Chapter 8 Exponents & Powers (Class 8 - Maths NCERT Exemplar Solutions)

Solution:

The multiplicative inverse of a number $n$ is a number which, when multiplied by $n$, gives the product $1$. It is also known as the reciprocal of the number.

The given number is $2^7$.

Multiplicative inverse of $2^7 = \frac{1}{2^7}$

According to the laws of exponents, $\frac{1}{a^n} = a^{-n}$.

Therefore, $\frac{1}{2^7} = 2^{-7}$.

Hence, the correct option is (a).

Solution:

First, let's understand the value of a billion in the decimal system. In the international system, $1$ billion is $1,000,000,000$ (one followed by nine zeros).

In the Indian perspective, $100$ billion is equivalent to $10,000$ crore ($1$ kharab).

Now, $100$ billion can be written as:

$100 \times 1,000,000,000$

$= 100,000,000,000$

Counting the zeros, there are $11$ zeros. Thus, it can be written as $10^{11}$.

Hence, the correct option is (b).

Solution:

We use the law of exponents: $a^n \times b^n = (a \times b)^n$.

Here, $a = -4$, $b = \frac{5}{4}$, and $n = 4$.

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

On simplifying the term inside the bracket:

$\left( -\cancel{4}^1 \times \frac{5}{\cancel{4}_1} \right)^4 = (-5)^4$

Since the exponent is an even number, $(-5)^4 = 5^4 = 5 \times 5 \times 5 \times 5 = 625$.

Hence, the blank should be filled with $625$ or $(-5)^4$.

Solution:

Using the law of exponents $(a^m)^n = a^{m \times n}$:

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

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

Now multiplying the results:

$2^{-6} \times 3^{-6}$

Using the law $a^n \times b^n = (ab)^n$:

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

We can also write this as $\frac{1}{6^6}$.

Hence, the blank should be filled with $6^{-6}$.

Solution:

First, we convert 150 million into numerical form. One million is $1,000,000$ ($10^6$).

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

To write this in standard exponential form (scientific notation), we move the decimal point 8 places to the left:

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

Hence, the distance is $1.5 \times 10^8$ km.

Solution:

The statement is False.

Very small numbers (numbers between 0 and 1) are expressed in standard form using negative exponents. For example, $0.00007$ is written as $7 \times 10^{-5}$. Positive exponents are used for very large numbers.

Solution:

The statement is False.

The product of $(-10)$ multiplied four times is $(-10)^4$.

Since the exponent is even, the result is positive:

$(-10) \times (-10) \times (-10) \times (-10) = 10,000 = 10^4$

Note that $10^4$ is not equal to $10^{-4}$ (which is $0.0001$).

Solution:

Given: $\frac{(-2)^3 \times (-2)^7}{3 \times 4^6}$

First, simplify the numerator using $a^m \times a^n = a^{m+n}$:

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

Since the exponent $10$ is even, $(-2)^{10} = 2^{10}$.

Next, express the denominator in base 2. We know $4 = 2^2$:

$4^6 = (2^2)^6 = 2^{2 \times 6} = 2^{12}$

The expression now becomes:

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

Using the law $\frac{a^m}{a^n} = a^{m-n}$ or $\frac{1}{a^{n-m}}$:

$\frac{1}{3 \times 2^{12-10}} = \frac{1}{3 \times 2^2}$

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

Hence, the simplified value is $\frac{1}{12}$.

Solution:

Given: $(-5)^{x+1} \times (-5)^5 = (-5)^7$

Using the law of exponents $a^m \times a^n = a^{m+n}$ on the Left Hand Side (LHS):

$(-5)^{(x+1) + 5} = (-5)^7$

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

Since the bases on both sides are the same and not equal to $0$, $1$, or $-1$, we can equate the exponents:

Transposing 6 to the RHS:

$x = 7 - 6$

$x = 1$

Hence, the value of $x$ is $1$.

Solution:

In the exponential expression $2^n$:

1. The number $2$ is called the base.

2. The number $n$ is called the exponent, index, or power.

In algebraic terms, if $n$ can take different values, it is a variable exponent. However, the standard mathematical name for the position of $n$ is the exponent.

Given the options provided:

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

(b) Constant - Incorrect (n is typically the variable)

(c) x - Incorrect

(d) Variable - Correct (as $n$ is the variable representing the power)

Hence, the correct option is (d).

Solution:

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

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

By the law of exponents:

This means the number becomes $\frac{1}{a}$ times the previous number.

In the context of the decimal number system (base $10$), if the exponent decreases by $1$, the number becomes $\frac{1}{10}$ (one-tenth) of the previous number.

Hence, the correct option is (a).

Solution:

According to the law of negative exponents:

Applying this to $3^{-2}$:

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

Hence, the correct option is (b).

Solution:

Using the law of exponents $\frac{1}{a^{-n}} = a^n$:

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

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

Hence, the correct option is (a).

Solution:

Using the law of exponents $a^m \div a^n = a^{m-n}$:

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

$= 3^{5 + 6}$

$= 3^{11}$

Hence, the correct option is (c).

Solution:

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

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

$= \frac{5^2}{2^2} = \frac{25}{4}$

Hence, the correct option is (c).

Solution:

First, let's find the value of $\left( \frac{2}{5} \right)^{-1}$:

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

Now, we need to find the reciprocal of $\frac{5}{2}$. The reciprocal of a fraction $\frac{a}{b}$ is $\frac{b}{a}$.

Reciprocal of $\frac{5}{2} = \frac{2}{5}$

Hence, the correct option is (a).

Solution:

The multiplicative inverse of a number $x$ is $\frac{1}{x}$.

Multiplicative inverse of $10^{-100} = \frac{1}{10^{-100}}$

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

$\frac{1}{10^{-100}} = 10^{100}$

Hence, the correct option is (c).

Solution:

First, we calculate the value of the exponent:

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

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

Since the exponent is an odd number, the result will be negative:

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

$(-2)^5 = -32$

Hence, the correct option is (c).

Solution:

We use the law of exponents $\left( \frac{a}{b} \right)^n = \frac{a^n}{b^n}$:

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

Since the exponent $4$ is an even number, the result of $(-2)^4$ will be positive:

$(-2)^4 = 16$

$3^4 = 3 \times 3 \times 3 \times 3 = 81$

Therefore, the value is $\frac{16}{81}$.

Hence, the correct option is (a).

Solution:

The multiplicative inverse of a number $z$ is $\frac{1}{z}$.

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

Using the law of exponents $\frac{1}{a^{-n}} = a^n$:

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

Hence, the correct option is (a).

Solution:

According to the Product Law of Exponents, for any non-zero integer $x$ and any integers $m$ and $n$ (whether positive or negative):

Hence, the correct option is (b).

Solution:

According to the Zero Exponent Rule, any non-zero base raised to the power of zero is always equal to $1$.

Hence, the correct option is (a).

Solution:

A negative exponent indicates the reciprocal of the base.

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

Hence, the correct option is (b).

Solution:

By the definition of negative exponents:

This rule states that $x^{-m}$ is the multiplicative inverse of $x^m$.

Hence, the correct option is (c).

Solution:

According to the Power of a Power Law of exponents, when a power is raised to another power, we multiply the exponents.

Hence, the correct option is (b).

Solution:

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

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

The base $-\frac{3}{4}$ is inverted to become $-\frac{4}{3}$, and the sign of the exponent changes from $-3$ to $3$.

Hence, the correct option is (d).

Solution:

Applying the reciprocal rule for negative exponents:

Substituting $a = -5$, $b = 7$, and $n = 5$:

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

Hence, the correct option is (d).

Solution:

According to the law of exponents, for any non-zero rational number $\frac{a}{b}$:

Given expression is $\left( \frac{-7}{5} \right)^{-1}$. Here $a = -7$ and $b = 5$.

Applying the rule:

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

Hence, the correct option is (b).

Solution:

Using the law of exponents for division with the same base: $a^m \div a^n = a^{m - n}$.

Here, the base $a = -9$, $m = 3$, and $n = 8$.

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

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

Hence, the correct option is (d).

Solution:

Using the law of exponents: $x^m \div x^n = x^{m - n}$.

Substituting the given values $m = 7$ and $n = 12$:

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

$= x^{-5}$

Hence, the correct option is (c).

Solution:

Using the power of a power law of exponents: $(a^m)^n = a^{m \times n}$.

Here, $a = x$, $m = 4$, and $n = -3$.

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

$= x^{-12}$

Hence, the correct option is (b).

Solution:

First, we simplify the terms inside the parentheses using the rule $a^{-1} = \frac{1}{a}$:

$( \frac{1}{7} - \frac{1}{8} )^{-1} - ( \frac{1}{3} - \frac{1}{4} )^{-1}$

Taking the LCM for the fractions:

$( \frac{8 - 7}{56} )^{-1} - ( \frac{4 - 3}{12} )^{-1}$

$( \frac{1}{56} )^{-1} - ( \frac{1}{12} )^{-1}$

Since $( \frac{1}{a} )^{-1} = a$:

$56 - 12 = 44$

Hence, the correct option is (a).

Solution:

To write a number in standard form ($m \times 10^n$), we move the decimal point so that there is only one non-zero digit to the left of the decimal point.

For $0.000064$, we move the decimal point 5 places to the right.

When the decimal moves to the right, the exponent of 10 is negative.

$0.000064 = 6.4 \times 10^{-5}$

Hence, the correct option is (d).

Solution:

In the Indian perspective, this number is 23 crore 40 lakh.

To express $234,000,000$ in standard form, we move the decimal point from the end of the number to the position between 2 and 3.

The decimal moves 8 places to the left.

When the decimal moves to the left, the exponent of 10 is positive.

$234,000,000 = 2.34 \times 10^8$

Hence, the correct option is (a).

Solution:

To convert from standard form to usual form with a negative exponent, we move the decimal point to the left by the number of places indicated by the exponent.

Here, the exponent is $-5$, so we move the decimal point 5 places to the left starting from the position between 2 and 0.

$2.03 \times 10^{-5} = 0.0000203$

Hence, the correct option is (d).

Solution:

According to the zero exponent identity, any non-zero number raised to the power of zero is equal to 1.

Substituting $a = \frac{1}{10}$:

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

Hence, the correct option is (c).

Solution:

According to the laws of exponents, for any non-zero rational numbers $a$ and $b$ and integer $n$:

Given the expression $\left( \frac{3}{4} \right)^{5} \div \left( \frac{5}{3} \right)^{5}$, we can see that the exponents are the same ($n = 5$).

Applying the law:

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

Hence, the correct option is (a).

Solution:

Using the law of exponents: $a^n \div b^n = (a \div b)^n$.

In the given expression $x^4 \div y^4$:

$a = x, b = y,$ and $n = 4$.

Therefore, $x^4 \div y^4 = (x \div y)^4$.

Hence, the correct option is (c).

Solution:

According to the division law of exponents for the same base:

Given $p^{13} \div p^8$, here $a = p, m = 13,$ and $n = 8$.

Substituting these values:

$p^{13} \div p^8 = p^{13 - 8} = p^5$

Hence, the correct option is (a).

Solution:

According to the power of a power law of exponents:

Applying this to $(z^{-2})^3$:

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

$= z^{-6}$

Hence, the correct option is (b).

Solution:

The cube of a number is the number multiplied by itself three times.

$\text{Cube of } \left( -\frac{1}{2} \right) = \left( -\frac{1}{2} \right)^3$

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

Since the product involves three negative signs (odd count), the result will be negative.

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

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

Hence, the correct option is (c).

Solution:

First, let's find the reciprocal of $\left( \frac{2}{3} \right)^4$. The reciprocal of any non-zero number $x$ is $\frac{1}{x}$.

Now let's check the options:

(a) $\left( \frac{3}{2} \right)^4$ — This is the same as (i).

(c) $\left( \frac{2}{3} \right)^{-4}$ — By the law $\left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^n$, this is equal to $\left( \frac{3}{2} \right)^4$.

(d) $\frac{3^4}{2^4}$ — This is just another way to write $\left( \frac{3}{2} \right)^4$.

However, option (b) is $\left( \frac{3}{2} \right)^{-4}$, which is equal to $\left( \frac{2}{3} \right)^4$. A number cannot be its own reciprocal (unless it is 1 or -1).

Hence, the correct option is (b).

Solution:

The multiplicative inverse of a number $a$ is $\frac{1}{a}$.

Multiplicative inverse of $10^{10} = \frac{1}{10^{10}}$

Using the negative exponent law $\frac{1}{a^n} = a^{-n}$:

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

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

Solution:

Using the product law of exponents: $a^m \times a^n = a^{m + n}$.

Here, the base is $a$ and the exponents are $3$ and $-10$.

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

$= a^{3 - 10}$

$= a^{-7}$

Hence, $a^3 \times a^{-10} = $ $a^{-7}$.

According to the Zero Exponent Rule, any non-zero number raised to the power of zero is always equal to 1.

Applying this to the given expression:

$5^0 = 1$

Hence, $5^0 = $ $1$.

Solution:

According to the product law of exponents:

Applying this to the given expression:

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

$= 5^0$

Hence, the blank should be filled with 1.

Solution:

Using the law of exponents $\left( \frac{1}{a^m} \right)^n = \frac{1}{a^{mn}}$:

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

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

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

Calculating the value: $2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$.

Hence, the value is $\frac{1}{64}$.

Solution:

First, we express the number 8 as a power of 2:

Now, substituting this value into the original expression $8^{-2}$:

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

Using the power of a power law $(a^m)^n = a^{mn}$:

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

$= 2^{-6}$

Hence, the expression is $2^{-6}$.

Solution:

In scientific notation or standard form, numbers much smaller than 1 (like $0.00004$) are written with the decimal point moved to the right, which results in a power of 10 with a negative index.

Hence, very small numbers can be expressed in standard form by using negative exponents.

Solution:

In standard form, numbers much larger than 1 (like $5,000,000$) are written with the decimal point moved to the left, which results in a power of 10 with a positive index.

Hence, very large numbers can be expressed in standard form by using positive exponents.

Solution:

We need to multiply $10^5$ by $10^{-10}$. Using the product law of exponents:

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

$= 10^{5 - 10}$

$= 10^{-5}$

Hence, we get $10^{-5}$.

Solution:

First, let us simplify the expression inside the square brackets:

$\left( \frac{2}{13} \right)^{-6} \div \left( \frac{2}{13} \right)^{3}$

Using $a^m \div a^n = a^{m-n}$:

$= \left( \frac{2}{13} \right)^{-6 - 3} = \left( \frac{2}{13} \right)^{-9}$

Now, applying the outer exponent of 3:

$\left[ \left( \frac{2}{13} \right)^{-9} \right]^3 = \left( \frac{2}{13} \right)^{-9 \times 3} = \left( \frac{2}{13} \right)^{-27}$

Now, multiply by the final term:

$\left( \frac{2}{13} \right)^{-27} \times \left( \frac{2}{13} \right)^{-9}$

$= \left( \frac{2}{13} \right)^{-27 + (-9)} = \left( \frac{2}{13} \right)^{-36}$

Hence, the result is $\left( \frac{2}{13} \right)^{-36}$.

Solution:

First, we simplify the expression inside the brackets:

$4^{-1} + 3^{-1} + 6^{-2}$

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

$= \frac{1}{4} + \frac{1}{3} + \frac{1}{36}$

To add these fractions, we find the LCM of 4, 3, and 36, which is 36.

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

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

Now, we apply the outer exponent of $-1$:

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

Hence, the value is $\frac{18}{11}$.

Solution:

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

First, check if the base is non-zero:

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

Since the sum of positive fractions is clearly not zero, we can apply the property directly.

Hence, $[2^{-1} + 3^{-1} + 4^{-1}]^0 = $ 1.

Solution:

First, we count the number of zeros in the denominator. The number $100,000,000$ has 8 zeros.

So, the expression can be written as:

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

Using the law of exponents $\frac{1}{a^n} = a^{-n}$:

In standard form ($m \times 10^n$), we write it as $1 \times 10^{-8}$.

Hence, the blank should be filled with $1 \times 10^{-8}$.

Solution:

In the Indian perspective, the number 12340000 is $1$ crore $23$ lakh and $40$ thousand.

To write this in standard form ($m \times 10^n$), where $1 \leq m < 10$, we move the decimal point from the end of the number 7 places to the left.

$12340000 = 1.2340000 \times 10^7$

$= 1.234 \times 10^7$

Hence, the blank should be filled with $1.234 \times 10^7$.

Solution:

To convert from standard form to usual form, we move the decimal point according to the power of 10.

Since the exponent is $+6$, we move the decimal point 6 places to the right.

$3.41 \times 10^6 = 3410000$

In the Indian perspective, this is $34$ lakh $10$ thousand.

Hence, the blank should be filled with $34,10,000$.

Solution:

The exponent of 10 is $+6$. Therefore, we move the decimal point 6 places to the right.

$2.39461 \times 10^6 = 2394610$

In the Indian perspective, this is $23$ lakh $94$ thousand $610$.

Hence, the blank should be filled with $23,94,610$.

Solution:

We are given $36 = 6^2$.

We need to find the value of $\frac{1}{36}$.

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

Using the law of exponents $\frac{1}{a^n} = a^{-n}$:

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

Hence, the blank should be filled with $6^{-2}$.

Solution:

Let the required number be $x$.

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

Using the law $\left( \frac{a}{b} \right)^n = \frac{a^n}{b^n}$:

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

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

$x = 3^4$

Value of $3^4 = 3 \times 3 \times 3 \times 3 = 81$.

Hence, the blank should be filled with $3^4$ or $81$.

Solution:

Using the quotient law of exponents $a^m \div a^n = a^{m-n}$:

Here, $a = 3, m = 5$ and $n = -6$.

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

$= 3^{5 + 6}$

$= 3^{11}$

Hence, the blank should be filled with $3^{11}$.

Solution:

To find the value, we convert the standard form into usual form.

Since the exponent is $-7$, we move the decimal point 7 places to the left starting from 3.

$3 \times 10^{-7} = 0.0000003$

Hence, the blank should be filled with $0.0000003$.

Solution:

When adding or subtracting numbers in standard form, we cannot perform the operation directly if the powers of 10 are different.

We must first express both numbers such that they have the same (equal) exponents so that we can take the power of 10 as a common factor.

Hence, the blank should be filled with same or equal.

Solution:

To convert a number into standard form ($m \times 10^n$), we move the decimal point so that there is only one non-zero digit to the left of the decimal point. For the number 32,50,00,00,000, we move the decimal point 10 places to the left.

In the Indian perspective, this number is 3250 crore or 32 arab 50 crore.

$32,50,00,00,000 = 3.25 \times 10^{10}$

Hence, the standard form is $3.25 \times 10^{10}$.

Solution:

To write 0.000000008 in standard form, we move the decimal point 9 places to the right to bring it after the digit 8.

When the decimal point is moved to the right, the exponent of 10 is negative.

$0.000000008 = 8 \times 10^{-9}$

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

Solution:

To convert $2.3 \times 10^{-10}$ into usual form, we move the decimal point 10 places to the left because the exponent is negative.

$2.3 \times 10^{-10} = 0.00000000023$

Hence, the usual form is $0.00000000023$.

Solution:

Let the number in the blank be $x$.

We know that $8 = 8^1$. Using the quotient law of exponents:

$x = 8^5 \div 8^1$

$x = 8^{5 - 1}$

$x = 8^4$

Since $8^4 = 8 \times 8 \times 8 \times 8 = 4096$.

Hence, the blank should be filled with $8^4$ or $4096$.

Solution:

Let the number in the blank be $x$.

To find $x$, we divide the product by the given multiplier:

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

Using the law $\frac{a^m}{a^n} = a^{m - n}$:

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

$x = 2^{5 + 5} = 2^{10}$

Value of $2^{10} = 1024$.

Hence, the blank should be filled with $2^{10}$ or $1024$.

Solution:

First, simplify the expression inside the bracket using the negative exponent rule $a^{-1} = \frac{1}{a}$:

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

Now, apply the outer square:

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

Alternatively, in exponential form: $12^{-2}$.

Hence, the value is $\frac{1}{144}$.

Solution:

First, simplify the expression inside the bracket:

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

Now, apply the outer exponent of $-1$:

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

According to the law $\left( \frac{1}{a} \right)^{-1} = a$:

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

Hence, the value is $6$.

Solution:

According to the Zero Exponent Rule, $a^0 = 1$ for any non-zero number $a$.

Substituting $6^0 = 1$ and $7^0 = 1$ in the expression:

$(1 - 1) \times (1 + 1)$

$0 \times 2 = 0$

Hence, the value is $0$.

Solution:

According to the law of exponents, $a^n = \frac{1}{a^{-n}}$.

Applying this to $3^5$:

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

Hence, the expression for $3^5$ with a negative exponent is $\frac{1}{3^{-5}}$.

Solution:

First, we consider the base $(-7)$ raised to an even power. According to the laws of exponents, $(-a)^n = a^n$ when $n$ is an even integer.

Now, substituting this back into the expression:

$7^6 \div 7^6$

Using the law of exponents $a^m \div a^n = a^{m-n}$:

$7^{6-6} = 7^0$

Hence, the blank should be filled with 1.

Solution:

First, we simplify the terms inside the square brackets using the property $a^{-n} = \frac{1}{a^n}$:

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

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

$3^{-2} = \frac{1}{3^2} = \frac{1}{9}$

Now, substitute these into the expression:

$\left[ 1 + \frac{1}{4} + \frac{1}{9} \right] \times 6^2$

$\left[ 1 + \frac{1}{4} + \frac{1}{9} \right] \times 36$

Using the distributive property:

$(1 \times 36) + \left( \frac{1}{4} \times 36 \right) + \left( \frac{1}{9} \times 36 \right)$

$36 + 9 + 4 = 49$

Hence, the blank should be filled with 49.

Solution:

The given statement is False.

First, find the value of $(-4)^{-2}$:

$(-4)^{-2} = \frac{1}{(-4)^2} = \frac{1}{16}$

The multiplicative inverse of a number is its reciprocal. The multiplicative inverse of $\frac{1}{16}$ is 16.

However, the statement claims the inverse is $(4)^{-2}$, which is $\frac{1}{4^2} = \frac{1}{16}$. Since $16 \neq \frac{1}{16}$, the statement is incorrect.

Solution:

The given statement is True.

Multiplicative inverse of $\left( \frac{3}{2} \right)^2$ is $\frac{1}{(3/2)^2} = \left( \frac{2}{3} \right)^2 = \frac{4}{9}$.

Now, let's find the value of the other expression $\left( \frac{2}{3} \right)^{-2}$:

$\left( \frac{2}{3} \right)^{-2} = \left( \frac{3}{2} \right)^2 = \frac{9}{4}$

Since $\frac{4}{9}$ is clearly not equal to $\frac{9}{4}$, the statement correctly says they are not equal.

Solution:

The given statement is True.

According to the negative exponent law:

Applying this to $10^{-2}$:

$10^{-2} = \frac{1}{10^2} = \frac{1}{10 \times 10} = \frac{1}{100}$

Solution:

The given statement is False.

The correct expanded form of the decimal $24.58$ is:

$24.58 = 2 \times 10^1 + 4 \times 10^0 + 5 \times 10^{-1} + 8 \times 10^{-2}$

$24.58 = 2 \times 10 + 4 \times 1 + \frac{5}{10} + \frac{8}{100}$

The statement incorrectly multiplies the decimal parts (5 and 8) by 10 and 100.

Solution:

The given statement is True.

Let's check each term of the expansion:

Hundreds place: $3 \times 10^2 = 300$

Tens place: $2 \times 10^1 = 20$

Units place: $9 \times 10^0 = 9 \times 1 = 9$

Tenths place: $2 \times 10^{-1} = \frac{2}{10} = 0.2$

Hundredths place: $5 \times 10^{-2} = \frac{5}{100} = 0.05$

Sum: $300 + 20 + 9 + 0.2 + 0.05 = 329.25$.

Solution:

The given statement is False.

According to the product law of exponents $a^m \times a^n = a^{m+n}$:

$(-5)^{-2} \times (-5)^{-3} = (-5)^{-2 + (-3)} = (-5)^{-5}$

The exponent should be added $(-2 - 3 = -5)$, but the statement shows a multiplication of exponents $(-2 \times -3 = 6)$.

Solution:

The given statement is False.

First, $(-4)^{-4}$ can be simplified because the exponent is even:

$(-4)^{-4} = 4^{-4}$

Now the expression becomes:

$4^{-4} \times 4^{-1} = 4^{-4 + (-1)} = 4^{-5}$

The result is $4^{-5}$ (which is $\frac{1}{4^5}$), which is not equal to $4^5$.

Solution:

The given statement is False.

According to the product law of exponents, when multiplying powers with the same base, we add the exponents:

On the Left Hand Side (LHS):

$\left( \frac{2}{3} \right)^{-2} \times \left( \frac{2}{3} \right)^{-5} = \left( \frac{2}{3} \right)^{-2 + (-5)} = \left( \frac{2}{3} \right)^{-7}$

The resulting exponent is $-7$, whereas the statement claims it is $10$. Therefore, the statement is incorrect.

Solution:

The given statement is False.

According to the zero exponent rule, any non-zero number raised to the power of zero is always equal to $1$.

So, $5^0 = 1$, not $5$.

Solution:

The given statement is False.

According to the zero exponent property, any non-zero integer (including negative integers) raised to the power zero is $1$.

Therefore, $(-2)^0 = 1$.

Solution:

The given statement is False.

First, let's simplify the base:

$-\frac{8}{2} = -4$

Now, applying the zero exponent rule:

$(-4)^0 = 1$

Any non-zero rational number raised to the power $0$ is $1$, not $0$.

Solution:

The given statement is False.

As per the laws of exponents, for any non-zero integer $a$, $a^0 = 1$.

Substituting $a = -6$:

$(-6)^0 = 1$

The result is positive $1$, not negative $1$.

Solution:

The given statement is True.

Using the law $a^m \times a^n = a^{m+n}$:

LHS = $(-7)^{-4} \times (-7)^2 = (-7)^{-4 + 2}$

LHS = $(-7)^{-2}$

Since the Left Hand Side (LHS) equals the Right Hand Side (RHS), the statement is correct.

Solution:

The given statement is True.

According to the law of negative exponents:

Applying this to the given expression:

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

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

The value is indeed $16$.

Solution:

The given statement is False.

First, express the base $4$ as a power of $2$:

$4 = 2^2$

Now, substitute this into the expression $4^{-3}$:

$4^{-3} = (2^2)^{-3}$

Using the power of a power law $(a^m)^n = a^{mn}$:

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

The correct expression is $2^{-6}$ (which is $\frac{1}{64}$), not $2^6$ (which is $64$).

Question 81. a^p × b^q = (ab)^pq

Solution:

The given statement is False.

There is no standard law of exponents that allows combining powers with different bases and different exponents in this manner. The laws that do exist are:

1. $a^n \times b^n = (ab)^n$ (Same exponents, different bases)

2. $a^p \times a^q = a^{p+q}$ (Same bases, different exponents)

The expression $(ab)^{pq}$ would actually be equal to $a^{pq} \times b^{pq}$.

Solution:

The given statement is True.

According to the laws of exponents, we have the power of a quotient law:

Also, according to the negative exponent law for fractions, $\left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^n$. Applying this to the right-hand side of the statement:

Since both sides are equal to $\left( \frac{x}{y} \right)^m$, the statement is correct.

Solution:

The given statement is True.

According to the negative exponent identity:

By cross-multiplying, we can also write:

Substituting $x = a$ and $n = m$, we get $a^m = \frac{1}{a^{-m}}$.

Solution:

The given statement is True.

We use the law of exponents $a^n \times b^n = (ab)^n$.

Here, the base $a = -2$ and $b = \frac{5}{2}$, and the exponent $n = 4$.

$(-2)^4 \times \left( \frac{5}{2} \right)^4 = \left( -2 \times \frac{5}{2} \right)^4$

Inside the bracket, the 2 in the numerator cancels the 2 in the denominator:

$\left( -\cancel{2}^1 \times \frac{5}{\cancel{2}_1} \right)^4 = (-5)^4$

Since the exponent 4 is an even number, $(-5)^4$ is equal to $5^4$.

Solution:

The given statement is True.

To convert $0.000037$ into standard form, we move the decimal point to the right until there is one non-zero digit to its left. We move it 5 places to the right.

When the decimal point is moved to the right, the exponent of 10 is negative.

$0.000037 = 3.7 \times 10^{-5}$

Solution:

The given statement is True.

In the Indian perspective, the number 2,03,000 is "Two Lakh Three Thousand".

To write this in standard form, we move the decimal point 5 places to the left from the end of the number.

When the decimal point is moved to the left, the exponent of 10 is positive.

$203000 = 2.03 \times 10^5$

Solution:

The given statement is False.

Let's calculate the value of $2 \times 10^{-2}$:

$2 \times 10^{-2} = \frac{2}{10^2} = \frac{2}{100}$

$\frac{2}{100} = 0.02$

Since the expression is equal to $0.02$, the statement that says it is "not equal" is incorrect.

Solution:

The given statement is False.

According to the negative exponent rule $a^{-n} = \frac{1}{a^n}$:

$5^{-2} = \frac{1}{5^2}$

$5^{-2} = \frac{1}{25}$

The value is $\frac{1}{25}$ (or $0.04$), not $25$.

Solution:

The given statement is True.

Very large numbers, such as the distance between stars or the population of a country, are written in standard form $m \times 10^n$ by moving the decimal point to the left. This results in a positive integer as the exponent $n$.

Solution:

The given statement is True.

This is a standard law of exponents known as the Power of a Product Law. It states that when two different bases are raised to the same power and multiplied, their product is raised to that same power.

(i) $100^{-10}$

Solution:

Using the negative exponent law $a^{-n} = \frac{1}{a^n}$:

We can also express the base 100 as $10^2$:

$(10^2)^{-10} = 10^{2 \times (-10)} = 10^{-20}$

Hence, the value is $10^{-20}$ or $\frac{1}{100^{10}}$.

(ii) $2^{-2} \times 2^{-3}$

Solution:

Using the product law of exponents $a^m \times a^n = a^{m+n}$:

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

$2^{-5}$

Converting to a positive exponent and calculating the value:

$\frac{1}{2^5} = \frac{1}{2 \times 2 \times 2 \times 2 \times 2} = \frac{1}{32}$

Hence, the result is $\frac{1}{32}$.

(iii) $\left( \frac{1}{2} \right)^{-2} \div \left( \frac{1}{2} \right)^{-3}$

Solution:

Using the quotient law of exponents $a^m \div a^n = a^{m-n}$:

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

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

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

Hence, the result is $\frac{1}{2}$.

Solution:

First, apply the product law of exponents:

$3^{-5} \times 3^{-4} = 3^{-5 + (-4)}$

$= 3^{-9}$

To express this with a positive exponent, we use the rule $a^{-n} = \frac{1}{a^n}$:

$3^{-9} = \frac{1}{3^9}$

Hence, the expression with a positive exponent is $\frac{1}{3^9}$.

Solution:

First, we express the base 16 as a power of 2:

Now, substitute this into the original expression:

$16^{-2} = (2^4)^{-2}$

Using the power of a power law $(a^m)^n = a^{mn}$:

$2^{4 \times (-2)} = 2^{-8}$

Hence, $16^{-2}$ expressed with base 2 is $2^{-8}$.

Solution:

For $\frac{27}{64}$:

We know that $27 = 3^3$ and $64 = 4^3$.

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

For $\frac{-27}{64}$:

We know that $(-3)^3 = -27$ and $4^3 = 64$.

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

Hence, the rational powers are $\left( \frac{3}{4} \right)^3$ and $\left( -\frac{3}{4} \right)^3$ respectively.

Solution:

For $\frac{16}{81}$:

We know that $16 = 2^4$ and $81 = 3^4$.

$\frac{16}{81} = \frac{2^4}{3^4} = \left( \frac{2}{3} \right)^4$

For $\frac{-16}{81}$:

The number is negative. Any rational number raised to an even power (like 4) results in a positive value. Therefore, we can express it with an odd power or by keeping the negative sign outside:

$\frac{-16}{81} = -\left( \frac{2}{3} \right)^4$

Or, as a power with base $-\frac{16}{81}$ and exponent 1:

$\left( -\frac{16}{81} \right)^1$

Hence, $\frac{16}{81}$ is $\left( \frac{2}{3} \right)^4$.

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

Solution:

Using the power of a power law $(a^m)^n = a^{mn}$:

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

To convert this into an expression with a negative exponent, we use the reciprocal rule $\left( \frac{a}{b} \right)^n = \left( \frac{b}{a} \right)^{-n}$:

$\left( \frac{-3}{2} \right)^6 = \left( \frac{2}{-3} \right)^{-6} = \left( -\frac{2}{3} \right)^{-6}$

Hence, the result is $\left( -\frac{2}{3} \right)^{-6}$.

(b) $(2^5 \div 2^8) \times 2^{-7}$

Solution:

First, simplify the division using $a^m \div a^n = a^{m-n}$:

$2^{5-8} = 2^{-3}$

Now, multiply the result by the next term using $a^m \times a^n = a^{m+n}$:

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

To write this as a power of a rational number ($\frac{2}{1}$):

$2^{-10} = \left( \frac{2}{1} \right)^{-10}$

Hence, the result is $2^{-10}$.

Solution:

First, we find the cube of $(-2)$:

$(-2)^3 = (-2) \times (-2) \times (-2) = -8$

Next, we find the square of $(+4)$:

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

Now, we find the product of these two results:

$\text{Product} = (-8) \times 16 = -128$

Hence, the product is $-128$.

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

Solution:

Using the negative exponent identity $\left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^n$:

$\left( \frac{1}{4} \right)^{-2} = 4^2 = 16$

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

$\left( \frac{1}{3} \right)^{-2} = 3^2 = 9$

Summing the values:

$16 + 4 + 9 = 29$

Hence, the simplified value is 29.

(ii) $\left( \left( \frac{-2}{3} \right)^{-2} \right)^{3} \times \left( \frac{1}{3} \right)^{-4} \times 3^{-1} \times \frac{1}{6}$

Solution:

First, simplify the power of a power:

$\left( \left( \frac{-2}{3} \right)^{-2} \right)^3 = \left( \frac{-2}{3} \right)^{-6} = \left( -\frac{3}{2} \right)^6 = \frac{3^6}{2^6}$

Now, simplify the other terms:

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

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

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

Now multiply all parts together using laws of exponents:

$\frac{3^6}{2^6} \times 3^4 \times 3^{-1} \times (2^{-1} \times 3^{-1})$

$\text{Power of 3} = 6 + 4 - 1 - 1 = 8$

$\text{Power of 2} = -6 - 1 = -7$

$\text{Result} = 3^8 \times 2^{-7} = \frac{6561}{128}$

Hence, the simplified result is $\frac{6561}{128}$.

(iii) $\frac{49 \times z^{-3}}{7^{-3} \times 10 \times z^{-5}}$ ($z \neq 0$)

Solution:

Express 49 as a power of 7 ($49 = 7^2$):

$\frac{7^2 \times z^{-3}}{7^{-3} \times 10 \times z^{-5}}$

Using the law $\frac{a^m}{a^n} = a^{m-n}$:

$\frac{7^{2 - (-3)} \times z^{-3 - (-5)}}{10}$

$\frac{7^{2 + 3} \times z^{-3 + 5}}{10} = \frac{7^5 \times z^2}{10}$

Since $7^5 = 16807$:

$\frac{16807 z^2}{10} = 1680.7 z^2$

Hence, the simplified form is $\frac{16807 z^2}{10}$.

(iv) $(2^5 \div 2^8) \times 2^{-7}$

Solution:

First, perform the division inside the bracket using $a^m \div a^n = a^{m-n}$:

$2^{5 - 8} = 2^{-3}$

Now multiply with the next term using $a^m \times a^n = a^{m+n}$:

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

$2^{-10} = \frac{1}{2^{10}} = \frac{1}{1024}$

Hence, the simplified value is $2^{-10}$ or $\frac{1}{1024}$.

(i) $\left( \frac{5}{3} \right)^{-2} \times \left( \frac{5}{3} \right)^{-14} = \left( \frac{5}{3} \right)^{8x}$

Solution:

On the Left Hand Side (LHS), use $a^m \times a^n = a^{m+n}$:

$\left( \frac{5}{3} \right)^{-2 + (-14)} = \left( \frac{5}{3} \right)^{8x}$

$\left( \frac{5}{3} \right)^{-16} = \left( \frac{5}{3} \right)^{8x}$

Since the bases are equal, we equate the exponents:

$-16 = 8x$

$x = \frac{-16}{8}$

$x = -2$

Hence, the value of $x$ is $-2$.

(ii) (–2)³ × (–2)^–6 = (–2)^{2x – 1}

Solution:

On the LHS, use $a^m \times a^n = a^{m+n}$:

$(-2)^{3 + (-6)} = (-2)^{2x - 1}$

$(-2)^{-3} = (-2)^{2x - 1}$

Equating the exponents:

$-3 = 2x - 1$

$-3 + 1 = 2x$

$-2 = 2x$

$x = -1$

Hence, the value of $x$ is $-1$.

(iii) (2^–1 + 4^–1 + 6^–1 + 8^–1)^x = 1

Solution:

We know that for any non-zero base $a$, $a^0 = 1$.

Let's check if the base $(2^{-1} + 4^{-1} + 6^{-1} + 8^{-1})$ is non-zero:

$\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8}$ is a sum of positive fractions, so it is clearly not zero.

For $(base)^x = 1$, the exponent $x$ must be $0$.

Hence, the value of $x$ is 0.

Solution:

The number 10,00,000 is 10 Lakh in the Indian perspective.

$\text{Division} = \frac{293}{10,00,000}$

$\text{Numerical result} = 0.000293$

To convert this into standard form ($m \times 10^n$):

We move the decimal point 4 places to the right to place it after the first non-zero digit (2).

$0.000293 = 2.93 \times 10^{-4}$

Hence, the result in standard form is $2.93 \times 10^{-4}$.

Given:

Solution:

First, simplify the expression for $x$:

The exponent $1 - 4 = -3$.

According to the zero exponent rule, any non-zero number raised to the power of zero is 1.

Substituting these into equation (i):

$x = (100)^{-3} \div 1$

$x = 100^{-3}$

Now, we need to find the value of $x^{-3}$:

$x^{-3} = (100^{-3})^{-3}$

Using the power of a power law $(a^m)^n = a^{mn}$:

$x^{-3} = 100^{(-3) \times (-3)}$

$x^{-3} = 100^9$

We can also write this in base 10 as $100 = 10^2$:

$x^{-3} = (10^2)^9 = 10^{18}$

Hence, the value of $x^{-3}$ is $10^{18}$.

Solution:

According to the zero exponent property, any non-zero integer raised to the power zero is always equal to 1.

Let the required number be $y$. According to the problem:

$(-29)^0 \times y = (29)^0$

$1 \times y = 1$

$y = 1$

Hence, we should multiply by 1.

Solution:

Let the required number be $k$.

According to the problem:

$\frac{(-15)^{-1}}{k} = (-15)^{-1}$

Multiplying both sides by $k$:

$(-15)^{-1} = (-15)^{-1} \times k$

Dividing both sides by $(-15)^{-1}$:

$k = \frac{(-15)^{-1}}{(-15)^{-1}}$

$k = 1$

Hence, the number is 1.

Solution:

First, let's simplify the given expression:

$(-7)^{-2} \div (90)^{-1}$

Using the negative exponent law $a^{-n} = \frac{1}{a^n}$:

$(-7)^{-2} = \frac{1}{(-7)^2} = \frac{1}{49}$

$(90)^{-1} = \frac{1}{90}$

Now, perform the division:

$\frac{1}{49} \div \frac{1}{90} = \frac{1}{49} \times 90 = \frac{90}{49}$

The multiplicative inverse of a number is its reciprocal.

Multiplicative inverse of $\frac{90}{49} = \frac{1}{90/49} = \frac{49}{90}$

Hence, the multiplicative inverse is $\frac{49}{90}$.

Solution:

Given: $5^{3x - 1} \div 25 = 125$

Express all terms with base 5:

$25 = 5^2$

$125 = 5^3$

Substitute these into the equation:

$5^{3x - 1} \div 5^2 = 5^3$

Using the quotient law $a^m \div a^n = a^{m - n}$:

$5^{(3x - 1) - 2} = 5^3$

$5^{3x - 3} = 5^3$

Since the bases on both sides are equal, we can equate the exponents:

$3x = 3 + 3$

$3x = 6$

$x = \frac{6}{3} = 2$

Hence, the value of $x$ is 2.

Solution:

In the Indian perspective, the number 39,00,00,000 is 39 Crore.

To write a number in standard form ($m \times 10^n$), we move the decimal point so that there is only one non-zero digit to its left.

For 390,000,000, we move the decimal point 8 places to the left.

When the decimal point moves to the left, the exponent of 10 is positive.

$39,00,00,000 = 3.9 \times 10^8$

Hence, the standard form is $3.9 \times 10^8$.

Solution:

To convert 0.000005678 into standard form, we move the decimal point to the right until it is after the first non-zero digit (5).

We move the decimal point 6 places to the right.

When the decimal point moves to the right, the exponent of 10 is negative.

$0.000005678 = 5.678 \times 10^{-6}$

Hence, the standard form is $5.678 \times 10^{-6}$.

Solution:

To find the product of two numbers given in scientific notation, we multiply the numerical coefficients and the powers of $10$ separately.

$\text{Product} = (3.2 \times 4.1) \times (10^6 \times 10^{-1})$

First, calculating the numerical product:

$3.2 \times 4.1 = 13.12$

Now, applying the product law of exponents $a^m \times a^n = a^{m+n}$ for the powers of $10$:

$10^6 \times 10^{-1} = 10^{6 + (-1)} = 10^5$

Combining the two results:

$\text{Product} = 13.12 \times 10^5$

To express this in standard form ($m \times 10^n$ where $1 \le m < 10$), we must move the decimal point one place to the left:

$13.12 = 1.312 \times 10^1$

Substituting this back:

$\text{Product} = 1.312 \times 10^1 \times 10^5$

$\text{Product} = 1.312 \times 10^{1+5}$

$\text{Product} = 1.312 \times 10^6$

Hence, the product in standard form is $1.312 \times 10^6$.

Solution:

We divide the numerical parts and the exponential parts separately.

$\frac{1.5 \times 10^6}{2.5 \times 10^{-4}} = \left( \frac{1.5}{2.5} \right) \times \left( \frac{10^6}{10^{-4}} \right)$

Simplifying the numerical fraction:

$\frac{1.5}{2.5} = \frac{15}{25} = \frac{3}{5} = 0.6$

Simplifying the powers of 10 using the law $\frac{a^m}{a^n} = a^{m-n}$:

$\frac{10^6}{10^{-4}} = 10^{6 - (-4)} = 10^{6 + 4} = 10^{10}$

Combining the parts:

$0.6 \times 10^{10}$

To convert this into standard form, we move the decimal point one place to the right:

$0.6 = 6.0 \times 10^{-1}$

$6.0 \times 10^{-1} \times 10^{10} = 6.0 \times 10^{-1 + 10}$

$6 \times 10^9$

Hence, the standard form is $6 \times 10^9$.

Given:

Distance = $15,000$ km

Solution:

First, we convert the distance from kilometres to metres.

Distance in metres $= 15,000 \times 1,000$

Distance in metres $= 1,50,00,000$ m

In the Indian perspective, this distance is 1 Crore 50 Lakh metres.

Now, we convert this into scientific notation (standard form):

Move the decimal point 7 places to the left to place it after the first non-zero digit (1).

$1,50,00,000 = 1.5 \times 10^7$

Hence, the distance in metres using scientific notation is $1.5 \times 10^7$ m.

Given:

Distance of Pluto $= 5,91,30,00,000$ m

Solution:

In the Indian perspective, the number 5,91,30,00,000 is 591 Crore 30 Lakh.

To express this in standard form ($m \times 10^n$):

We need to move the decimal point from the end of the number to the position immediately after the first non-zero digit (5).

Counting the places to move the decimal to the left:

$\underbrace{5 . 9 1 3 0 0 0 0 0 0}_{9 \text{ places}}$

Since we moved the decimal 9 places to the left, the exponent of 10 is $+9$.

$5,91,30,00,000 = 5.913 \times 10^9$

Hence, the distance in standard form is $5.913 \times 10^9$ m.

Solution:

To express the number $0.00000001$ in standard form ($m \times 10^n$), we move the decimal point to the right until there is one non-zero digit to its left.

We move the decimal point 8 places to the right.

Since the decimal moves to the right, the exponent of 10 will be negative.

$0.00000001 = 1.0 \times 10^{-8}$

Hence, the weight in standard form is $1 \times 10^{-8}$ g.

Solution:

First, we convert the descriptive number into a numerical figure. We know that:

So, 3 billion 720 million can be written as:

$3,000,000,000 + 720,000,000 = 3,720,000,000$

In the Indian perspective, this is 372 Crore kilograms.

To convert this into standard form, we move the decimal point 9 places to the left.

$3,720,000,000 = 3.72 \times 10^9$

Hence, the number in standard form is $3.72 \times 10^9$ kg.

Given:

RBC count per $mm^3 = 5.5 \text{ million} = 5.5 \times 10^6$

Total blood volume = 5 litres

Conversion: $1 \text{ litre} = 1,00,000 \text{ mm}^3$ (1 Lakh $mm^3$)

Solution:

First, calculate the total volume of blood in cubic millimetres:

$\text{Volume} = 5 \times 1,00,000 = 5,00,000 \text{ mm}^3$

$\text{Volume in standard form} = 5 \times 10^5 \text{ mm}^3$

Now, calculate the total number of red blood cells:

$\text{Total RBC} = \text{Count per } mm^3 \times \text{Total Volume}$

$\text{Total RBC} = (5.5 \times 10^6) \times (5 \times 10^5)$

$\text{Total RBC} = (5.5 \times 5) \times 10^{6+5}$

$\text{Total RBC} = 27.5 \times 10^{11}$

To write this in standard form:

$27.5 = 2.75 \times 10^1$

$\text{Total RBC} = 2.75 \times 10^1 \times 10^{11} = 2.75 \times 10^{12}$

Hence, the total number of red blood cells is $2.75 \times 10^{12}$.

(a) The mass of a proton in gram is $\frac{1673}{1000000000000000000000000000}$

Counting the zeros in the denominator, there are 27 zeros.

Mass $= \frac{1673}{10^{27}} = 1673 \times 10^{-27}$

Standard form: $1.673 \times 10^3 \times 10^{-27} = 1.673 \times 10^{-24}$

Hence, the mass is $1.673 \times 10^{-24}$ g.

(b) A Helium atom has a diameter of 0.000000022 cm.

Move the decimal point 8 places to the right.

Standard form: $2.2 \times 10^{-8}$ cm.

(c) Mass of a molecule of hydrogen gas is about 0.00000000000000000000334 tons.

Counting the decimal places to move it after the first '3': 21 places to the right.

Standard form: $3.34 \times 10^{-21}$ tons.

(d) Human body has 1 trillion of cells which vary in shapes and sizes.

$1 \text{ trillion} = 1,000,000,000,000$ (1 followed by 12 zeros).

In the Indian perspective, this is 1 Lakh Crore.

Standard form: $1 \times 10^{12}$ cells.

(e) Express 56 km in m.

$56 \text{ km} = 56 \times 1000 \text{ m} = 56,000 \text{ m}$

Standard form: $5.6 \times 10^4$ m.

(f) Express 5 tons in g.

$1 \text{ ton} = 1000 \text{ kg}$ and $1 \text{ kg} = 1000 \text{ g}$.

$5 \text{ tons} = 5 \times 1000 \times 1000 \text{ g} = 5,000,000 \text{ g}$

Standard form: $5 \times 10^6$ g.

(g) Express 2 years in seconds.

$2 \text{ years} = 2 \times 365 \times 24 \times 60 \times 60 \text{ seconds}$

$2 \text{ years} = 6,30,72,000 \text{ seconds}$

Standard form: $6.3072 \times 10^7$ seconds.

(h) Express 5 hectares in cm² (1 hectare = 10,000 m²)

$1 \text{ m}^2 = 100 \text{ cm} \times 100 \text{ cm} = 10,000 \text{ cm}^2$

$5 \text{ hectares} = 5 \times 10,000 \text{ m}^2 = 50,000 \text{ m}^2$

$50,000 \text{ m}^2 = 50,000 \times 10,000 \text{ cm}^2 = 50,00,00,000 \text{ cm}^2$

Standard form: $5 \times 10^8$ cm².

Given:

$\left( \frac{2}{9} \right)^{3} \times \left( \frac{2}{9} \right)^{-6} = \left( \frac{2}{9} \right)^{2x-1}$

Solution:

On the Left Hand Side (LHS), we use the product law of exponents:

Applying this law to the LHS:

$\left( \frac{2}{9} \right)^{3 + (-6)} = \left( \frac{2}{9} \right)^{2x-1}$

$\left( \frac{2}{9} \right)^{-3} = \left( \frac{2}{9} \right)^{2x-1}$

Since the bases on both sides are the same, we can equate the exponents:

Transposing $-1$ to the LHS:

$-3 + 1 = 2x$

$-2 = 2x$

$x = \frac{-2}{2}$

$x = -1$

Hence, the value of $x$ is $-1$.

Given:

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

Quotient = $\left( \frac{4}{27} \right)^{-2}$

To Find:

The divisor.

Solution:

Let the required number (divisor) be $x$. According to the problem:

$\text{Dividend} \div \text{Divisor} = \text{Quotient}$

$\text{Divisor} = \text{Dividend} \div \text{Quotient}$

First, let's simplify the Dividend:

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

$= \frac{2^3}{(-3)^3} = \frac{8}{-27} = -\frac{8}{27}$

Next, let's simplify the Quotient:

$\left( \frac{4}{27} \right)^{-2} = \left( \frac{27}{4} \right)^2$

$= \frac{27^2}{4^2} = \frac{729}{16}$

Now, substitute these simplified values back into equation (i):

$x = -\frac{8}{27} \div \frac{729}{16}$

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

$x = -\frac{8}{27} \times \frac{16}{729}$

$x = -\frac{8 \times 16}{27 \times 729}$

$x = -\frac{128}{19683}$

Hence, the required number is $-\frac{128}{19683}$.

Solution:

Given: $\frac{6^{n}}{6^{-2}} = 6^{3}$

Using the law of exponents $\frac{a^m}{a^n} = a^{m - n}$ on the Left Hand Side (LHS):

$6^{n - (-2)} = 6^3$

$6^{n + 2} = 6^3$

Since the bases on both sides are the same, we can equate the exponents:

Transposing 2 to the Right Hand Side:

$n = 3 - 2$

$n = 1$

Hence, the value of $n$ is 1.

Solution:

Given: $\frac{2^{n} \times 2^{6}}{2^{-3}} = 2^{18}$

First, simplify the numerator using the product law $a^m \times a^n = a^{m+n}$:

$\frac{2^{n+6}}{2^{-3}} = 2^{18}$

Now, apply the quotient law $\frac{a^m}{a^n} = a^{m-n}$:

$2^{(n+6) - (-3)} = 2^{18}$

$2^{n + 6 + 3} = 2^{18}$

$2^{n + 9} = 2^{18}$

Equating the exponents since the bases are identical:

$n = 18 - 9$

$n = 9$

Hence, the value of $n$ is 9.

Solution:

Given expression: $\frac{125 \times x^{-3}}{5^{-3} \times 25 \times x^{-6}}$

First, express the numerical constants as powers of 5:

Substitute these into the expression:

$\frac{5^3 \times x^{-3}}{5^{-3} \times 5^2 \times x^{-6}}$

Simplify the denominator using the product law for the base 5:

$\frac{5^3 \times x^{-3}}{5^{-3 + 2} \times x^{-6}} = \frac{5^3 \times x^{-3}}{5^{-1} \times x^{-6}}$

Now, use the quotient law $\frac{a^m}{a^n} = a^{m-n}$ for both 5 and $x$:

$5^{3 - (-1)} \times x^{-3 - (-6)}$

$5^{3 + 1} \times x^{-3 + 6}$

$5^4 \times x^3$

Since $5^4 = 625$:

$625x^3$

Hence, the simplified form is $625x^3$.

Solution:

Given expression: $\frac{16 \times 10^{2} \times 64}{2^{4} \times 4^{2}}$

Let us express the terms in powers of 2 where possible:

$16 = 2^4$

$64 = 2^6$

$4^2 = (2^2)^2 = 2^4$

Substituting these back into the expression:

$\frac{2^4 \times 10^2 \times 2^6}{2^4 \times 2^4}$

Simplifying using exponent laws:

$\frac{2^{4+6} \times 10^2}{2^{4+4}} = \frac{2^{10} \times 10^2}{2^8}$

$2^{10-8} \times 10^2 = 2^2 \times 10^2$

$4 \times 100 = 400$

Hence, the value of the expression is 400.

Solution:

Given: $\frac{5^m \times 5^{3} \times 5^{-2}}{5^{-5}} = 5^{12}$

Simplify the numerator using the product law:

$\frac{5^{m + 3 + (-2)}}{5^{-5}} = 5^{12}$

$\frac{5^{m + 1}}{5^{-5}} = 5^{12}$

Now apply the quotient law:

$5^{(m+1) - (-5)} = 5^{12}$

$5^{m + 1 + 5} = 5^{12}$

$5^{m + 6} = 5^{12}$

Since the bases are the same, equating the exponents:

$m = 12 - 6$

$m = 6$

Hence, the value of $m$ is 6.

Solution:

Given:

Weight of newborn bear = 4 kg

Growth condition: Weight increases by the power of 2 in 5 years.

According to the problem, if the initial weight is $W$, the weight after 5 years becomes $W^2$.

Weight after 5 years $= (4)^2$ kg

Weight after 5 years $= 4 \times 4 = 16$ kg

Hence, a five year old bear might weigh 16 kg.

Solution:

The bacteria population grows exponentially according to the formula:

Where $N_0 = 1$ (starting cell) and $k$ is the number of doubling periods.

Since the cells double every 30 minutes, there are 2 doublings in 1 hour.

(a) After 12 hours:

Number of doubling periods ($k$) $= 12 \times 2 = 24$

Number of cells $= 1 \times 2^{24} = 2^{24}$

(b) After 24 hours:

Number of doubling periods ($k$) $= 24 \times 2 = 48$

Number of cells $= 1 \times 2^{48} = 2^{48}$

Hence, there will be $2^{24}$ cells after 12 hours and $2^{48}$ cells after 24 hours.

Solution:

To compare distances in standard form, we should ideally express them with the same power of 10.

Distance of Planet A $= 9.35 \times 10^6$ km

Distance of Planet B $= 6.27 \times 10^7$ km

Let's convert Planet B's distance to the power of $10^6$:

$6.27 \times 10^7 = 6.27 \times 10 \times 10^6 = 62.7 \times 10^6$ km

Now comparing the numerical coefficients of $10^6$:

Therefore, $9.35 \times 10^6$ km is a smaller distance than $6.27 \times 10^7$ km.

Hence, Planet A is nearer to Earth.

Solution:

Initially, at $t = 0$, the number of cells is 1, which can be written as $2^0$.

After 1 hour, the cells double: $1 \times 2 = 2^1$ cells.

After 2 hours, they double again: $2 \times 2 = 2^2 = 4$ cells.

Following this pattern, after $n$ hours, the number of cells will be $2^n$.

Therefore, after 8 hours, the number of cells will be $2^8$.

Hence, the number of cells after 8 hours is $2^8$ (or 256).

Solution (a): Make a table showing the insect’s location for the first 10 hops.

In each hop, the insect covers half of the remaining distance to 1. The location after $n$ hops is given by $1 - (\frac{1}{2})^n$.

Solution (b): Where will the insect be after n hops?

As observed from the pattern in the table, the location of the insect after $n$ hops will be $1 - (\frac{1}{2})^n$.

Solution (c): Will the insect ever get to 1? Explain.

No, the insect will never reach exactly 1. Mathematically, for the insect to reach 1, the term $(\frac{1}{2})^n$ must become 0. However, for any finite number of hops $n$, $(\frac{1}{2})^n$ will always be a very small positive number, not zero. The insect gets infinitely close to 1, but there is always a tiny remaining distance to cover.

Completed Table of Powers:

(a) Describe patterns you see in the ones digits of the powers.

The patterns observed in the ones digits are known as cyclicity. The cycles for different bases are:

1. Bases 1, 5, 6, 10: These have a cyclicity of 1. The ones digit never changes regardless of the power.

2. Bases 4, 9: These have a cyclicity of 2. The ones digits alternate (4 $\rightarrow$ 6 and 9 $\rightarrow$ 1).

3. Bases 2, 3, 7, 8: These have a cyclicity of 4. The sequence of ones digits repeats every four steps.

(b) Predict the ones digit in the following:

1. 4¹²: Cyclicity of 4 is (4, 6). Since 12 is an even power, the ones digit is 6.

2. 9²⁰: Cyclicity of 9 is (9, 1). Since 20 is an even power, the ones digit is 1.

3. 3¹⁷: Cyclicity of 3 is (3, 9, 7, 1). $17 \div 4$ gives remainder 1. So, ones digit is the same as 3¹, which is 3.

4. 5¹⁰⁰: Base is 5. Any power of 5 results in a ones digit of 5.

5. 10⁵⁰⁰: Base is 10. Any positive power results in a ones digit of 0.

(c) Predict the ones digit in the following:

1. 31¹⁰: Base ends in 1. Ones digit is 1.

2. 12¹⁰: Base ends in 2. Cyclicity is (2, 4, 8, 6). $10 \div 4$ gives remainder 2. The 2nd digit in cycle is 4.

3. 17²¹: Base ends in 7. Cyclicity is (7, 9, 3, 1). $21 \div 4$ gives remainder 1. The 1st digit in cycle is 7.

4. 29¹⁰: Base ends in 9. Since the power 10 is even, the ones digit 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

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.

Solution:

To compare numbers in scientific notation, we first look at the power of 10. The number with the higher exponent (closer to zero for negative exponents) is larger. If the exponents are the same, we compare the numerical coefficients.

(a) Which is the heaviest element?

The exponents are $-25, -26,$ and $-27$. Among these, $-25$ is the largest power.

Comparing elements with $10^{-25}$:

Lead: $3.44 \times 10^{-25}$ kg

Silver: $1.44 \times 10^{-25}$ kg

Since $3.44 > 1.44$, Lead is the heaviest element.

(b) Which element is lighter, Silver or Titanium?

Silver: $1.44 \times 10^{-25}$ kg

Titanium: $7.95 \times 10^{-26}$ kg

To compare, let's write them with the same exponent:

Silver: $1.44 \times 10^{-25}$ kg

Titanium: $0.795 \times 10^{-25}$ kg

Since $0.795 < 1.44$, Titanium is lighter than Silver.

(c) List all five elements in order from lightest to heaviest.

Let's convert all masses to the same exponent ($10^{-27}$) for easy comparison:

1. Hydrogen: $1.674 \times 10^{-27}$ kg

2. Lithium: $11.5 \times 10^{-27}$ kg ($1.15 \times 10^{-26}$)

3. Titanium: $79.5 \times 10^{-27}$ kg ($7.95 \times 10^{-26}$)

4. Silver: $144 \times 10^{-27}$ kg ($1.44 \times 10^{-25}$)

5. Lead: $344 \times 10^{-27}$ kg ($3.44 \times 10^{-25}$)

The order from lightest to heaviest is: Hydrogen < Lithium < Titanium < Silver < Lead.

Given:

Distance = $2,896,819,200,000$ m

Solution:

In the Indian perspective, this distance is 2 Lakh 89 Thousand 681 Crore 92 Lakh metres.

To express a number in standard form ($m \times 10^n$), we move the decimal point so that there is exactly one non-zero digit to the left of the decimal point.

For the number $2,896,819,200,000$, the decimal point is currently at the very end. We move it 12 places to the left to place it between the first 2 and the 8.

Since the decimal is moved to the left, the exponent of 10 is positive.

$2,896,819,200,000 = 2.8968192 \times 10^{12}$

Hence, the distance in standard form is $2.8968192 \times 10^{12}$ m.

To Find: The standard form of the distance $0.02543$ metres.

To express a number in standard form, we write it as $a \times 10^n$, where $1 \leq a < 10$.

The distance in standard form is $2.543 \times 10^{-2}$ metres.

To Find: Express $7.67 \times 10^{-7}$ in usual form.

Since the exponent is $-7$, we move the decimal point 7 places to the left.

The usual form is $0.000000767$.

Given:

Mass of an electron $\approx 9.1093826 \times 10^{-31}$ kilograms.

To Find: The mass in grams.

We know that:

Now, we multiply the given mass by $10^3$ to convert it to grams:

$\text{Mass in grams} = 9.1093826 \times 10^{-31} \times 10^3$

Using the law of exponents $a^m \times a^n = a^{m+n}$:

The mass of an electron is $9.1093826 \times 10^{-28}$ grams.

Given: World population $\approx 6.1 \times 10^9$.

To Find: Usual form and words (Indian perspective).

(i) Usual Form:

$6.1 \times 10^9 = 6,100,000,000$

(ii) In Words:

In the Indian numbering system, the number $6,10,00,00,000$ is expressed as:

Six hundred and ten crore.

Alternatively, in the International system, it is 6.1 billion or Six billion one hundred million.

(a) Table of ancestors:

Shikha has 2 parents (Gen 1), 4 grandparents (Gen 2), and 8 great-grandparents (Gen 3). The number of ancestors doubles with each generation.

Graph:

The graph will be an exponential curve starting at $(1, 2)$ and rising sharply to $(12, 4096)$ on a Cartesian plane where the x-axis represents the generation number and the y-axis represents the number of ancestors.

(b) Equation:

Let $A$ be the number of ancestors and $n$ be the generation number.

Given: Water flow per day = 230 billion litres.

First, express the daily flow in standard notation:

1 billion = $10^9$

Daily flow = $230 \times 10^9 = 2.3 \times 10^{11}$ litres

(i) Water flow in a week:

There are 7 days in a week.

Weekly flow = $7 \times (2.3 \times 10^{11}) = 16.1 \times 10^{11}$

(ii) Water flow in a year:

Taking a non-leap year of 365 days:

Yearly flow = $365 \times (2.3 \times 10^{11}) = 839.5 \times 10^{11}$

Evaluation:

The quantity of the substance remaining after 3 half-lives is $300 \times 2^{-3}$ grams.

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

$300 \times 2^{-3} = 300 \times \frac{1}{2^3}$

$300 \times 2^{-3} = \frac{300}{8}$

Explanation:

A half-life period means that the quantity of the substance is reduced to half ($\frac{1}{2}$) of its current amount. Mathematically, multiplying by $\frac{1}{2}$ is the same as multiplying by $2^{-1}$.

After 1 half-life, amount = $300 \times (\frac{1}{2})^1 = 300 \times 2^{-1}$

After 2 half-lives, amount = $300 \times (\frac{1}{2}) \times (\frac{1}{2}) = 300 \times (\frac{1}{2})^2 = 300 \times 2^{-2}$

Following this logic, after $n$ half-lives, the amount is multiplied by $(\frac{1}{2})$ a total of $n$ times.

Given:

The fraction of a radioactive substance remaining after $t$ half-lives is given by the expression $3^{-t}$.

(a) To Find: Fraction of substance remaining after 7 half-lives.

Here, $t = 7$. Substituting this value in the expression:

Calculating $3^7$:

$3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 2187$

(b) To Find: Number of half-lives ($t$) when the fraction is $\frac{1}{243}$.

We set the expression equal to the given fraction:

We can write $243$ as a power of $3$:

$243 = 3 \times 3 \times 3 \times 3 \times 3 = 3^5$

So, the equation becomes:

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

Comparing the exponents on both sides:

Thus, after 5 half-lives, the fraction will be $\frac{1}{243}$.

Given:

1 Fermi = $10^{-15}$ metre

Radius of a proton = $1.3$ Fermis

To Find: Radius of a proton in metres in standard form.

To convert Fermis to metres, we multiply the given radius by the value of 1 Fermi:

$\text{Radius} = 1.3 \times 10^{-15} \text{ metres}$

Standard form is $a \times 10^n$ where $1 \leq a < 10$. Since $1.3$ is already between 1 and 10, the expression is already in standard form.

The radius of a proton is $1.3 \times 10^{-15}$ metres.

Given:

Length of the paper clip = $0.05$ m

To Find: Length in standard form.

To write $0.05$ in standard form, we move the decimal point two places to the right:

The length of the paper clip in standard form is $5 \times 10^{-2}$ m.

To Prove: Verification of exponential statements.

(a) $\frac{1}{4} (2^n) = 2^{n - 2}$

L.H.S. $= \frac{1}{4} \times 2^n$

We know that $4 = 2^2$. So,

L.H.S. $= \frac{1}{2^2} \times 2^n$

Using law of exponents $\frac{a^m}{a^n} = a^{m-n}$:

L.H.S. $= 2^{n-2}$

(b) $4^{n - 1} = \frac{1}{4} (4)^n$

R.H.S. $= \frac{1}{4} \times 4^n$

R.H.S. $= \frac{4^n}{4^1}$

Using law of exponents $\frac{a^m}{a^n} = a^{m-n}$:

R.H.S. $= 4^{n-1}$

(c) $25(5^{n - 2}) = 5^n$

L.H.S. $= 25 \times 5^{n-2}$

We know that $25 = 5^2$. So,

L.H.S. $= 5^2 \times 5^{n-2}$

Using law of exponents $a^m \times a^n = a^{m+n}$:

L.H.S. $= 5^{2 + (n - 2)}$

To Find: Fill in the blank circles in the diagram.

Let the blanks be $B_1, B_2,$ and $B_3$.

Step 1: Calculate the first blank

$B_1 = 144 \times 2^{-3}$

$B_1 = 144 \times \frac{1}{2^3} = \frac{144}{8}$

Step 2: Calculate the second blank

$B_2 = B_1 \times 12^{-1}$

$B_2 = 18 \times \frac{1}{12} = \frac{18}{12}$

Step 3: Calculate the third blank

$B_3 = B_2 \times 3^{-2}$

$B_3 = \frac{3}{2} \times \frac{1}{3^2} = \frac{3}{2} \times \frac{1}{9}$

$B_3 = \frac{3}{18}$

The values for the circles are 18, 1.5, and 1/6 respectively.

Given:

Number of seconds in a day = $86,400$

To Find:

The length of a second in terms of days, expressed in scientific notation.

Solution:

To find how many days long a second is, we need to find the reciprocal of the total number of seconds in a day.

First, let us write $86,400$ in scientific notation:

Now, we calculate the value:

$\text{Length} = \frac{1}{8.64 \times 10^4}$

$\text{Length} \approx 0.000011574 \text{ days}$

To express this in scientific notation (standard form $a \times 10^n$ where $1 \leq a < 10$):

The length of a second is approximately $1.1574 \times 10^{-5}$ days.

(a) For which crop(s) did the production decrease?

Based on the "Increase/Decrease" column in the table, a negative sign indicates a decrease.

Therefore, the production decreased for Bajra, Jowar, and Rice.

(b) Write the production of all the crops in 2009 in their standard form.

We calculate the 2009 harvest by adding/subtracting the change from the 2008 harvest.

1. Bajra:

$2008 \text{ Harvest} = 1.4 \times 10^3 = 1400$

$2009 \text{ Harvest} = 1400 - 100 = 1300$

Standard form: $1.3 \times 10^3$

2. Jowar:

$2008 \text{ Harvest} = 1.7 \times 10^6 = 17,00,000$

$2009 \text{ Harvest} = 17,00,000 - 4,40,000 = 12,60,000$

Standard form: $1.26 \times 10^6$

3. Rice:

$2008 \text{ Harvest} = 3.7 \times 10^3 = 3700$

$2009 \text{ Harvest} = 3700 - 100 = 3600$

Standard form: $3.6 \times 10^3$

4. Wheat:

$2008 \text{ Harvest} = 5.1 \times 10^5 = 5,10,000$

$2009 \text{ Harvest} = 5,10,000 + 1,90,000 = 7,00,000$

Standard form: $7.0 \times 10^5$

(c) Rice production in 2015 assuming same annual decrease.

Given:

Initial production (2008) = $3700$ hectares.

Annual decrease = $100$ hectares.

Number of years from 2008 to 2015 = $2015 - 2008 = 7$ years.

To Find:

Production in 2015 in standard form.

Solution:

Total decrease over 7 years = $7 \times 100 = 700$ hectares.

Production in 2015 = $3700 - 700$

Converting to standard form:

The production in 2015 will be $3.0 \times 10^3$ hectares.

Given:

Input length of carrot = $10 \text{ cm}$

Stretching factor of the machine = $\times 4$

To Find:

The output length of the carrot.

Solution:

The stretching machine multiplies the length of the input object by the stretching factor indicated on the machine.

Substituting the given values:

The carrot will be 40 cm long when it comes out of the machine.

(a) Solution:

A $(\times 10^2)$ machine is equivalent to a $(\times 100)$ machine.

When two machines are hooked together, their stretching factors are multiplied. To find which two machines hooked together equal a $(\times 100)$ machine, we look for factors of 100.

One possible arrangement is hooking two $(\times 10)$ machines together.

Is there more than one arrangement?

Yes, there are several arrangements possible because there are many pairs of factors for 100. Other arrangements include:

1. A $(\times 2)$ machine and a $(\times 50)$ machine.

2. A $(\times 4)$ machine and a $(\times 25)$ machine.

3. A $(\times 5)$ machine and a $(\times 20)$ machine.

(b) Solution:

To find the single machine equivalent to two $(\times 2)$ machines hooked together, we multiply the factors of both machines.

Therefore, a $(\times 4)$ stretching machine does the same work as two $(\times 2)$ machines hooked together.

Given:

Initial length of the strip = $4 \text{ cm}$

Repeater machine factor = $(\times 2^4)$

To Find:

The new length of the strip after passing through the machine.

Solution:

The repeater machine $(\times 2^4)$ applies the base stretch of 2, four times consecutively. This means the total stretching factor is $2^4$.

Now, we calculate the final length of the strip by multiplying the input length with the total stretch factor:

The new length of the strip will be 64 cm.

To Find:

Number of times the base machine is applied and the total stretch for machines (a), (b), and (c).

Solution:

In a repeater machine $(\times \text{base}^{\text{power}})$, the base represents the multiplier of the single machine, and the power (exponent) represents how many times that base machine is applied.

For machine (a), the base 100 is applied 2 times for a total stretch of 10,000.

For machine (b), the base 7 is applied 5 times for a total stretch of 16,807.

For machine (c), the base 5 is applied 7 times for a total stretch of 78,125.

To Find:

Three repeater machines equivalent to a $(\times 64)$ machine.

Solution:

To find equivalent repeater machines, we need to express $64$ as a power of different bases (where the base and exponent are integers).

1. Using Base 8:

Since $8 \times 8 = 64$, we can write this as $8^2$.

2. Using Base 4:

Since $4 \times 4 \times 4 = 64$, we can write this as $4^3$.

3. Using Base 2:

Since $2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$, we can write this as $2^6$.

Thus, the three repeater machines are $(\times 8^2)$, $(\times 4^3)$, and $(\times 2^6)$.

Given:

Initial length of the piece of chalk = $2 \text{ cm}$

Repeater machine factor = $(\times 1^{100})$

To Find:

The length of the chalk after it passes through the machine.

Solution:

The repeater machine $(\times 1^{100})$ applies the base stretch of 1, a total of 100 times. We know that the number 1 raised to any power remains 1.

To find the output length, we multiply the input length by the total stretching factor:

Therefore, the machine will not change the length of the chalk. The output remains 2 cm.

(a) Solution:

In a repeater machine, the exponent tells us how many times the stretching factor is applied. If the exponent is $0$, it means the stretching action is applied zero times.

If you do not apply the stretching machine at all, the item remains exactly as it was. Thus, these machines do nothing to the piece of chalk; the piece of chalk stays the same length as its input length.

(b) Solution:

Since applying a repeater machine zero times leaves the object's length unchanged, it is mathematically equivalent to multiplying the length by 1.

Following this logic for any base $a$ (where $a \neq 0$), the value of the base raised to the power of 0 must be 1.

In general, according to the laws of exponents:

Therefore, the value of $6^0$ is 1.

Given:

Input length of the sandwich = $9 \text{ cm}$

Machine base stretching factor (from image) = $\times 3$

Exponent (number of times applied) = $-1$

To Find:

The final length of the sandwich after it emerges from the machine.

Solution:

The repeater machine follows the rule where the total stretching factor is $(\text{base})^{\text{exponent}}$. In this case, the total factor is $3^{-1}$.

Using the law of exponents:

Therefore, the stretching factor is:

To find the output length, we multiply the input length by the total stretching factor:

The sandwich will be 3 cm long when it emerges from the machine.

Given:

Initial length of the worms = $1 \text{ cm}$

Solution for Hook-up (i):

The first machine is a $(\times 2)$ machine and the second machine is a $(\times 2^{-1})$ machine.

When machines are hooked together, their total effect is the product of their factors:

Using the law of exponents $a^m \times a^n = a^{m+n}$:

Now, calculate the final length:

For hook-up (i), the worm remains 1 cm long.

Solution for Hook-up (ii):

The first machine is a $(\times 2^{-1})$ machine and the second machine is a $(\times 2^{-2})$ machine.

Evaluating the factor:

Now, calculate the final length:

For hook-up (ii), the worm shrinks to 0.125 cm.

Given:

Initial length of the gum = $1 \text{ cm}$

Machine stretching factor = $3^{-2}$

To Find:

The final length of the stick when it emerges from the machine.

Solution:

The length of the object coming out of a repeater machine is the product of its initial length and the stretching factor.

The stretching factor is $3^{-2}$. We can rewrite this using the law of exponents:

Now, we calculate the final length:

In decimal form, this is approximately $0.11 \text{ cm}$.

Given:

Initial length = $1 \text{ cm}$

Final length = $\frac{1}{1,00,000} \text{ cm}$

Base machine factor = $\frac{1}{10}$

To Find:

The missing value (exponent), let it be $n$.

Solution:

The machine operates by raising the base to the power of the exponent. The relationship is:

We can express $1,00,000$ as a power of 10:

So, the equation becomes:

By comparing the exponents on both sides:

The missing value in the machine is 5.

To Find:

A single equivalent machine factor for each hook-up.

Solution:

When machines are hooked together, the total stretching factor is found by multiplying the individual factors of each machine.

(a) $(\times 2^3)$ machine followed by $(\times 2^{-2})$ machine:

Single machine: $(\times 2)$.

(b) $(\times 2^4)$ machine followed by $(\times (\frac{1}{2})^2)$ machine:

First, express $(\frac{1}{2})^2$ as a power of 2:

Single machine: $(\times 4)$ or $(\times 2^2)$.

(c) $(\times 5^{99})$ machine followed by a $(5^{-100})$ machine:

Single machine: $(\times 5^{-1})$ or $(\times 1/5)$.

To Find:

A single equivalent repeater machine factor for the given hook-ups (a) through (f).

Solution:

When repeater machines are hooked together, the total stretching factor is the product of the stretching factors of the individual machines. We use the law of exponents: $a^m \times a^n = a^{m+n}$ and $a^m \times b^m = (ab)^m$.

(a) Hook-up of $(\times 2^2)$, $(\times 2^3)$, and $(\times 2^4)$:

Single machine: $(\times 2^9)$.

(b) Hook-up of $(\times 100^2)$ and $(\times 100^{10})$:

Single machine: $(\times 100^{12})$.

(c) Hook-up of $(\times 7^{10})$, $(\times 7^{50})$, and $(\times 7^1)$:

Single machine: $(\times 7^{61})$.

(d) Hook-up of $(\times 3^y)$ and $(\times 3^y)$:

Single machine: $(\times 3^{2y})$.

(e) Hook-up of $(\times 2^2)$, $(\times (\frac{1}{2})^3)$, and $(\times 2^4)$:

First, rewrite $(\frac{1}{2})^3$ as $2^{-3}$:

Single machine: $(\times 2^3)$ or $(\times 8)$.

(f) Hook-up of $(\times (\frac{1}{2})^2)$ and $(\times (\frac{1}{3})^2)$:

Using the law $a^m \times b^m = (ab)^m$:

Single machine: $(\times (\frac{1}{6})^2)$ or $(\times \frac{1}{36})$.

To Find:

A single equivalent repeater machine for hook-ups (a) to (e).

Solution:

A single repeater machine $(\times \text{base}^{\text{exponent}})$ is possible if the combined stretching factor of the hook-up can be expressed as a single base raised to an integer power.

(a) Hook-up of $(\times 7^3)$ and $(\times 7^2)$:

Since the bases are the same, we add the exponents:

Yes, a single repeater machine $(\times 7^5)$ will do the same work.

(b) Hook-up of $(\times 2^3)$ and $(\times 3^2)$:

The total factor is the product of the two machines:

Since 72 cannot be written as a single integer base raised to a power greater than 1, there is no single repeater machine with a common base for this hook-up.

(c) Hook-up of $(\times 2^2)$, $(\times (\frac{1}{3})^3)$, and $(\times 5^4)$:

The total factor is:

As the bases 2, 3, and 5 are different primes, the result cannot be expressed as a single base raised to a power. Thus, no single repeater machine exists for this hook-up.

(d) Hook-up of $(\times 0.5^2)$ and $(\times 0.5^3)$:

Since the bases are the same, we add the exponents:

Yes, a single repeater machine $(\times 0.5^5)$ will do the same work.

(e) Hook-up of $(\times 12^2)$ and $(\times 12^3)$:

Since the bases are the same, we add the exponents:

Yes, a single repeater machine $(\times 12^5)$ will do the same work.

Given:

First machine stretching factor = $2^3$

Second machine stretching factor = $3^3$

To Find:

A single repeater machine equivalent to the hook-up of these two machines.

Solution:

When two machines are hooked together, the total stretching factor is the product of the individual factors.

According to the law of exponents, if the exponents are the same but the bases are different, we can multiply the bases:

Applying this rule to the given problem:

Therefore, the single repeater machine Shikha should use is $(\times 6^3)$.

To Find:

A hook-up of two repeater machines equivalent to $(\times 25^2)$.

Solution:

We need a total stretch of $25^2$. Since the $(\times 25)$ machine is broken, we should use a different base to achieve the same result. We know that:

Therefore, the total required stretch can be expressed with base 5:

Using the law $(a^m)^n = a^{m \times n}$:

To achieve a total stretch of $5^4$ using a hook-up of two machines, we can divide the exponent 4 into two parts (such as $2 + 2$ or $1 + 3$).

One possible arrangement is using two $(\times 5^2)$ machines hooked together:

Alternatively, if base 25 is strictly unavailable, she can use machines with base 5. Another hook-up could be $(\times 5^1)$ followed by $(\times 5^3)$.

(a) Solution:

Input length = $5 \text{ cm}$, Output length = $5 \text{ cm}$.

Since the length remains unchanged, the stretching factor is 1.

In terms of exponents, any non-zero base raised to the power of 0 is 1. The missing machine is $(\times 1)$ or any $(\times a^0)$ machine.

(b) Solution:

Input length = $3 \text{ cm}$, Output length = $15 \text{ cm}$.

The missing machine information is $(\times 5)$.

(c) Solution:

Input length = $1.25 \text{ cm}$, Machine factor = $(\times 4)$.

The missing output value is 5 cm.

(d) Solution:

Two machines are hooked together: $(\times 4)$ and $(\times 3)$. Final Output = $36 \text{ cm}$.

First, find the total stretching factor:

Now, find the input length:

The missing input value is 3 cm.

To Find:

A hook-up of prime base machines for each given stretching factor.

Solution:

To find a hook-up of prime base machines, we perform prime factorisation of the given numbers.

(a) Stretching machine $(\times 100)$:

Prime factorisation of $100$:

The hook-up can be $(\times 2^2)$ machine followed by $(\times 5^2)$ machine.

(b) Stretching machine $(\times 99)$:

Prime factorisation of $99$:

The hook-up can be $(\times 3^2)$ machine followed by $(\times 11^1)$ machine.

(c) Stretching machine $(\times 37)$:

The number $37$ is already a prime number. Since we are not allowed to use $(\times 1)$ machines, it is not possible to create a hook-up of multiple machines to achieve exactly 37.

(d) Stretching machine $(\times 1,111)$:

Prime factorisation of $1,111$:

Both 11 and 101 are prime numbers. The hook-up can be $(\times 11^1)$ machine followed by $(\times 101^1)$ machine.

To Find:

Two repeater machines that equivalent to a $(\times 81)$ machine when hooked together.

Solution:

We need to find two factors of $81$ that can be expressed as powers of the same or different bases. We know that:

Since $9 = 3^2$, we can also write:

Possible hook-ups of two repeater machines are:

1. Two $(\times 9^1)$ machines.

2. Two $(\times 3^2)$ machines.

3. A $(\times 3^1)$ machine followed by a $(\times 3^3)$ machine.

To Find:

A repeater machine equivalent to $(\times \frac{1}{8})$.

Solution:

A repeater machine is usually expressed in the form $(\times \text{base}^{\text{exponent}})$. We can rewrite $\frac{1}{8}$ as a power of 2 or $\frac{1}{2}$.

We know that $8 = 2^3$. Using the law of exponents $a^{-n} = \frac{1}{a^n}$:

Alternatively, using base $\frac{1}{2}$:

The repeater machine can be described as $(\times 2^{-3})$ or $(\times (\frac{1}{2})^3)$.

To Find:

Three machines that are equivalent to hook-ups of $(\times 5)$ machines.

Solution:

Hook-ups of $(\times 5)$ machines mean we are using base 5 with different positive integer exponents. We can choose any three powers of 5.

1. First Machine:

A hook-up of two $(\times 5)$ machines:

This can be replaced by a $(\times 25)$ machine.

2. Second Machine:

A hook-up of three $(\times 5)$ machines:

This can be replaced by a $(\times 125)$ machine.

3. Third Machine:

A hook-up of four $(\times 5)$ machines:

This can be replaced by a $(\times 625)$ machine.

Solution:

To complete the chart, we use the relationship: $\text{Input Length} \times \text{Machine Factor} = \text{Output Length}$.

Step 1: Finding Machine Factors and Inputs

From the second row: $\text{Input} \times 2 = 1$, so the first Input Length is $0.5$.

Using this input for the second column: $0.5 \times \text{Machine 2} = 5$, so Machine 2 is $\times 10$.

From the third row: $3 \times \text{Machine 4} = 15$, so Machine 4 is $\times 5$.

From the fourth row: $\text{Input} \times 2 = 14$, so the third Input Length is $7$.

Using this input for the third column: $7 \times \text{Machine 3} = 7$, so Machine 3 is $\times 1$.

Completed Chart:

Solution:

Step 1: Finding Missing Values

From the second row: $\text{Input} \times 2^3 = 40 \implies \text{Input} \times 8 = 40 \implies \text{Input} = 5$.

Using Input 5 for the third column: $5 \times \text{Machine 3} = 125 \implies \text{Machine 3} = 25$, which is $5^2$.

From the fourth row: The output is 162 in the second machine column. If we assume the input is 2 (from the third row), then $2 \times \text{Machine 2} = 162 \implies \text{Machine 2} = 81$, which is $3^4$.

Step 2: Calculating Remaining Outputs

For Input 5, Machine 2: $5 \times 3^4 = 5 \times 81 = 405$.

For Input 2, Machine 1: $2 \times 2^3 = 2 \times 8 = 16$.

For Input 2, Machine 3: $2 \times 5^2 = 2 \times 25 = 50$.

Completed Chart:

Note: In the given table, the output 162 was listed in a separate row from input 2, but based on the common factors, it corresponds to input 2 passing through a $3^4$ machine.

Given:

Total squares on a chessboard = 64

Amount on the 1st square = $\textsf{₹} 1$

The amount doubles on each subsequent square.

To Find:

1. The total amount of money earned by the farmer.

2. The first square on which the king will place at least $\textsf{₹} 10 \text{ lakh}$.

Solution:

Step 1: Observe the pattern of money on each square

Let us express the amount on each square as a power of 2:

1st square = $1 = 2^0$

2nd square = $2 = 2^1$

3rd square = $4 = 2^2$

4th square = $8 = 2^3$

Following this pattern, the amount on the $n^{th}$ square is $2^{n-1}$.

Step 2: Calculate the total money earned

The total money is the sum of the amounts on all 64 squares:

$\text{Total Amount} = 2^0 + 2^1 + 2^2 + 2^3 + ... + 2^{63}$

In mathematics, the sum of such a doubling sequence is always one less than the next power of 2.

The farmer earned $\textsf{₹} (2^{64} - 1)$. This is an incredibly large value (more than 18 quintillion rupees).

Step 3: Find the square for at least $\textsf{₹} 10 \text{ lakh}$

In the Indian numbering system, $10 \text{ lakh} = 10,00,000$.

We need to find the smallest power of 2 that is greater than or equal to $10,00,000$.

We know that:

Now, let us calculate $2^{20}$:

Since $10,48,576$ is just over $10,00,000$, we can see that $2^{20}$ is the required power.

As we found in Step 1, the amount on the $n^{th}$ square is $2^{n-1}$. Therefore:

The 21st square is the first square on which the king will place at least $\textsf{₹} 10 \text{ lakh}$.

Given:

Diameter of the Sun ($D_s$) = $1.4 \times 10^9 \text{ m}$

Diameter of the Earth ($D_e$) = $1.2756 \times 10^7 \text{ m}$

Solution:

To compare the diameters, we divide $D_s$ by $D_e$:

Using law of exponents $\frac{a^m}{a^n} = a^{m-n}$:

The diameter of the Sun is approximately 110 times the diameter of the Earth.

Given:

Mass of Mars = $6.42 \times 10^{29} \text{ kg}$

Mass of the Sun = $1.99 \times 10^{30} \text{ kg}$

Solution:

To find the total mass, we must express both masses with the same exponent of 10.

Total Mass = Mass of Sun + Mass of Mars

The total mass is $2.632 \times 10^{30} \text{ kg}$.

Given:

Distance (Sun to Earth) = $1.496 \times 10^8 \text{ km} = 14,96,00,000 \text{ km}$

Distance (Earth to Moon) = $3.84 \times 10^8 \text{ m} = 3,84,000 \text{ km}$

Solution:

During a solar eclipse, the Moon is between the Sun and the Earth.

Distance (Moon to Sun) = Distance (Sun to Earth) - Distance (Earth to Moon)

First, we convert both to standard notation in km:

Subtracting the values:

The distance between the Moon and the Sun is $1.49216 \times 10^8 \text{ km}$.

Given:

Distance ($d$) = $8.1 \times 10^{13} \text{ km}$

Speed of light ($v$) = $3 \times 10^8 \text{ m/s}$

Solution:

First, convert distance into metres ($1 \text{ km} = 10^3 \text{ m}$):

We use the formula $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$:

Light takes $2.7 \times 10^8 \text{ seconds}$ to reach the Earth.

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, we have:

Using the law of exponents $a^{-1} = \frac{1}{a}$:

By cross-multiplying:

Therefore, $(-15)^{-1}$ should be divided by $1/3$.

To Find:

The number by which $(-8)^{-3}$ should be multiplied to get $(-6)^{-3}$.

Solution:

Let the required number be $x$.

According to the problem:

Using the law of exponents $a^{-n} = \frac{1}{a^n}$:

Using the law $\frac{a^n}{b^n} = (\frac{a}{b})^n$:

Therefore, the number is $64/27$.

Solution (1):

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

Using law $a^m \div a^n = a^{m-n}$:

Since $(-\frac{1}{7}) = (-7)^{-1}$:

Comparing exponents: $x = -2$.

Solution (2):

$\left( \frac{2}{5} \right)^{2x+6} \times \left( \frac{2}{5} \right)^{3} = \left( \frac{2}{5} \right)^{x+2}$

Using law $a^m \times a^n = a^{m+n}$:

Solution (3):

$2^x + 2^x + 2^x = 192$

Since $64 = 2^6$:

Therefore, $x = 6$.

Solution (4):

$\left( \frac{-6}{7} \right)^{x-7} = 1$

We know that $a^0 = 1$ for any non-zero $a$. So:

Therefore, $x = 7$.

Solution (5):

$2^{3x} = 8^{2x+1}$

Expressing 8 as a power of 2:

Equating exponents:

Solution (6):

$5^x + 5^{x-1} = 750$

Taking $5^x$ as common:

Since $625 = 5^4$:

Therefore, $x = 4$.

Given:

$a = -1$

$b = 2$

To Find:

The numerical value of the given exponential expressions.

Solution:

(1) Value of $a^b + b^a$:

Substituting the values of $a$ and $b$:

$(-1)^2 + (2)^{-1}$

The value is 1.5 or $3/2$.

(2) Value of $a^b - b^a$:

Substituting the values of $a$ and $b$:

$(-1)^2 - (2)^{-1}$

The value is 0.5 or $1/2$.

(3) Value of $a^b \times b^2$:

Substituting the values of $a$ and $b$:

$(-1)^2 \times (2)^2$

The value is 4.

(4) Value of $a^b \div b^a$:

Substituting the values of $a$ and $b$:

$(-1)^2 \div (2)^{-1}$

The value is 2.

To Find:

The exponential form of the given rational numbers.

Solution:

(1) Exponential form of $\frac{-1296}{14641}$

Let us find the prime factors of 1296 and 14641:

For 1296:

So, $1296 = 2^4 \times 3^4 = (2 \times 3)^4 = 6^4$.

For 14641:

So, $14641 = 11^4$.

(2) Exponential form of $\frac{-125}{343}$

Factorizing numerator and denominator:

$125 = 5 \times 5 \times 5 = 5^3$

$343 = 7 \times 7 \times 7 = 7^3$

(3) Exponential form of $\frac{400}{3969}$

Factorizing numerator and denominator:

$400 = 20 \times 20 = 20^2$

$3969 = 63 \times 63 = 63^2$

(4) Exponential form of $\frac{-625}{10000}$

Simplifying the fraction first:

Now, $16 = 2^4$ and $1 = 1^4$.

Solution (1):

Expression: $\left[ \left( \frac{1}{2} \right)^{2} - \left( \frac{1}{4} \right)^{3} \right]^{-1} \times 2^{-3}$

First, we evaluate the terms inside the square brackets:

Subtracting these fractions:

Now, applying the exponent $-1$ to the result:

Finally, multiplying by $2^{-3}$:

The simplified value is $8/15$.

Solution (2):

Expression: $\left[ \left( \frac{4}{3} \right)^{-2} - \left( \frac{3}{4} \right)^{2} \right]^{(-2)}$

First, we simplify the terms inside the square brackets:

The expression inside the bracket becomes:

Now, applying the outer exponent:

Since division by zero is not possible, the value of this expression is undefined.

Solution (3):

Expression: $\left( \frac{4}{13} \right)^{4} \times \left( \frac{13}{7} \right)^{2} \times \left( \frac{7}{4} \right)^{3}$

We expand the powers:

Regrouping the terms with same bases:

Using the law $a^m \div a^n = a^{m-n}$:

The simplified value is $28/169$.

Solution (4):

Expression: $\left( \frac{1}{5} \right)^{45} \times \left( \frac{1}{5} \right)^{-60} - \left( \frac{1}{5} \right)^{28} \times \left( \frac{1}{5} \right)^{-43}$

Using the law $a^m \times a^n = a^{m+n}$ for each part:

Subtracting the two parts:

The result is 0.

Solution (5):

Expression: $\frac{(9)^3 \times 27 \times t^4}{(3)^{-2} \times (3)^4 \times t^2}$

Convert everything to base 3 ($9 = 3^2$ and $27 = 3^3$):

Now, divide the numerator by the denominator:

The result is $2187 t^2$.

Solution (6):

Expression: $\frac{(3^{-2})^{2} \times (5^{2})^{-3} \times (t^{-3})^{2}}{(3^{-2})^{5} \times (5^{3})^{-2} \times (t^{-4})^{3}}$

Apply the power of a power law $(a^m)^n = a^{mn}$ to simplify:

Using the division law $a^m \div a^n = a^{m-n}$:

Since $5^0 = 1$ and $3^6 = 729$:

The result is $729 t^6$.