as a (× $25^2$) machine is: a (× $5^2$) machine followed by a (× $5^2$) machine.

Solution:

When machines are hooked together, the output is obtained by multiplying the input by the factor of the first machine, and then multiplying the result by the factor of the second machine. Effectively, the total multiplication factor is the product of the factors of the individual machines.

Diagram 1:

Input length = 3 cm

First machine factor = $\times 2^3$

Second machine factor = $\times 2^2$

The total multiplication factor for the hook-up is the product of the individual factors:

Total factor $= 2^3 \times 2^2$

Using the product of powers property, $a^m \times a^n = a^{m+n}$:

Total factor $= 2^{3+2} = 2^5$

Now, calculate the value of the total factor:

$2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32$

The output length is the input length multiplied by the total factor:

Output length = Input length $\times$ Total factor

$= 3 \text{ cm} \times 32$

$= 96$ cm

The missing information in the first diagram is the output length, which is $\mathbf{96}$ cm.

Diagram 2:

Input length = 3 cm

First machine factor = $\times 3^2$

Second machine factor = $\times 5^2$

The total multiplication factor for the hook-up is the product of the individual factors:

Total factor $= 3^2 \times 5^2$

Using the property of exponents $(ab)^m = a^m b^m$, we can rewrite the expression:

Total factor $= (3 \times 5)^2 = 15^2$

Now, calculate the value of the total factor:

$15^2 = 15 \times 15 = 225$

The output length is the input length multiplied by the total factor:

Output length = Input length $\times$ Total factor

$= 3 \text{ cm} \times 225$

$= 675$ cm

The missing information in the second diagram is the output length, which is $\mathbf{675}$ cm.

Given:

A stretching machine with factor (× $10^3$).

To Find:

A hook-up of machines with prime number bases that does the same work.

Solution:

The given stretching machine has a multiplication factor of $10^3$. We need to find a hook-up of machines whose bases are prime numbers and whose combined multiplication factor is equal to $10^3$.

When machines are hooked together, their multiplication factors are multiplied. So, we need to express $10^3$ as a product of powers of prime numbers.

First, let's find the prime factorization of the base, 10.

$10 = 2 \times 5$

Now substitute this prime factorization back into the given factor:

$10^3 = (2 \times 5)^3$

Using the power of a product property of exponents, $(a \times b)^m = a^m \times b^m$, we can distribute the exponent 3 to the prime factors 2 and 5:

$10^3 = 2^3 \times 5^3$

This expression $2^3 \times 5^3$ represents the total multiplication factor as the product of two factors: $2^3$ and $5^3$.

A machine with a factor of $2^3$ is a repeater machine with a prime base 2, applied 3 times.

A machine with a factor of $5^3$ is a repeater machine with a prime base 5, applied 3 times.

Hooking these two machines together will result in the total multiplication factor of $2^3 \times 5^3 = 10^3$. The bases 2 and 5 are prime numbers, and we are not using a (× 1) machine (where the base is 1).

Therefore, it is possible to find such a hook-up.

The hook-up of prime base number machines that will do the same work as the (× $10^3$) machine is: a (× $2^3$) machine hooked up with a (× $5^3$) machine.

(The order of the machines in the hook-up does not affect the total multiplication factor).

Solution:

A repeater machine with factor (× $a^n$) has a multiplication factor of $a^n$. The question asks for two repeater machines whose combined effect, when hooked together, is the same as a (× 81) machine. This means we are looking for two repeater machines with factors $M_1$ and $M_2$ such that $M_1 \times M_2 = 81$. A repeater machine factor is typically expressed in the form $a^n$, where $a$ is the base and $n$ is the exponent (number of times the base machine is applied). Based on the context of previous questions (like Q164), it is implied that the base of the repeater machine should be a prime number unless otherwise stated.

First, let's find the prime factorization of 81:

$\begin{array}{c|cc} 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $81 = 3 \times 3 \times 3 \times 3 = 3^4$.

We are looking for two repeater machines with prime bases whose factors, when multiplied, equal $3^4$. Since the only prime factor of 81 is 3, the bases of our repeater machines must be 3.

Let the factors of the two repeater machines be $3^n$ and $3^m$. When hooked together, the total factor is the product of their factors: $3^n \times 3^m$.

Using the product of powers property, $a^m \times a^n = a^{m+n}$, the total factor is $3^{n+m}$.

We need this total factor to be equal to 81, which is $3^4$.

$3^{n+m} = 3^4$

Since the bases are equal, the exponents must be equal:

$n+m = 4$

Also, the question states not to use (× 1) machines. A repeater machine (× $a^n$) has a factor of 1 if $a^n = 1$. Since the base is 3 (not equal to 1), $3^n = 1$ only if $n=0$. Thus, the exponents $n$ and $m$ must be non-zero ($n \neq 0$ and $m \neq 0$). In the context of repeater machines (applying the base machine $n$ times), the exponent $n$ usually represents a positive integer ($n \geq 1$). Let's assume $n$ and $m$ are positive integers.

We need to find pairs of positive integers $(n, m)$ such that $n+m = 4$. The possible pairs are:

1. $n=1, m=3$. The factors are $3^1$ and $3^3$. These correspond to repeater machines (× $3^1$) and (× $3^3$).

2. $n=2, m=2$. The factors are $3^2$ and $3^2$. These correspond to repeater machines (× $3^2$) and (× $3^2$).

3. $n=3, m=1$. The factors are $3^3$ and $3^1$. These correspond to repeater machines (× $3^3$) and (× $3^1$). This hook-up uses the same two machines as in case 1, just in a different order. The question asks for two repeater machines, implying the factors used. We can list the pairs of machine factors.

Two pairs of repeater machines that can be hooked up to do the same work as a (× 81) machine are:

Pair 1:

Machine 1: A repeater machine with base 3 applied 1 time, i.e., a (× $3^1$) machine.

Machine 2: A repeater machine with base 3 applied 3 times, i.e., a (× $3^3$) machine.

Hooking these two machines gives a total factor of $3^1 \times 3^3 = 3^{1+3} = 3^4 = 81$.

Pair 2:

Machine 1: A repeater machine with base 3 applied 2 times, i.e., a (× $3^2$) machine.

Machine 2: A repeater machine with base 3 applied 2 times, i.e., a (× $3^2$) machine.

Hooking these two machines gives a total factor of $3^2 \times 3^2 = 3^{2+2} = 3^4 = 81$.

Given:

A machine with a multiplication factor of $\frac{1}{8}$.

To Find:

A single repeater machine that does the same work.

Solution:

A repeater machine is typically represented in the form (× $a^n$), where the base is $a$ and the exponent is $n$. This machine multiplies the input length by the factor $a^n$.

We are given a machine with a factor of $\frac{1}{8}$. We need to express this factor in the form $a^n$ to identify the equivalent repeater machine.

Let's consider expressing $\frac{1}{8}$ as a power of some number.

We know that $8 = 2 \times 2 \times 2 = 2^3$.

So, $\frac{1}{8} = \frac{1}{2^3}$.

Using the property of negative exponents, $\frac{1}{a^m} = a^{-m}$, we can write $\frac{1}{2^3}$ as $2^{-3}$.

Thus, the factor $\frac{1}{8}$ can be written as $2^{-3}$.

This is in the form $a^n$, where the base $a=2$ and the exponent $n=-3$.

Therefore, the repeater machine that does the same work as a ($\times \frac{1}{8}$) machine is a machine with the factor (× $2^{-3}$).

The equivalent single repeater machine is the (× $2^{-3}$) machine.

Alternate Solution:

Alternatively, we can express $\frac{1}{8}$ as a power of $\frac{1}{2}$.

$\frac{1}{8} = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \left(\frac{1}{2}\right)^3$

This is also in the form $a^n$, where the base $a = \frac{1}{2}$ and the exponent $n=3$.

This corresponds to a repeater machine (× $(\frac{1}{2})^3$), meaning a base machine (× $\frac{1}{2}$) is applied 3 times.

Depending on the allowed bases for repeater machines, the (× $(\frac{1}{2})^3$) machine is also a valid answer.

A hook-up of $( \times 5 )$ machines means multiplying by 5 multiple times. A machine that multiplies by a number $N$ can be replaced by a hook-up of $( \times 5 )$ machines if $N$ can be expressed as a power of 5, i.e., $N = 5^n$ for some positive integer $n$. The number of $( \times 5 )$ machines needed would be $n$.

Here are three examples of such machines:

1. A machine that multiplies by 25: Since $25 = 5 \times 5 = 5^2$, this machine can be replaced by hooking up two $( \times 5 )$ machines.

2. A machine that multiplies by 125: Since $125 = 5 \times 5 \times 5 = 5^3$, this machine can be replaced by hooking up three $( \times 5 )$ machines.

3. A machine that multiplies by 625: Since $625 = 5 \times 5 \times 5 \times 5 = 5^4$, this machine can be replaced by hooking up four $( \times 5 )$ machines.

Let the four machines be $M_A$, $M_B$, $M_C$, and $M_D$. From the table header, we know $M_A$ is $(\times 2)$. The table provides outputs for certain inputs when passed through these machines.

For the second row, let the input length be $I_2$. We are given the output for $M_A$ is 1 and for $M_B$ is 5.

Using $M_A$: $I_2 \times 2 = 1$. Solving for $I_2$, we get $I_2 = \frac{1}{2} = 0.5$.

Now we know the input for the second row is 0.5. Using this input with the output for $M_B$: $0.5 \times (\text{Factor of } M_B) = 5$. Solving for the factor of $M_B$, we get $\frac{5}{0.5} = 10$. So, $M_B$ is $(\times 10)$.

For the fourth row, let the input length be $I_4$. We are given the output for $M_A$ is 14 and for $M_C$ is 7.

Using $M_A$: $I_4 \times 2 = 14$. Solving for $I_4$, we get $I_4 = \frac{14}{2} = 7$.

Now we know the input for the fourth row is 7. Using this input with the output for $M_C$: $7 \times (\text{Factor of } M_C) = 7$. Solving for the factor of $M_C$, we get $\frac{7}{7} = 1$. So, $M_C$ is $(\times 1)$.

For the third row, the input length is given as 3. We are given the output for $M_D$ is 15.

Using the input 3 with the output for $M_D$: $3 \times (\text{Factor of } M_D) = 15$. Solving for the factor of $M_D$, we get $\frac{15}{3} = 5$. So, $M_D$ is $(\times 5)$.

The machines are: $M_A: (\times 2)$, $M_B: (\times 10)$, $M_C: (\times 1)$, and $M_D: (\times 5)$. The input lengths are 0.5, 3, and 7.

Now we can complete the table by calculating the missing outputs:

• For input 0.5:

  $0.5 \times 2 = 1$ (Given)

  $0.5 \times 10 = 5$ (Given)

  $0.5 \times 1 = 0.5$

  $0.5 \times 5 = 2.5$

• For input 3:

  $3 \times 2 = 6$

  $3 \times 10 = 30$

  $3 \times 1 = 3$

  $3 \times 5 = 15$ (Given)

• For input 7:

  $7 \times 2 = 14$ (Given)

  $7 \times 10 = 70$

  $7 \times 1 = 7$ (Given)

  $7 \times 5 = 35$

Here is the completed chart:

A repeater machine multiplies the input by a number that is a power of some base number, such as $\times a^b$. From the table, the first repeater machine listed is $\times 2^3$.

Let the three repeater machines be $R_1$, $R_2$, and $R_3$. Based on the table structure, $R_1 = \times 2^3 = \times 8$.

Let's examine the first data row (the third row in the provided table). Let the input length be $I_1$. The output for $R_1$ is 40 and the output for $R_3$ is 125.

Using $R_1$: $I_1 \times 8 = 40$. Solving for $I_1$, we get $I_1 = \frac{40}{8} = 5$. The input length for the first data row is 5.

Using the input $I_1=5$ with $R_3$: $5 \times R_3 = 125$. Solving for $R_3$, we get $R_3 = \frac{125}{5} = 25$. Since $R_3$ is a repeater machine, its factor must be a power. We recognize that $25 = 5^2$. So, $R_3 = \times 5^2$.

Now we know $R_1 = \times 2^3$ and $R_3 = \times 5^2$. We need to find the second machine, $R_2$, and the missing input lengths.

Let's look at the second data row (the fourth row in the provided table). The input length is given as $I_2=2$. We can calculate the outputs for this row:

Output for $R_1$: $2 \times 2^3 = 2 \times 8 = 16$.

Output for $R_3$: $2 \times 5^2 = 2 \times 25 = 50$.

Output for $R_2$: $2 \times R_2$.

Let's look at the third data row (the fifth row in the provided table). Let the input length be $I_3$. The output for $R_2$ is 162.

Using $R_2$: $I_3 \times R_2 = 162$.

From the previous row, the output for $R_2$ for input 2 is $2 \times R_2$. If we assume that the input length for the third data row ($I_3$) is the same as the input length for the second data row ($I_2$), i.e., $I_3 = 2$, let's see if it works.

If $I_3 = 2$, then $2 \times R_2 = 162$. Solving for $R_2$, we get $R_2 = \frac{162}{2} = 81$. Since $R_2$ is a repeater machine, its factor must be a power. We recognize that $81 = 3^4$. So, $R_2 = \times 3^4$.

Let's check if this is consistent. If $I_2=2$ and $I_3=2$, then the outputs for these rows must be identical.

For input 2 with $R_1 (\times 8)$: $2 \times 8 = 16$. This would be the missing output in the second data row (column 2) and the missing output in the third data row (column 2).

For input 2 with $R_2 (\times 81)$: $2 \times 81 = 162$. This matches the given output in the third data row (column 3). This would be the missing output in the second data row (column 3).

For input 2 with $R_3 (\times 25)$: $2 \times 25 = 50$. This would be the missing output in the second data row (column 4) and the missing output in the third data row (column 4).

This confirms that the input for the third data row is indeed 2, and the second repeater machine is $\times 3^4$.

The inputs are 5 (first data row) and 2 (second and third data rows). The repeater machines are $\times 2^3$, $\times 3^4$, and $\times 5^2$. We can now complete the chart by filling in the missing input lengths and outputs.

Here is the completed chart:

Given:

The amount of money on the first square is $\textsf{₹}1$.

The amount on each subsequent square is twice the amount on the previous square.

There are 64 squares on the chessboard.

To Find:

1. The total amount of money the farmer earned.

2. The first square on which the king will place at least $\textsf{₹}10$ lakh ($\textsf{₹}1,000,000$).

Solution:

The amounts of money placed on the squares form a geometric progression (GP). The amounts on the squares are $1, 2, 4, 8, \dots$.

The first term of this GP is $a = 1$.

The common ratio is $r = \frac{2}{1} = \frac{4}{2} = 2$.

The number of terms (squares) is $n = 64$.

The amount of money on the $k$-th square is given by the formula for the $k$-th term of a GP:

$A_k = a \times r^{k-1}$

Substituting $a=1$ and $r=2$, the amount on the $k$-th square is:

$A_k = 1 \times 2^{k-1} = 2^{k-1}$

To find the total amount of money the farmer earned, we need to calculate the sum of the amounts on all 64 squares. This is the sum of the first 64 terms of the GP. The formula for the sum of the first $n$ terms of a GP is:

$S_n = a \frac{r^n - 1}{r - 1}$

Substituting the values $a=1$, $r=2$, and $n=64$:

$S_{64} = 1 \times \frac{2^{64} - 1}{2 - 1}$

$S_{64} = \frac{2^{64} - 1}{1}$

$S_{64} = 2^{64} - 1$

Now, we need to calculate the value of $2^{64}$. This is a very large number.

$2^{64} = 18,446,744,073,709,551,616$

The total amount earned is $S_{64} = 2^{64} - 1$:

$S_{64} = 18,446,744,073,709,551,616 - 1$

$S_{64} = 18,446,744,073,709,551,615$

The total amount earned by the farmer is $\textsf{₹}18,446,744,073,709,551,615$.

Now, let's find the first square on which the amount is at least $\textsf{₹}10$ lakh ($\textsf{₹}1,000,000$).

Let $k$ be the number of the square. The amount on the $k$-th square is $A_k = 2^{k-1}$.

We want to find the smallest positive integer $k$ such that $A_k \geq 1,000,000$.

$2^{k-1} \geq 1,000,000$

$2^{k-1} \geq 10^6$

To solve this inequality for $k$, we can take the logarithm base 10 on both sides:

$\log_{10}(2^{k-1}) \geq \log_{10}(10^6)$

Using the logarithm property $\log(a^b) = b \log(a)$ and $\log_{10}(10^x) = x$:

$(k-1) \log_{10}(2) \geq 6$

Now, we isolate $k-1$ by dividing by $\log_{10}(2)$. Since $\log_{10}(2)$ is positive, the inequality direction remains the same.

$k-1 \geq \frac{6}{\log_{10}(2)}$

Using the approximate value $\log_{10}(2) \approx 0.30103$:

$k-1 \geq \frac{6}{0.30103}$

$k-1 \geq 19.9315 \dots$

Adding 1 to both sides:

$k \geq 19.9315 \dots + 1$

$k \geq 20.9315 \dots$

Since $k$ must be an integer (representing the square number), the smallest integer $k$ that satisfies this inequality is $k = 21$.

Let's verify the amounts on the 20th and 21st squares:

Amount on 20th square: $A_{20} = 2^{20-1} = 2^{19}$.

$2^{19} = 524,288$.

Since $524,288 < 1,000,000$, the amount on the 20th square is less than $\textsf{₹}10$ lakh.

Amount on 21st square: $A_{21} = 2^{21-1} = 2^{20}$.

$2^{20} = 1,048,576$.

Since $1,048,576 \geq 1,000,000$, the amount on the 21st square is at least $\textsf{₹}10$ lakh.

Therefore, the first square on which the king will place at least $\textsf{₹}10$ lakh is the 21st square.

Given:

Diameter of the Sun, $D_{\text{Sun}} = 1.4 \times 10^9$ m

Diameter of the Earth, $D_{\text{Earth}} = 1.2756 \times 10^7$ m

To Find:

The ratio of the diameter of the Sun to the diameter of the Earth by division.

Solution:

To compare the diameters by division, we will calculate the ratio $\frac{D_{\text{Sun}}}{D_{\text{Earth}}}$.

$\frac{D_{\text{Sun}}}{D_{\text{Earth}}} = \frac{1.4 \times 10^9 \text{ m}}{1.2756 \times 10^7 \text{ m}}$

We can separate the decimal parts and the power of 10 parts:

$\frac{D_{\text{Sun}}}{D_{\text{Earth}}} = \frac{1.4}{1.2756} \times \frac{10^9}{10^7}$

First, let's calculate the ratio of the decimal parts:

$\frac{1.4}{1.2756} \approx 1.09759$

Next, let's calculate the ratio of the powers of 10 using the rule $\frac{a^m}{a^n} = a^{m-n}$:

$\frac{10^9}{10^7} = 10^{9-7} = 10^2 = 100$

Now, multiply the results:

$\frac{D_{\text{Sun}}}{D_{\text{Earth}}} \approx 1.09759 \times 100$

$\frac{D_{\text{Sun}}}{D_{\text{Earth}}} \approx 109.759$

Rounding to one decimal place for a simpler comparison, the ratio is approximately 110.

This means the diameter of the Sun is approximately 110 times the diameter of the Earth.

Conclusion:

By dividing the diameter of the Sun by the diameter of the Earth, we find that the ratio is approximately 110.

This indicates that the diameter of the Sun is about 110 times larger than the diameter of the Earth.

Given:

Mass of Mars, $M_{\text{Mars}} = 6.42 \times 10^{29}$ kg

Mass of the Sun, $M_{\text{Sun}} = 1.99 \times 10^{30}$ kg

To Find:

The total mass of Mars and the Sun.

Solution:

To find the total mass, we need to add the mass of Mars and the mass of the Sun:

Total Mass = $M_{\text{Mars}} + M_{\text{Sun}}$

Total Mass = $(6.42 \times 10^{29}) + (1.99 \times 10^{30})$ kg

To add numbers in scientific notation, the exponents of 10 must be the same. We can convert $6.42 \times 10^{29}$ to a term with $10^{30}$:

$6.42 \times 10^{29} = 6.42 \times 10^{-1} \times 10^{30} = 0.642 \times 10^{30}$

Now, the addition becomes:

Total Mass = $(0.642 \times 10^{30}) + (1.99 \times 10^{30})$ kg

Total Mass = $(0.642 + 1.99) \times 10^{30}$ kg

Let's add the decimal parts:

$\begin{array}{cc} & 0 & . & 6 & 4 & 2 \\ + & 1 & . & 9 & 9 & 0 \\ \hline & 2 & . & 6 & 3 & 2 \\ \hline \end{array}$

The sum of the decimal parts is 2.632.

Therefore, the total mass is $2.632 \times 10^{30}$ kg.

Conclusion:

The total mass of Mars and the Sun is $2.632 \times 10^{30}$ kg.

Given:

Distance between the Sun and the Earth, $D_{\text{SE}} = 1.496 \times 10^8$ km

Distance between the Earth and the Moon, $D_{\text{EM}} = 3.84 \times 10^8$ m

To Find:

The distance between the Moon and the Sun ($D_{\text{SM}}$) during a solar eclipse.

Solution:

During a solar eclipse, the Moon is positioned between the Earth and the Sun. This means the three celestial bodies are approximately aligned in the order Sun - Moon - Earth.

The distance from the Sun to the Earth is the sum of the distance from the Sun to the Moon and the distance from the Moon to the Earth.

$D_{\text{SE}} = D_{\text{SM}} + D_{\text{ME}}$

Since the distance from the Moon to the Earth ($D_{\text{ME}}$) is the same as the distance from the Earth to the Moon ($D_{\text{EM}}$), we can write:

$D_{\text{SE}} = D_{\text{SM}} + D_{\text{EM}}$

We want to find the distance between the Moon and the Sun, $D_{\text{SM}}$. We can rearrange the equation:

$D_{\text{SM}} = D_{\text{SE}} - D_{\text{EM}}$

Before subtracting, we need to ensure that both distances are in the same unit. The distance $D_{\text{SE}}$ is in kilometers (km), and $D_{\text{EM}}$ is in meters (m).

Let's convert the distance $D_{\text{EM}}$ from meters to kilometers. We know that 1 kilometer = 1000 meters, or $1 \text{ km} = 10^3 \text{ m}$. Therefore, $1 \text{ m} = 10^{-3} \text{ km}$.

$D_{\text{EM}} = 3.84 \times 10^8 \text{ m}$

$D_{\text{EM}} = (3.84 \times 10^8) \times 10^{-3} \text{ km}$

$D_{\text{EM}} = 3.84 \times 10^{8-3} \text{ km}$

$D_{\text{EM}} = 3.84 \times 10^5 \text{ km}$

Now we can substitute the values of $D_{\text{SE}}$ and $D_{\text{EM}}$ (in km) into the equation for $D_{\text{SM}}$:

$D_{\text{SM}} = (1.496 \times 10^8 \text{ km}) - (3.84 \times 10^5 \text{ km})$

To subtract numbers in scientific notation, their powers of 10 must be the same. Let's express $3.84 \times 10^5$ with a power of $10^8$:

$3.84 \times 10^5 = 3.84 \times 10^{5-8} \times 10^8 = 3.84 \times 10^{-3} \times 10^8 = 0.00384 \times 10^8$

Now, the subtraction is:

$D_{\text{SM}} = (1.496 \times 10^8) - (0.00384 \times 10^8)$ km

$D_{\text{SM}} = (1.496 - 0.00384) \times 10^8$ km

Perform the subtraction of the decimal parts:

$\begin{array}{cc} & 1 & . & 4 & 9 & 6 & 0 & 0 \\ - & 0 & . & 0 & 0 & 3 & 8 & 4 \\ \hline & 1 & . & 4 & 9 & 2 & 1 & 6 \\ \hline \end{array}$

So, $1.496 - 0.00384 = 1.49216$.

Therefore, the distance between the Moon and the Sun during a solar eclipse is:

$D_{\text{SM}} = 1.49216 \times 10^8$ km

Conclusion:

At the time of a solar eclipse, when the Moon is between the Earth and the Sun, the distance between the Moon and the Sun is approximately $1.49216 \times 10^8$ km.

Given:

Distance from the star to Earth, $d = 8.1 \times 10^{13}$ km

Speed of light, $v = 3 \times 10^8$ m/s

To Find:

Time taken for light to travel from the star to the Earth, $t$.

Solution:

We know the relationship between distance, speed, and time is given by:

$d = v \times t$

To find the time $t$, we rearrange the formula:

$t = \frac{d}{v}$

Before substituting the values, we need to ensure that the units are consistent. The distance is given in kilometers (km), and the speed is given in meters per second (m/s).

Let's convert the distance from kilometers to meters. We know that $1 \text{ km} = 1000 \text{ m} = 10^3 \text{ m}$.

So, the distance in meters is:

$d = 8.1 \times 10^{13} \text{ km} = 8.1 \times 10^{13} \times 10^3 \text{ m}$

$d = 8.1 \times 10^{13+3} \text{ m} = 8.1 \times 10^{16} \text{ m}$

Now, substitute the distance in meters and the speed into the formula for time:

$t = \frac{8.1 \times 10^{16} \text{ m}}{3 \times 10^8 \text{ m/s}}$

We can separate the numbers and the powers of 10 for easier calculation:

$t = \left(\frac{8.1}{3}\right) \times \left(\frac{10^{16}}{10^8}\right)$ seconds

Calculate the numerical part:

$\frac{8.1}{3} = 2.7$

Calculate the power of 10 part using the exponent rule $\frac{a^m}{a^n} = a^{m-n}$:

$\frac{10^{16}}{10^8} = 10^{16-8} = 10^8$

Multiply the results:

$t = 2.7 \times 10^8$ seconds

This time can also be expressed in years. First, find the number of seconds in one year:

$1 \text{ year} = 365.25 \text{ days} \times 24 \text{ hours/day} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute}$

$1 \text{ year} = 31,557,600$ seconds $\approx 3.156 \times 10^7$ seconds

Now, divide the time in seconds by the number of seconds in a year:

$t_{\text{years}} = \frac{2.7 \times 10^8 \text{ seconds}}{3.15576 \times 10^7 \text{ seconds/year}}$

$t_{\text{years}} \approx \left(\frac{2.7}{3.15576}\right) \times \left(\frac{10^8}{10^7}\right)$ years

$t_{\text{years}} \approx 0.8556 \times 10^1$ years

$t_{\text{years}} \approx 8.556$ years

Conclusion:

The time taken for light to travel from that star to the Earth is $2.7 \times 10^8$ seconds, which is approximately 8.56 years.

Given:

The number to be divided is $(–15)^{-1}$.

The desired quotient is $(–5)^{-1}$.

To Find:

The number by which $(–15)^{-1}$ should be divided to get $(–5)^{-1}$.

Solution:

Let the required number be $x$.

According to the problem statement, when $(–15)^{-1}$ is divided by $x$, the result is $(–5)^{-1}$. We can write this as an equation:

$\frac{(-15)^{-1}}{x} = (-5)^{-1}$

We use the property of negative exponents which states that $a^{-1} = \frac{1}{a}$ for any non-zero number $a$.

Applying this property, we can rewrite the terms with negative exponents:

$(-15)^{-1} = \frac{1}{-15}$

$(-5)^{-1} = \frac{1}{-5}$

Substitute these values back into the equation:

$\frac{\frac{1}{-15}}{x} = \frac{1}{-5}$

The left side of the equation can be simplified as $\frac{1}{-15} \div x = \frac{1}{-15} \times \frac{1}{x} = \frac{1}{-15x}$.

So the equation becomes:

$\frac{1}{-15x} = \frac{1}{-5}$

To solve for $x$, we can take the reciprocal of both sides of the equation:

$-15x = -5$

Now, divide both sides of the equation by $-15$ to find the value of $x$:

$x = \frac{-5}{-15}$

Simplify the fraction:

$x = \frac{5}{15}$

$x = \frac{1}{3}$

Conclusion:

Therefore, $(–15)^{-1}$ should be divided by $\frac{1}{3}$ so that the quotient is equal to $(–5)^{-1}$.

The required number is $\frac{1}{3}$.

Given:

The number to be multiplied is $(–8)^{-3}$.

The desired product is $(–6)^{-3}$.

To Find:

The number by which $(–8)^{-3}$ should be multiplied to get $(–6)^{-3}$.

Solution:

Let the required number be $y$.

According to the problem statement, when $(–8)^{-3}$ is multiplied by $y$, the product is $(–6)^{-3}$. We can write this as an equation:

$(-8)^{-3} \times y = (-6)^{-3}$

We use the property of negative exponents which states that $a^{-n} = \frac{1}{a^n}$ for any non-zero number $a$ and positive integer $n$.

Applying this property, we can rewrite the terms with negative exponents:

$(-8)^{-3} = \frac{1}{(-8)^3} = \frac{1}{-8 \times -8 \times -8} = \frac{1}{-512}$

$(-6)^{-3} = \frac{1}{(-6)^3} = \frac{1}{-6 \times -6 \times -6} = \frac{1}{-216}$

Substitute these values back into the equation:

$\frac{1}{-512} \times y = \frac{1}{-216}$

To solve for $y$, multiply both sides of the equation by $-512$:

$y = \frac{1}{-216} \times (-512)$

$y = \frac{-512}{-216}$

Simplify the fraction by dividing the numerator and the denominator by their greatest common divisor. Both numbers are negative, so the fraction is positive. Both are divisible by 8.

$512 \div 8 = 64$

$216 \div 8 = 27$

So the fraction simplifies to:

$y = \frac{64}{27}$

Conclusion:

Therefore, $(–8)^{-3}$ should be multiplied by $\frac{64}{27}$ so that the product is equal to $(–6)^{-3}$.

The required number is $\frac{64}{27}$.

Part (1):

We are given the equation:

$\left( -\frac{1}{7} \right)^{-5}$ ÷ $\left( -\frac{1}{7} \right)^{-7}$ = $(-7)^x$

Using the exponent rule $a^m \div a^n = a^{m-n}$ on the left side:

$\left( -\frac{1}{7} \right)^{-5 - (-7)} = (-7)^x$

$\left( -\frac{1}{7} \right)^{-5 + 7} = (-7)^x$

$\left( -\frac{1}{7} \right)^{2} = (-7)^x$

Now, let's evaluate the left side:

$\left( -\frac{1}{7} \right)^{2} = \left( -\frac{1}{7} \right) \times \left( -\frac{1}{7} \right) = \frac{(-1) \times (-1)}{7 \times 7} = \frac{1}{49}$

So the equation becomes:

$\frac{1}{49} = (-7)^x$

We can express $\frac{1}{49}$ as a power of 7. Since $49 = 7^2$, we have $\frac{1}{49} = \frac{1}{7^2}$.

Using the property $\frac{1}{a^n} = a^{-n}$, we get $\frac{1}{7^2} = 7^{-2}$.

The equation is now:

$7^{-2} = (-7)^x$

Notice that $(-7)^2 = (-7) \times (-7) = 49 = 7^2$. So, $7^{-2} \neq (-7)^{-2}$. However, if the base is negative, like $(-7)^x$, the sign depends on the exponent $x$.

Let's rewrite $\frac{1}{49}$ as a power of $(-7)$.

Since $(-7)^2 = 49$, we have $\frac{1}{49} = \frac{1}{(-7)^2} = (-7)^{-2}$.

So the equation is:

$(-7)^{-2} = (-7)^x$

Since the bases are equal, the exponents must be equal.

$x = -2$

Part (2):

We are given the equation:

$\left( \frac{2}{5} \right)^{2x+6}$ × $\left( \frac{2}{5} \right)^{3}$ = $\left( \frac{2}{5} \right)^{x+2}$

Using the exponent rule $a^m \times a^n = a^{m+n}$ on the left side:

$\left( \frac{2}{5} \right)^{(2x+6) + 3} = \left( \frac{2}{5} \right)^{x+2}$

$\left( \frac{2}{5} \right)^{2x+9} = \left( \frac{2}{5} \right)^{x+2}$

Since the bases are equal $\left( \frac{2}{5} \right)$ and are not 0 or 1, the exponents must be equal.

$2x+9 = x+2$

Now, we solve for $x$. Subtract $x$ from both sides:

$2x - x + 9 = x - x + 2$

$x + 9 = 2$

Subtract 9 from both sides:

$x + 9 - 9 = 2 - 9$

$x = -7$

Part (3):

We are given the equation:

$2^x + 2^x + 2^x = 192$

Combine the terms on the left side:

$3 \times 2^x = 192$

Divide both sides by 3 to isolate $2^x$:

$\frac{3 \times 2^x}{3} = \frac{192}{3}$

$2^x = 64$

Now, express 64 as a power of 2:

$64 = 2 \times 32 = 2 \times 2 \times 16 = 2 \times 2 \times 2 \times 8 = 2 \times 2 \times 2 \times 2 \times 4 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6$

So the equation is:

$2^x = 2^6$

Since the bases are equal, the exponents must be equal.

$x = 6$

Part (4):

We are given the equation:

$\left( \frac{-6}{7} \right)^{x-7} = 1$

We know that any non-zero number raised to the power of 0 is equal to 1. The base $\frac{-6}{7}$ is not zero.

So, for the equation to be true, the exponent must be 0.

$x-7 = 0$

Add 7 to both sides:

$x - 7 + 7 = 0 + 7$

$x = 7$

Part (5):

We are given the equation:

$2^{3x} = 8^{2x+1}$

To solve this equation, we need to express both sides with the same base. We know that $8 = 2^3$.

Substitute $8$ with $2^3$ on the right side:

$2^{3x} = (2^3)^{2x+1}$

Using the exponent rule $(a^m)^n = a^{mn}$ on the right side:

$2^{3x} = 2^{3 \times (2x+1)}$

$2^{3x} = 2^{6x+3}$

Since the bases are equal, the exponents must be equal.

$3x = 6x + 3$

Subtract $6x$ from both sides:

$3x - 6x = 6x - 6x + 3$

$-3x = 3$

Divide both sides by -3:

$\frac{-3x}{-3} = \frac{3}{-3}$

$x = -1$

Part (6):

We are given the equation:

$5^x + 5^{x–1} = 750$

We can rewrite $5^{x-1}$ as $5^x \times 5^{-1}$, which is $5^x \times \frac{1}{5}$.

Substitute this into the equation:

$5^x + 5^x \times \frac{1}{5} = 750$

Factor out $5^x$ from the terms on the left side:

$5^x \left( 1 + \frac{1}{5} \right) = 750$

Simplify the expression in the parenthesis:

$1 + \frac{1}{5} = \frac{5}{5} + \frac{1}{5} = \frac{5+1}{5} = \frac{6}{5}$

So the equation becomes:

$5^x \left( \frac{6}{5} \right) = 750$

To isolate $5^x$, multiply both sides by the reciprocal of $\frac{6}{5}$, which is $\frac{5}{6}$:

$5^x = 750 \times \frac{5}{6}$

Now, calculate the right side:

$750 \times \frac{5}{6} = \frac{750 \times 5}{6}$

We can simplify by dividing 750 by 6:

$\frac{750}{6} = 125$

So, $5^x = 125 \times 5$

$5^x = 625$

Now, express 625 as a power of 5:

$625 = 5 \times 125 = 5 \times 5 \times 25 = 5 \times 5 \times 5 \times 5 = 5^4$

So the equation is:

$5^x = 5^4$

Since the bases are equal, the exponents must be equal.

$x = 4$

Given:

$a = -1$

$b = 2$

To Find:

The values of the given expressions.

Solution:

First, let's calculate the values of the individual terms $a^b$, $b^a$, and $b^2$ using the given values of $a$ and $b$.

$a^b = (-1)^2$

$(-1)^2 = (-1) \times (-1) = 1$

So, $a^b = 1$.

$b^a = 2^{-1}$

Using the property $x^{-n} = \frac{1}{x^n}$, we have:

$2^{-1} = \frac{1}{2^1} = \frac{1}{2}$

So, $b^a = \frac{1}{2}$.

$b^2 = 2^2$

$2^2 = 2 \times 2 = 4$

So, $b^2 = 4$.

Now, let's evaluate each expression:

(1) $a^b + b^a$

$a^b + b^a = 1 + \frac{1}{2}$

To add these, find a common denominator:

$1 + \frac{1}{2} = \frac{2}{2} + \frac{1}{2} = \frac{2+1}{2} = \frac{3}{2}$

The value of $a^b + b^a$ is $\frac{3}{2}$.

(2) $a^b – b^a$

$a^b – b^a = 1 - \frac{1}{2}$

To subtract these, find a common denominator:

$1 - \frac{1}{2} = \frac{2}{2} - \frac{1}{2} = \frac{2-1}{2} = \frac{1}{2}$

The value of $a^b – b^a$ is $\frac{1}{2}$.

(3) $a^b \times b^2$

$a^b \times b^2 = 1 \times 4$

$1 \times 4 = 4$

The value of $a^b \times b^2$ is 4.

(4) $a^b \div b^a$

$a^b \div b^a = 1 \div \frac{1}{2}$

Dividing by a fraction is the same as multiplying by its reciprocal:

$1 \div \frac{1}{2} = 1 \times \frac{2}{1} = 1 \times 2 = 2$

The value of $a^b \div b^a$ is 2.

Part (1): $\frac{-1296}{14641}$

We need to express the numerator and the denominator as powers of integers.

$1296 = 6 \times 6 \times 6 \times 6 = 6^4$

$14641 = 11 \times 11 \times 11 \times 11 = 11^4$

So the fraction can be written as $\frac{-6^4}{11^4}$.

Since the numerator is negative and the denominator is positive, the fraction is negative. We can write this as $-\frac{6^4}{11^4}$.

Using the property $\frac{a^n}{b^n} = \left(\frac{a}{b}\right)^n$, we have:

$-\frac{6^4}{11^4} = -\left(\frac{6}{11}\right)^4$

Since the exponent (4) is even, a positive base raised to this power is positive, and a negative base raised to this power is also positive. To get a negative result with an even power, the negative sign must be outside the power.

The exponential form is $-\left(\frac{6}{11}\right)^4$.

Part (2): $\frac{-125}{343}$

We need to express the numerator and the denominator as powers of integers.

$125 = 5 \times 5 \times 5 = 5^3$

$343 = 7 \times 7 \times 7 = 7^3$

So the fraction can be written as $\frac{-5^3}{7^3}$.

Using the property $\frac{a^n}{b^n} = \left(\frac{a}{b}\right)^n$ and noting that for an odd exponent $(-a)^n = -a^n$, we can move the negative sign into the base of the numerator:

$\frac{-5^3}{7^3} = \frac{(-5)^3}{7^3} = \left(\frac{-5}{7}\right)^3$

The exponential form is $\left(\frac{-5}{7}\right)^3$.

Part (3): $\frac{400}{3969}$

We need to express the numerator and the denominator as powers of integers.

$400 = 20 \times 20 = 20^2$

$3969 = 63 \times 63 = 63^2$

So the fraction can be written as $\frac{20^2}{63^2}$.

Using the property $\frac{a^n}{b^n} = \left(\frac{a}{b}\right)^n$:

$\frac{20^2}{63^2} = \left(\frac{20}{63}\right)^2$

The exponential form is $\left(\frac{20}{63}\right)^2$.

Part (4): $\frac{-625}{10000}$

We need to express the numerator and the denominator as powers of integers.

$625 = 5 \times 5 \times 5 \times 5 = 5^4$

$10000 = 10 \times 10 \times 10 \times 10 = 10^4$

So the fraction can be written as $\frac{-5^4}{10^4}$.

Since the numerator is negative and the denominator is positive, the fraction is negative. We can write this as $-\frac{5^4}{10^4}$.

Using the property $\frac{a^n}{b^n} = \left(\frac{a}{b}\right)^n$, we have:

$-\frac{5^4}{10^4} = -\left(\frac{5}{10}\right)^4$

We can simplify the fraction inside the parenthesis:

$\frac{5}{10} = \frac{1}{2}$

So the expression becomes:

$-\left(\frac{1}{2}\right)^4$

Since the exponent (4) is even, a positive base raised to this power is positive. To get a negative result with an even power, the negative sign must be outside the power.

The exponential form is $-\left(\frac{1}{2}\right)^4$.

Part (1): Simplify $\left( \frac{1}{2} \right)^{2} - \left[ \left( \frac{1}{4} \right)^{3} \right]^{-1} \times 2^{-3}$

First term: $\left( \frac{1}{2} \right)^{2} = \frac{1^2}{2^2} = \frac{1}{4}$

Second term: $\left[ \left( \frac{1}{4} \right)^{3} \right]^{-1} = \left( \frac{1}{4} \right)^{3 \times -1} = \left( \frac{1}{4} \right)^{-3}$

Using the property $(a/b)^{-n} = (b/a)^n$: $\left( \frac{1}{4} \right)^{-3} = \left( \frac{4}{1} \right)^{3} = 4^3 = 4 \times 4 \times 4 = 64$

Third term: $2^{-3}$

Using the property $a^{-n} = \frac{1}{a^n}$: $2^{-3} = \frac{1}{2^3} = \frac{1}{8}$

Now substitute these values back into the expression:

$\frac{1}{4} - (64) \times \left( \frac{1}{8} \right)$

Perform the multiplication first:

$64 \times \frac{1}{8} = \frac{64}{8} = 8$

Now perform the subtraction:

$\frac{1}{4} - 8$

Find a common denominator:

$\frac{1}{4} - \frac{8 \times 4}{4} = \frac{1}{4} - \frac{32}{4} = \frac{1 - 32}{4} = \frac{-31}{4}$

The simplified value is $\frac{-31}{4}$.

Part (2): Simplify $\left( \frac{4}{3} \right)^{-2} - \left[ \left( \frac{3}{4} \right)^{2} \right]^{(-2)}$

First term: $\left( \frac{4}{3} \right)^{-2}$

Using the property $(a/b)^{-n} = (b/a)^n$: $\left( \frac{4}{3} \right)^{-2} = \left( \frac{3}{4} \right)^{2} = \frac{3^2}{4^2} = \frac{9}{16}$

Second term: $\left[ \left( \frac{3}{4} \right)^{2} \right]^{(-2)}$

Using the property $(a^m)^n = a^{mn}$: $\left[ \left( \frac{3}{4} \right)^{2} \right]^{-2} = \left( \frac{3}{4} \right)^{2 \times (-2)} = \left( \frac{3}{4} \right)^{-4}$

Using the property $(a/b)^{-n} = (b/a)^n$: $\left( \frac{3}{4} \right)^{-4} = \left( \frac{4}{3} \right)^{4} = \frac{4^4}{3^4} = \frac{256}{81}$

Now substitute these values back into the expression:

$\frac{9}{16} - \frac{256}{81}$

To subtract, find a common denominator, which is the product of 16 and 81: $16 \times 81 = 1296$.

$\frac{9}{16} = \frac{9 \times 81}{16 \times 81} = \frac{729}{1296}$

$\frac{256}{81} = \frac{256 \times 16}{81 \times 16} = \frac{4096}{1296}$

Now perform the subtraction:

$\frac{729}{1296} - \frac{4096}{1296} = \frac{729 - 4096}{1296}$

$729 - 4096 = -3367$

The simplified value is $\frac{-3367}{1296}$.

Part (3): Simplify $\left( \frac{4}{13} \right)^{4} \times \left( \frac{13}{7} \right)^{2} \times \left( \frac{7}{4} \right)^{3}$

Rewrite the expression by separating the bases:

$\frac{4^4}{13^4} \times \frac{13^2}{7^2} \times \frac{7^3}{4^3}$

Group terms with the same base:

$\left( \frac{4^4}{4^3} \right) \times \left( \frac{13^2}{13^4} \right) \times \left( \frac{7^3}{7^2} \right)$

Using the property $\frac{a^m}{a^n} = a^{m-n}$:

$4^{4-3} \times 13^{2-4} \times 7^{3-2}$

$4^1 \times 13^{-2} \times 7^1$

$4 \times \frac{1}{13^2} \times 7$

$4 \times \frac{1}{169} \times 7$