Chapter 11 Exponents & Powers (Class 7 - Maths NCERT Exemplar Solutions)

Given Number: $\frac{-8}{27}$

Solution:

Let us evaluate each option one by one:

(a) $-\left( \frac{2}{3} \right)^3 = -\left( \frac{2 \times 2 \times 2}{3 \times 3 \times 3} \right) = -\frac{8}{27}$ (Equal to given number)

(b) $\left( \frac{-2}{3} \right)^3 = \frac{(-2) \times (-2) \times (-2)}{3 \times 3 \times 3} = \frac{-8}{27}$ (Equal to given number)

(c) $-\left( \frac{-2}{3} \right)^3 = -\left( \frac{-8}{27} \right) = \frac{8}{27}$ (Not equal to given number)

(d) $\left( \frac{-2}{3} \right) \times \left( \frac{-2}{3} \right) \times \left( \frac{-2}{3} \right) = \frac{-8}{27}$ (Equal to given number)

The calculation for option (c) results in a positive value $\frac{8}{27}$, whereas the original number is negative.

Hence, the correct option is (c).

Solution:

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

Given expression: $(-7)^5 \times (-7)^3$

Here, base $a = -7$, $m = 5$, and $n = 3$.

Applying the law:

$(-7)^5 \times (-7)^3 = (-7)^{5+3}$

$= (-7)^8$

Hence, the correct option is (a).

Solution:

According to the laws of exponents, for any non-zero integers $a$ and $b$ and whole number $m$:

Given expression: $x^3 \div y^3$

Here, $a = x$, $b = y$, and $m = 3$.

Applying the law:

$x^3 \div y^3 = \left( \frac{x}{y} \right)^3$

Hence, the correct option is (b).

Solution:

First, we simplify the expression inside the parentheses using the law $a^m \div a^n = a^{m-n}$:

Now, substitute this result into the original expression:

$(5^1)^2$

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

$= 5^{1 \times 2}$

$= 5^2 = 25$

Therefore, $(5^7 \div 5^6)^2 = 25$ (or $5^2$).

Solution:

We can simplify the expression by applying the law $\frac{a^m}{a^n} = a^{m-n}$ to each variable separately:

For variable $a$:

$\frac{a^7}{a^5} = a^{7-5} = a^2$

For variable $b$ (note that $b = b^1$):

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

Multiplying these results together:

$\frac{a^7b^3}{a^5b} = a^2b^2$

Therefore, the answer is $a^2b^2$ or $(ab)^2$.

Statement: In the number $7^5$, 5 is the base and 7 is the exponent.

Solution:

The given statement is False.

In power notation $a^n$, $a$ is known as the base and $n$ is known as the exponent (or index).

In $7^5$:

The base is $7$.

The exponent is $5$.

Statement: $\frac{a^4}{b^3} = \frac{a + a + a + a}{b + b + b}$

Solution:

The given statement is False.

Exponents represent repeated multiplication, not repeated addition.

1. $a^4 = a \times a \times a \times a$, whereas $a + a + a + a = 4a$.

2. $b^3 = b \times b \times b$, whereas $b + b + b = 3b$.

Therefore, $\frac{a^4}{b^3}$ is not equal to $\frac{4a}{3b}$.

Statement: $a^b > b^a$ is true, if $a = 3$ and $b = 4$; but false, if $a = 2$ and $b = 3$.

Solution:

The given statement is True.

Case 1: When $a = 3$ and $b = 4$

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

$b^a = 4^3 = 4 \times 4 \times 4 = 64$

Since $81 > 64$, the condition $a^b > b^a$ is true.

Case 2: When $a = 2$ and $b = 3$

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

$b^a = 3^2 = 3 \times 3 = 9$

Since $8$ is not greater than $9$, the condition $a^b > b^a$ is false.

Since both parts of the statement correctly describe the results of the evaluation, the overall statement is true.

To Find:

The number which when multiplied by $3^3$ gives $3^7$.

Solution:

Let the required number be $x$.

According to the question:

$x \times 3^3 = 3^7$

$x = \frac{3^7}{3^3}$

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

$x = 3^{7-3}$

$x = 3^4$

$x = 3 \times 3 \times 3 \times 3 = 81$

Therefore, we should multiply $3^3$ by $3^4$ (or $81$) to get $3^7$.

Given Equation:

$\left( \frac{5}{3} \right)^5 \times \left( \frac{5}{3} \right)^{11} = \left( \frac{5}{3} \right)^{8x}$

Solution:

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

$\left( \frac{5}{3} \right)^{5 + 11} = \left( \frac{5}{3} \right)^{8x}$

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

Since the bases are equal on both sides, their exponents must also be equal:

Thus, the value of $x$ is $2$.

To Find:

The exponential notation of $648$.

Solution:

First, we find the prime factorisation of $648$:

$\begin{array}{c|cc} 2 & 648 \\ \hline 2 & 324 \\ \hline 2 & 162 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

Now, we write $648$ as a product of its prime factors:

$648 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$

$648 = 2^3 \times 3^4$

Therefore, the exponential notation of $648$ is $2^3 \times 3^4$.

To Find:

Standard form of $2,36,00,000$.

Solution:

A number is said to be in standard form if it is expressed as $k \times 10^n$, where $1 \leq k < 10$ and $n$ is an integer.

Given number: $2,36,00,000$

To make the first part between $1$ and $10$, we place the decimal point after the first digit '$2$':

$2.3600000$

We moved the decimal point $7$ places to the left. Therefore, we multiply by $10^7$.

$2,36,00,000 = 2.36 \times 10^7$

The standard form is $2.36 \times 10^7$.

To Compare:

$3^{12}$ and $6^6$.

Solution:

To compare these two numbers, we can try to make their exponents the same.

We can rewrite $3^{12}$ as follows:

$3^{12} = 3^{(2 \times 6)}$

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

$3^{12} = (3^2)^6$

$3^{12} = 9^6$

Now we compare $9^6$ and $6^6$. Since the exponents are the same ($6$), we compare the bases.

Therefore:

Thus, $3^{12}$ is larger than $6^6$.

Given Equation:

$\left( \frac{1}{5} \right)^5 \times \left( \frac{1}{5} \right)^{19} = \left( \frac{1}{5} \right)^{8x}$

Solution:

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

$\left( \frac{1}{5} \right)^{5 + 19} = \left( \frac{1}{5} \right)^{8x}$

$\left( \frac{1}{5} \right)^{24} = \left( \frac{1}{5} \right)^{8x}$

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

Therefore, the value of $x$ is $3$.

Given:

The exponential expression is $[(-3)^2]^3$.

Solution:

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

Here, $a = -3$, $m = 2$, and $n = 3$.

Applying the law:

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

$[(-3)^2]^3 = (-3)^6$

Therefore, the correct option is (b).

Given:

The expression is $x^8 \div x^2$, where $x$ is a non-zero rational number.

Solution:

According to the laws of exponents, for any non-zero integer $a$ and whole numbers $m$ and $n$ (where $m > n$):

Applying this law to the given expression:

$x^8 \div x^2 = x^{8 - 2}$

$x^8 \div x^2 = x^6$

Therefore, the correct option is (b).

Given:

1. Square of $x = x^2$

2. Cube of $x = x^3$

Solution:

We need to find the product of these two terms:

$\text{Product} = x^2 \times x^3$

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

$\text{Product} = x^{2 + 3}$

$\text{Product} = x^5$

This is the fifth power of x.

Therefore, the correct option is (c).

Given:

The expression is $x^5 \div y^5$.

Solution:

According to the laws of exponents, if two different bases have the same exponent, their quotient is the quotient of the bases raised to that same exponent:

Applying this law to the given expression:

$x^5 \div y^5 = (x \div y)^5$

Therefore, the correct option is (c).

Solution:

This is a fundamental law of exponents. When multiplying two powers with the same base, we keep the base and add the exponents.

Therefore, the correct option is (c).

Solution:

According to the law of exponents, any non-zero number raised to the power of zero is $1$.

Applying this to each term in the expression:

$1^0 = 1$

$2^0 = 1$

$3^0 = 1$

Now, finding the sum:

$\text{Sum} = 1 + 1 + 1$

$\text{Sum} = 3$

Therefore, the correct option is (c).

Given:

The expression is $\frac{10^{22} + 10^{20}}{10^{20}}$.

Solution:

We can simplify the expression by splitting the fraction:

$\frac{10^{22} + 10^{20}}{10^{20}} = \frac{10^{22}}{10^{20}} + \frac{10^{20}}{10^{20}}$

Using the law $\frac{a^m}{a^n} = a^{m - n}$ for the first part and the fact that $\frac{x}{x} = 1$ for the second part:

$= 10^{22 - 20} + 1$

$= 10^2 + 1$

$= 100 + 1$

$= 101$

Therefore, the correct option is (c).

Given Number: $12345$

Solution:

A number is in standard form (or scientific notation) when it is written as $k \times 10^n$, where $1 \leq k < 10$.

To convert $12345$ into standard form, we move the decimal point from its current position (at the end of the number) to after the first non-zero digit ($1$).

Starting with $12345.0$:

Move decimal 1 place left: $1234.5 \times 10^1$

Move decimal 2 places left: $123.45 \times 10^2$

Move decimal 3 places left: $12.345 \times 10^3$

Move decimal 4 places left: $1.2345 \times 10^4$

Now, $1.2345$ lies between $1$ and $10$.

Therefore, the correct option is (d).

Given:

To Find:

The value of $K$.

Solution:

We can rewrite the terms on the Left Hand Side (LHS) by expressing each power of 2 in terms of $2^{1995}$ using the law $a^m = a^{m-n} \cdot a^n$:

$2^{1998} = 2^3 \cdot 2^{1995}$

$2^{1997} = 2^2 \cdot 2^{1995}$

$2^{1996} = 2^1 \cdot 2^{1995}$

Now, substitute these into the original equation:

$(2^3 \cdot 2^{1995}) - (2^2 \cdot 2^{1995}) - (2^1 \cdot 2^{1995}) + (1 \cdot 2^{1995}) = K \cdot 2^{1995}$

Taking $2^{1995}$ as a common factor on the LHS:

$2^{1995} (2^3 - 2^2 - 2^1 + 1) = K \cdot 2^{1995}$

Simplifying the expression inside the bracket:

$2^{1995} (8 - 4 - 2 + 1) = K \cdot 2^{1995}$

$2^{1995} (3) = K \cdot 2^{1995}$

Comparing both sides, we get:

Hence, the correct option is (c).

Solution:

We know from the laws of exponents that for any non-zero integer $a$, $a^0 = 1$. Let us evaluate each option:

(a) $2^0 + 3^0 + 4^0 = 1 + 1 + 1 = 3$

(b) $2^0 \times 3^0 \times 4^0 = 1 \times 1 \times 1 = 1$

(c) $(3^0 - 2^0) \times 4^0 = (1 - 1) \times 1 = 0 \times 1 = 0$

(d) $(3^0 - 2^0) \times (3^0 + 2^0) = (1 - 1) \times (1 + 1) = 0 \times 2 = 0$

Only option (b) results in the value 1.

Hence, the correct option is (b).

Given Number: $72105.4$

Solution:

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

Original position: $72105.4$

To get $7.21054$, we move the decimal point 4 places to the left.

Since the decimal is moved 4 places to the left, the exponent $n$ of 10 will be positive 4.

Hence, the correct option is (c).

Solution:

The square of a number means raising that number to the power of 2.

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

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

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

$= \frac{4}{9}$

Note: The square of any negative rational number is always positive.

Hence, the correct option is (d).

Solution:

The cube of a number means raising that number to the power of 3.

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

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

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

$= \frac{-1}{64}$

Note: The cube of a negative rational number is always negative.

Hence, the correct option is (c).

Solution:

Given expression: $\left( \frac{-5}{4} \right)^4$

Since the exponent (4) is an even number, the result will be a positive value.

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

Let us check the options:

(a) $\frac{(-5)^4}{4^4} = \frac{625}{256}$ (Positive)

(b) $\frac{5^4}{(-4)^4} = \frac{625}{256}$ (Positive)

(c) $-\frac{5^4}{4^4} = -\frac{625}{256}$ (Negative)

(d) $\left( -\frac{5}{4} \right) \times \left( -\frac{5}{4} \right) \times \left( -\frac{5}{4} \right) \times \left( -\frac{5}{4} \right)$ = Positive product.

Option (c) represents a negative value, while the original expression is positive.

Hence, the correct option is (c).

Solution:

Let us evaluate each expression:

(a) $\frac{2^3 \times 3^2}{4 \times 18} = \frac{8 \times 9}{4 \times 18} = \frac{72}{72} = 1$

(b) $[(-2)^3 \times (-2)^4] \div (-2)^7 = (-2)^{3+4} \div (-2)^7 = (-2)^7 \div (-2)^7 = 1$

(c) $\frac{3^0 \times 5^3}{5 \times 25} = \frac{1 \times 125}{5 \times 25} = \frac{125}{125} = 1$

(d) $\frac{2^4}{(7^0 + 3^0)^3} = \frac{16}{(1 + 1)^3} = \frac{16}{2^3} = \frac{16}{8} = 2$

The value of expression (d) is 2, not 1.

Hence, the correct option is (d).

Solution:

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

Given expression: $\left( \frac{2}{3} \right)^3 \times \left( \frac{5}{7} \right)^3$

Here, $a = \frac{2}{3}$, $b = \frac{5}{7}$, and $m = 3$. Applying the law:

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

Therefore, the correct option is (c).

Given:

The number is $82,90,30,000$ (using the Indian system of numeration).

Solution:

A number is in standard form if it is written in the form $K \times 10^n$, where $1 \leq K < 10$.

To convert $829030000$ into standard form, we move the decimal point from the end of the number to the left until there is only one digit to the left of the decimal.

We move the decimal point 8 places to the left:

$8.29030000 \times 10^8$

This can be simplified as $8.2903 \times 10^8$.

Comparing this with $K \times 10^8$, we find:

Therefore, the correct option is (d).

Solution:

To determine the largest value, let us convert all options into decimal form or powers of 10:

(a) $0.0001 = 10^{-4}$

(b) $\frac{1}{1000} = 0.001 = 10^{-3}$

(c) $\frac{1}{10^6} = 0.000001 = 10^{-6}$

(d) $\frac{1}{10^6} \div 0.1 = \frac{1}{10^6} \div \frac{1}{10} = \frac{1}{10^6} \times 10 = \frac{1}{10^5} = 0.00001 = 10^{-5}$

Comparing the decimal values:

$0.001$ (Option b) is the largest among $0.0001, 0.001, 0.000001,$ and $0.00001$.

Therefore, the correct option is (b).

Given:

The number is $72$ crore.

Solution:

According to the Indian system of numeration:

So, $72$ crore = $72 \times 10^7$.

To express this in standard form ($K \times 10^n$), the value of $K$ must be between $1$ and $10$.

$72 \times 10^7 = (7.2 \times 10^1) \times 10^7$

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

Therefore, the correct option is (c).

Solution:

This follows the Quotient Law of Exponents, which states that when dividing powers with the same base, we subtract the exponent of the divisor from the exponent of the dividend.

In this problem, the base is $\left( \frac{a}{b} \right)$. Applying the law:

$\left( \frac{a}{b} \right)^m \div \left( \frac{a}{b} \right)^n = \left( \frac{a}{b} \right)^{m-n}$

Therefore, the correct option is (c).

Solution:

Let us evaluate each statement to find the incorrect one:

(a) $3^2 = 9$ and $2^3 = 8$. Since $9 > 8$, the statement is True.

(b) $4^3 = 64$ and $2^6 = 64$. Since $64 = 64$, the statement is True.

(c) $3^3 = 3 \times 3 \times 3 = 27$. The statement says $3^3 = 9$, which is False.

(d) $2^5 = 32$ and $5^2 = 25$. Since $32 > 25$, the statement is True.

Therefore, the correct option is (c).

To Find:

The power $n$ such that $8^n = 2^6$.

Solution:

We know that $8$ can be expressed as a power of $2$:

Substituting this into the equation $8^n = 2^6$:

$(2^3)^n = 2^6$

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

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

Since the bases are equal, we equate the exponents:

Therefore, the correct option is (b).

Solution:

Using the Product Law of Exponents $a^m \times a^n = a^{m+n}$:

$(-2)^{31} \times (-2)^{13} = (-2)^{31 + 13}$

The blank should be filled with 44.

Solution:

Using the Quotient Law of Exponents $a^m \div a^n = a^{m-n}$:

$(-3)^8 \div (-3)^5 = (-3)^{8-5}$

The blank should be filled with 3.

Solution:

Let the missing value in the blank be $x$.

The equation is: $\left( \frac{11}{15} \right)^4 \times (x)^5 = \left( \frac{11}{15} \right)^9$

Using the law of exponents $a^m \times a^n = a^{m+n}$, we know that for the bases to stay the same, $x$ must be $\frac{11}{15}$.

Then the sum of the exponents on the Left Hand Side (LHS) must equal the exponent on the Right Hand Side (RHS):

Since the condition is satisfied when the base is $\frac{11}{15}$, the blank is filled by this base.

Therefore, the blank should be filled with $\frac{11}{15}$.

Solution:

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

Let the exponent in the blank be $n$.

Equating the exponents since the bases are the same:

$3 + n = 11$

$n = 11 - 3$

$n = 8$

The blank should be filled with 8.

Solution:

According to the Power of a Power Law $(a^m)^n = a^{m \times n}$:

$\left[ \left( \frac{7}{11} \right)^3 \right]^4 = \left( \frac{7}{11} \right)^{3 \times 4}$

The blank should be filled with 12.

Solution:

First, simplify the denominator/divisor using $(a^m)^n = a^{mn}$:

$\left[ \left( \frac{6}{13} \right)^5 \right]^2 = \left( \frac{6}{13} \right)^{5 \times 2} = \left( \frac{6}{13} \right)^{10}$

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

$\left( \frac{6}{13} \right)^{10} \div \left( \frac{6}{13} \right)^{10} = \left( \frac{6}{13} \right)^{10-10}$

The blank should be filled with 0.

Solution:

Applying the law $(a^m)^n = a^{m \times n}$:

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

The blank should be filled with 32.

Solution:

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

To obtain a result with base $\frac{13}{14}$, the divisor must also have the same base.

Let the exponent in the divisor be $n$. Then:

$5 - n = 3$

$n = 5 - 3 = 2$

Since the exponent given in the blank is already $2$, the base must be $\frac{13}{14}$.

The blank should be filled with $\frac{13}{14}$.

Solution:

Applying the product law $a^m \times a^n \times a^p = a^{m+n+p}$:

$a^6 \times a^5 \times a^0 = a^{6 + 5 + 0}$

The blank should be filled with 11.

Solution (Indian Perspective):

In the Indian system of numeration, 1 lakh is written as:

1 lakh = $1,00,000$

To express this as a power of 10, we count the number of zeros.

Since there are 5 zeros:

$1,00,000 = 10 \times 10 \times 10 \times 10 \times 10$

The blank should be filled with 5.

Solution:

In the International system, 1 million is represented as $1,000,000$.

In the Indian system, this is equivalent to 10 lakhs ($10,00,000$).

To express this as a power of 10, we count the number of zeros:

$10,00,000 = 10 \times 10 \times 10 \times 10 \times 10 \times 10$

The blank should be filled with 6.

Solution:

To find the power of 3, we perform the prime factorisation of 729:

$\begin{array}{c|cc} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

Counting the number of times 3 is multiplied:

$729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3$

The blank should be filled with 6.

Solution:

We find the prime factorisation of 432:

$\begin{array}{c|cc} 2 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

From the factorisation, we have:

$432 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$

The blank should be filled with 3.

Solution:

Given number: $5,37,00,000$ (5 crore 37 lakhs).

We need to express this with $10^7$ as a factor. To do this, we move the decimal point 7 places to the left:

$5,37,00,000.0 \xrightarrow{\text{7 places left}} 5.37$

So, $53700000 = 5.37 \times 10^7$.

The blank should be filled with 5.37.

Solution:

Given number: $88,88,00,00,000$.

We need to move the decimal point 10 places to the left to match the power $10^{10}$:

$88880000000.0 \xrightarrow{\text{10 places left}} 8.888$

So, $88880000000 = 8.888 \times 10^{10}$.

The blank should be filled with 8.888.

Solution:

To convert $27500000$ to $2.75$, we count how many places the decimal point moves to the left:

$27500000.0 \xrightarrow{\text{7 places left}} 2.75$

Since we moved the decimal 7 places, the power of 10 is 7.

The blank should be filled with 7.

Solution:

To convert $340900000$ to $3.409$, we move the decimal point to the left:

$340900000.0 \xrightarrow{\text{8 places left}} 3.409$

Since the decimal moved 8 places, the exponent is 8.

The blank should be filled with 8.

Solution:

(a) $3^2$ ______ $15$

$3^2 = 3 \times 3 = 9$

Since $9 < 15$, the sign is $<$.

(b) $2^3$ ______ $3^2$

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

$3^2 = 3 \times 3 = 9$

Since $8 < 9$, the sign is $<$.

(c) $7^4$ ______ $5^4$

Since the exponents are the same ($4$) and the base $7 > 5$:

$7^4 > 5^4$. The sign is $>$.

(d) $10,000$ ______ $10^5$

$10,000 = 10^4$

$10^5 = 1,00,000$ (1 lakh)

Since $10,000 < 1,00,000$, the sign is $<$.

(e) $6^3$ _____ $4^4$

$6^3 = 6 \times 6 \times 6 = 216$

$4^4 = 4 \times 4 \times 4 \times 4 = 256$

Since $216 < 256$, the sign is $<$.

Statement: One million = $10^7$

Solution:

The given statement is False.

According to the International system of numeration, 1 million is written as $1,000,000$.

In power notation, $1,000,000$ is expressed as $10^6$.

From an Indian Perspective:

Since 1 million is equal to 10 lakhs ($10^6$), the statement $10^7$ is incorrect.

Statement: One hour = $60^2$ seconds

Solution:

The given statement is True.

We know that:

To find the number of seconds in an hour, we multiply:

1 hour = $60 \times 60$ seconds

Using exponential notation, $60 \times 60$ is written as $60^2$.

Total seconds = $3600$ seconds.

Statement: $1^0 \times 0^1 = 1$

Solution:

The given statement is False.

Evaluating the terms on the Left Hand Side (LHS):

Multiplying these values:

LHS = $1 \times 0 = 0$

Since the RHS is 1 and LHS is 0, the statement is incorrect.

Statement: $(-3)^4 = -12$

Solution:

The given statement is False.

Evaluating $(-3)^4$:

$(-3)^4 = (-3) \times (-3) \times (-3) \times (-3)$

$(-3)^4 = 81$

Note that when a negative number is raised to an even power, the result is always positive. Also, $3^4$ is $81$, not $12$ (which is $3 \times 4$).

Statement: $3^4 > 4^3$

Solution:

The given statement is True.

Evaluating both sides:

Comparing the results:

$81 > 64$

Therefore, $3^4 > 4^3$ is a correct statement.

Statement: $\left( \frac{-3}{5} \right)^{100} = \frac{-3^{100}}{-5^{100}}$

Solution:

The given statement is True.

To verify the equality, let us evaluate the Left Hand Side (LHS) and the Right Hand Side (RHS) separately using the laws of exponents.

Evaluating LHS:

LHS = $\left( \frac{-3}{5} \right)^{100}$

According to the law of exponents $\left( \frac{a}{b} \right)^n = \frac{a^n}{b^n}$, we have:

LHS = $\frac{(-3)^{100}}{5^{100}}$

Since the exponent $100$ is an even number, any negative base raised to this power becomes positive. Therefore, $(-3)^{100} = 3^{100}$.

LHS = $\frac{3^{100}}{5^{100}}$

Evaluating RHS:

RHS = $\frac{-3^{100}}{-5^{100}}$

In the expression $-3^{100}$, the negative sign is a coefficient to the power, which means $-(3^{100})$. Similarly, $-5^{100}$ means $-(5^{100})$.

RHS = $\frac{-(3^{100})}{-(5^{100})}$

The negative signs in the numerator and the denominator cancel each other out:

RHS = $\frac{3^{100}}{5^{100}}$

Since LHS = RHS, both sides represent the same numerical value. Thus, the statement is True.

Statement: $(10 + 10)^{10} = 10^{10} + 10^{10}$

Solution:

The given statement is False.

Evaluating the LHS:

LHS = $(20)^{10}$

Evaluating the RHS:

RHS = $10^{10} + 10^{10} = 2 \times 10^{10}$

We can rewrite $(20)^{10}$ as $(2 \times 10)^{10} = 2^{10} \times 10^{10} = 1024 \times 10^{10}$.

Clearly, $1024 \times 10^{10} \neq 2 \times 10^{10}$. Powers do not distribute over addition.

Statement: $x^0 \times x^0 = x^0 \div x^0$ is true for $x \neq 0$.

Solution:

The given statement is True.

We know that for any non-zero value $x$, $x^0 = 1$.

Evaluating LHS:

LHS = $1 \times 1 = 1$

Evaluating RHS:

RHS = $1 \div 1 = 1$

Since LHS = RHS = 1, the statement holds true for all non-zero values of $x$.

Statement: In the standard form, a large number can be expressed as a decimal number between $0$ and $1$, multiplied by a power of $10$.

Solution:

The given statement is False.

A number is said to be in standard form (or scientific notation) if it is expressed as $k \times 10^n$, where $n$ is an integer and $k$ is a decimal number such that:

This means the decimal number $k$ must be greater than or equal to $1$ and less than $10$. It cannot be between $0$ and $1$.

Statement: $4^2$ is greater than $2^4$.

Solution:

The given statement is False.

Let us evaluate both expressions:

Since both values are equal ($16 = 16$), $4^2$ is not greater than $2^4$.

Statement: $x^m + x^m = x^{2m}$

Solution:

The given statement is False.

When we add like terms in algebra:

The law $x^m \times x^m = x^{m+m} = x^{2m}$ applies to multiplication, not addition. For example, if $x=2$ and $m=3$:

$2^3 + 2^3 = 8 + 8 = 16$

$2^{2 \times 3} = 2^6 = 64$

Since $16 \neq 64$, the statement is incorrect.

Statement: $x^m \times y^m = (x \times y)^{2m}$

Solution:

The given statement is False.

According to the laws of exponents, when the powers are the same but the bases are different, we multiply the bases and keep the exponent the same:

The exponent remains $m$, it does not become $2m$.

Statement: $x^m \div y^m = (x \div y)^m$

Solution:

The given statement is True.

This is a fundamental law of exponents. When dividing two powers with the same exponent but different non-zero bases, the result is the quotient of the bases raised to that common exponent.

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

Solution:

The given statement is True.

This is the Product Law of Exponents. It states that when multiplying two powers with the same base, we keep the base and add the exponents.

Statement: $4^9$ is greater than $16^3$.

Solution:

The given statement is True.

Let us express both numbers with the same base to compare them. We know that $16 = 4^2$.

So, $16^3 = (4^2)^3$

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

Now, comparing $4^9$ and $4^6$:

Since the base $4$ is greater than $1$ and the exponent $9$ is greater than $6$, it follows that $4^9 > 4^6$.

Therefore, $4^9 > 16^3$.

Statement: $\left( \frac{2}{5} \right)^3 \div \left( \frac{5}{2} \right)^3 = 1$

Solution:

The given statement is False.

Evaluating the Left Hand Side (LHS):

LHS = $\left( \frac{2}{5} \right)^3 \div \left( \frac{5}{2} \right)^3$

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

LHS = $\left( \frac{2/5}{5/2} \right)^3$

LHS = $\left( \frac{2}{5} \times \frac{2}{5} \right)^3$

LHS = $\left( \frac{4}{25} \right)^3$

LHS = $\frac{64}{15625}$

Since $\frac{64}{15625} \neq 1$, the statement is incorrect. For the result to be $1$, the dividend and divisor must be identical.

Statement: $\left( \frac{4}{3} \right)^{5} \times \left( \frac{5}{7} \right)^{5} = \left( \frac{4}{3} + \frac{5}{7} \right)^{5}$

Solution:

The given statement is False.

To check the equality, let us evaluate the Left Hand Side (LHS) and the Right Hand Side (RHS) using the laws of exponents.

Evaluating LHS:

LHS = $\left( \frac{4}{3} \right)^{5} \times \left( \frac{5}{7} \right)^{5}$

According to the law of exponents $a^m \times b^m = (a \times b)^m$:

LHS = $\left( \frac{4}{3} \times \frac{5}{7} \right)^{5}$

LHS = $\left( \frac{20}{21} \right)^{5}$

Evaluating RHS:

RHS = $\left( \frac{4}{3} + \frac{5}{7} \right)^{5}$

First, we solve the addition within the parentheses by taking the LCM of the denominators 3 and 7, which is 21:

RHS = $\left( \frac{4 \times 7 + 5 \times 3}{21} \right)^{5}$

RHS = $\left( \frac{28 + 15}{21} \right)^{5}$

RHS = $\left( \frac{43}{21} \right)^{5}$

Conclusion:

Comparing the two results:

LHS = $\left( \frac{20}{21} \right)^{5}$ and RHS = $\left( \frac{43}{21} \right)^{5}$

Since $\frac{20}{21} \neq \frac{43}{21}$, it follows that $\left( \frac{20}{21} \right)^{5} \neq \left( \frac{43}{21} \right)^{5}$.

The power of a product is equal to the product of the powers, not the power of a sum. Therefore, the statement is incorrect.

Statement: $\left( \frac{5}{8} \right)^9 \div \left( \frac{5}{8} \right)^4 = \left( \frac{5}{8} \right)^4$

Solution:

The given statement is False.

According to the Quotient Law of Exponents, for any non-zero rational number $a$:

Applying this to the LHS:

LHS = $\left( \frac{5}{8} \right)^9 \div \left( \frac{5}{8} \right)^4 = \left( \frac{5}{8} \right)^{9-4}$

Since the result is $\left( \frac{5}{8} \right)^5$ and not $\left( \frac{5}{8} \right)^4$, the statement is incorrect.

Statement: $\left( \frac{7}{3} \right)^2 \times \left( \frac{7}{3} \right)^5 = \left( \frac{7}{3} \right)^{10}$

Solution:

The given statement is False.

According to the Product Law of Exponents, when multiplying powers with the same base, we add the exponents:

Applying this to the LHS:

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

The statement incorrectly multiplied the exponents ($2 \times 5 = 10$) instead of adding them.

Statement: $5^0 \times 25^0 \times 125^0 = (5^0)^6$

Solution:

The given statement is True.

According to the law of exponents, any non-zero number raised to the power of $0$ is $1$ ($a^0 = 1$).

Evaluating LHS:

LHS = $1 \times 1 \times 1 = 1$

Evaluating RHS:

RHS = $(1)^6 = 1$

Since LHS = RHS, the statement is correct.

Statement: $876543$ in expanded form using powers of $10$.

Solution:

The given statement is True.

Let us write the number $8,76,543$ in its expanded form according to its place values (Indian Perspective):

$8,76,543 = 8,00,000 + 70,000 + 6,000 + 500 + 40 + 3$

Now, expressing these as powers of $10$:

$= 8 \times 10^5 + 7 \times 10^4 + 6 \times 10^3 + 5 \times 10^2 + 4 \times 10^1 + 3 \times 10^0$

Since $10^0 = 1$, the final term $3 \times 10^0$ equals $3$. The expansion matches the statement.

Statement: $600060 = 6 \times 10^5 + 6 \times 10^2$

Solution:

The given statement is False.

Let us evaluate the expression on the Right Hand Side:

RHS = $6 \times 10^5 + 6 \times 10^2$

RHS = $6,00,000 + 600$

RHS = $6,00,600$

The given number is $6,00,060$ (Six lakh sixty). The correct expanded form for $6,00,060$ should be:

$6 \times 10^5 + 6 \times 10^1$

Statement: $4 \times 10^5 + 3 \times 10^4 + 2 \times 10^3 + 1 \times 10^0 = 432010$

Solution:

The given statement is False.

Let us solve the Left Hand Side:

LHS = $4,00,000 + 30,000 + 2,000 + 1$

LHS = $4,32,001$

The statement claims the result is $4,32,010$. The error lies in the units and tens place; $1 \times 10^0$ is $1$, not $10$.

Statement: $8 \times 10^6 + 2 \times 10^4 + 5 \times 10^2 + 9 \times 10^0 = 8020509$

Solution:

The given statement is True.

Evaluating the sum:

$8 \times 1,000,000 = 80,00,000$

$2 \times 10,000 = 20,000$

$5 \times 100 = 500$

$9 \times 1 = 9$

The resulting sum $80,20,509$ matches the statement.

Statement: $4^0 + 5^0 + 6^0 = (4 + 5 + 6)^0$

Solution:

The given statement is False.

According to the law of exponents, any non-zero number raised to the power of $0$ is equal to $1$. That is, $a^0 = 1$ for $a \neq 0$.

Evaluating the Left Hand Side (LHS):

LHS = $4^0 + 5^0 + 6^0$

LHS = $1 + 1 + 1$

LHS = $3$

Evaluating the Right Hand Side (RHS):

RHS = $(4 + 5 + 6)^0$

RHS = $(15)^0$

RHS = $1$

Conclusion:

Since the LHS value is $3$ and the RHS value is $1$, $3 \neq 1$. Therefore, the statement is incorrect.

Given expressions: $2^5$, $3^3$, $2^3 \times 2$, $(3^3)^2$, $3^5$, $4^0$, $2^3 \times 3^1$

Step 1: Evaluate the value of each expression:

1. $2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32$

2. $3^3 = 3 \times 3 \times 3 = 27$

3. $2^3 \times 2 = 2^4 = 16$

4. $(3^3)^2 = 3^{3 \times 2} = 3^6 = 729$

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

6. $4^0 = 1$

7. $2^3 \times 3^1 = 8 \times 3 = 24$

Step 2: List the calculated values:

$32, 27, 16, 729, 243, 1, 24$

Step 3: Arrange the values in ascending order:

$1 < 16 < 24 < 27 < 32 < 243 < 729$

Step 4: Map the values back to their original expressions:

Ascending Order: $4^0$, $2^3 \times 2$, $2^3 \times 3^1$, $3^3$, $2^5$, $3^5$, $(3^3)^2$

First, we evaluate the value of each given term:

$2^{2+3} = 2^5 = 32$

$(2^2)^3 = 2^{2 \times 3} = 2^6 = 64$

$2 \times 2^2 = 2^{1+2} = 2^3 = 8$

$\frac{3^5}{3^2} = 3^{5-2} = 3^3 = 27$

$3^2 \times 3^0 = 3^{2+0} = 3^2 = 9$

$2^3 \times 5^2 = 8 \times 25 = 200$

The values of the given terms are: $32, 64, 8, 27, 9, 200$.

Arranging these values in descending order, we get:

$200, 64, 32, 27, 9, 8$.

Now, we replace these values with their original expressions:

$200$ corresponds to $2^3 \times 5^2$

$64$ corresponds to $(2^2)^3$

$32$ corresponds to $2^{2+3}$

$27$ corresponds to $\frac{3^5}{3^2}$

$9$ corresponds to $3^2 \times 3^0$

$8$ corresponds to $2 \times 2^2$

Therefore, the terms arranged in descending order are:

$2^3 \times 5^2$, $(2^2)^3$, $2^{2+3}$, $\frac{3^5}{3^2}$, $3^2 \times 3^0$, $2 \times 2^2$.

Let the number by which $(-4)^5$ should be divided be $x$.

According to the question, we have:

$(-4)^5 \div x = (-4)^3$

We can write this equation as:

$\frac{(-4)^5}{x} = (-4)^3$

To find $x$, we can rearrange the equation:

$x = \frac{(-4)^5}{(-4)^3}$

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

$x = (-4)^{5-3}$

$x = (-4)^2$

Now, we calculate the value of $(-4)^2$:

$(-4)^2 = (-4) \times (-4) = 16$

So, the number is 16.

Thus, $(-4)^5$ should be divided by 16 so that the quotient is equal to $(-4)^3$.

Given the equation:

$\left( \frac{2}{9} \right)^{3}$ × $\left( \frac{2}{9} \right)^{6}$ = $\left( \frac{2}{9} \right)^{2m \;-\; 1}$

Using the property of exponents $a^m \times a^n = a^{m+n}$ on the left side of the equation, we combine the terms with the same base:

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

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

Since the bases on both sides of the equation are equal and non-zero ($2/9 \neq 0$) and not equal to 1 or -1, their exponents must be equal.

Equating the exponents, we get:

$9 = 2m - 1$

Now, we solve this linear equation for $m$:

Add 1 to both sides of the equation:

$9 + 1 = 2m - 1 + 1$

$10 = 2m$

Divide both sides by 2:

$\frac{10}{2} = \frac{2m}{2}$

$m = 5$

Thus, the value of $m$ is 5.

Given the expression for $\frac{p}{q}$:

$\frac{p}{q} = \left( \frac{3}{2} \right)^{2} \div \left( \frac{9}{4} \right)^{0}$

First, we evaluate the terms on the right side.

Calculate $\left( \frac{3}{2} \right)^{2}$:

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

Calculate $\left( \frac{9}{4} \right)^{0}$. Using the property that any non-zero number raised to the power of 0 is 1, we have:

$\left( \frac{9}{4} \right)^{0} = 1$

Now substitute these values back into the expression for $\frac{p}{q}$:

$\frac{p}{q} = \frac{9}{4} \div 1$

$\frac{p}{q} = \frac{9}{4}$

Next, we need to find the value of $\left( \frac{p}{q} \right)^{3}$.

Substitute the value of $\frac{p}{q} = \frac{9}{4}$:

$\left( \frac{p}{q} \right)^{3} = \left( \frac{9}{4} \right)^{3}$

Calculate $\left( \frac{9}{4} \right)^{3}$:

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

$9^3 = 9 \times 9 \times 9 = 81 \times 9 = 729$

$4^3 = 4 \times 4 \times 4 = 16 \times 4 = 64$

So,

$\left( \frac{p}{q} \right)^{3} = \frac{729}{64}$

Thus, the value of $\left( \frac{p}{q} \right)^{3}$ is $\frac{729}{64}$.

Let the given rational number be $N$.

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

First, we evaluate the powers:

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

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

Now, substitute these values back into the expression for $N$:

$N = \frac{1}{4} \div \frac{8}{27}$

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

$N = \frac{1}{4} \times \frac{27}{8}$

Multiply the numerators and the denominators:

$N = \frac{1 \times 27}{4 \times 8} = \frac{27}{32}$

We need to find the reciprocal of this rational number $N = \frac{27}{32}$.

The reciprocal of a rational number $\frac{a}{b}$ is $\frac{b}{a}$, provided $a \neq 0$ and $b \neq 0$.

The reciprocal of $\frac{27}{32}$ is $\frac{32}{27}$.

Thus, the reciprocal of the given rational number is $\frac{32}{27}$.

We need to find the value of each expression.

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

(a) Evaluate $7^0$:

$7^0 = 1$

(b) Evaluate $7^7 \div 7^7$:

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

$7^7 \div 7^7 = 7^{7-7} = 7^0 = 1$

Alternatively, any non-zero number divided by itself is 1.

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

(c) Evaluate $(-7)^{2 \times 7 – 6 – 8}$:

First, evaluate the exponent:

$2 \times 7 - 6 - 8 = 14 - 6 - 8 = 8 - 8 = 0$

So the expression is $(-7)^0$.

$(-7)^0 = 1$

(d) Evaluate $(2^0 + 3^0 + 4^0) (4^0 – 3^0 – 2^0)$:

Evaluate each term with exponent 0:

$2^0 = 1$

$3^0 = 1$

$4^0 = 1$

Substitute these values into the expression:

$(1 + 1 + 1) (1 - 1 - 1)$

$(3)(-1)$

$3 \times (-1) = -3$

(e) Evaluate $2 \times 3 \times 4 \div 2^0 \times 3^0 \times 4^0$:

Evaluate the terms with exponent 0:

$2^0 = 1$

$3^0 = 1$

$4^0 = 1$

Substitute these values into the expression and follow the order of operations (division and multiplication from left to right):

$2 \times 3 \times 4 \div 1 \times 1 \times 1$

$6 \times 4 \div 1 \times 1 \times 1$

$24 \div 1 \times 1 \times 1$

$24 \times 1 \times 1$

$24$

(f) Evaluate $(8^0 – 2^0) \times (8^0 + 2^0)$:

Evaluate the terms with exponent 0:

$8^0 = 1$

$2^0 = 1$

Substitute these values into the expression:

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

$(0) \times (2)$

$0 \times 2 = 0$

Given:

To Find:

The value of the integer $n$.

Solution:

To solve for $n$, we need to express both sides of the equation in terms of prime bases, specifically bases $2$ and $3$.

First, let us simplify the Left Hand Side (LHS):

Using the law of exponents $(ab)^m = a^m \times b^m$:

$\text{LHS} = 2^{n - 5} \times 2^{2n - 4} \times 3^{2n - 4}$

Using the product law $a^m \times a^n = a^{m+n}$ for the base $2$:

$\text{LHS} = 2^{(n - 5) + (2n - 4)} \times 3^{2n - 4}$

Now, let us simplify the Right Hand Side (RHS):

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

$\text{RHS} = \frac{1}{2^8 \times 3^4 \times 2^1}$

Combining the powers of $2$ in the denominator:

$\text{RHS} = \frac{1}{2^9 \times 3^4}$

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

Equating LHS and RHS from equations (i) and (ii):

$2^{3n - 9} \times 3^{2n - 4} = 2^{-9} \times 3^{-4}$

Since the bases on both sides are the same, we can compare the exponents of corresponding bases.

Comparing exponents of base 3:

$2n = -4 + 4$

$2n = 0$

Verification using exponents of base 2:

Substituting $n = 0$ into the exponent of base $2$ on the LHS:

$3n - 9 = 3(0) - 9 = -9$

This matches the exponent of base $2$ on the RHS ($-9$).

Therefore, the value of the integer $n$ is $0$.

To express a number from standard form ($a \times 10^n$) to usual form:

If the exponent $n$ is positive, move the decimal point $n$ places to the right.

If the exponent $n$ is negative, move the decimal point $|n|$ places to the left.

(a) Express $8.01 \times 10^7$ in usual form:

The exponent is $7$, which is positive.

We move the decimal point in $8.01$ seven places to the right.

$8.01 \times 10^7 = 80,100,000$

(b) Express $1.75 \times 10^{-3}$ in usual form:

The exponent is $-3$, which is negative.

We move the decimal point in $1.75$ three places to the left.

$1.75 \times 10^{-3} = 0.00175$

(a) To find the value of $2^5$, we multiply 2 by itself 5 times:

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

$2 \times 2 = 4$

$4 \times 2 = 8$

$8 \times 2 = 16$

$16 \times 2 = 32$

So, the value of $2^5$ is 32.

(b) To find the value of $(-3^5)$, the exponent applies only to the base 3. The negative sign is outside the power.

First, calculate $3^5$:

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

$3 \times 3 = 9$

$9 \times 3 = 27$

$27 \times 3 = 81$

$81 \times 3 = 243$

So, $3^5 = 243$.

Now, apply the negative sign:

$-3^5 = -(243) = -243$

So, the value of $(-3^5)$ is -243.

(c) To find the value of $-(-4)^4$, the exponent 4 applies to the base $(-4)$. The negative sign outside is applied after calculating the power.

First, calculate $(-4)^4$:

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

$(-4) \times (-4) = 16$

$16 \times (-4) = -64$

$(-64) \times (-4) = 256$

So, $(-4)^4 = 256$.

Now, apply the negative sign outside:

$-(-4)^4 = -(256) = -256$

So, the value of $-(-4)^4$ is -256.

To express a product in exponential form, we identify the base being multiplied repeatedly and count the number of times it is multiplied. The number of times the base is multiplied becomes the exponent.

(a) Given expression: $3 \times 3 \times 3 \times a \times a \times a \times a$

The base 3 is multiplied 3 times. Its exponential form is $3^3$.

The base $a$ is multiplied 4 times. Its exponential form is $a^4$.

Combining these, the exponential form of the given expression is:

$3^3 a^4$

(b) Given expression: $a \times a \times b \times b \times b \times c \times c \times c \times c$

The base $a$ is multiplied 2 times. Its exponential form is $a^2$.

The base $b$ is multiplied 3 times. Its exponential form is $b^3$.

The base $c$ is multiplied 4 times. Its exponential form is $c^4$.

Combining these, the exponential form of the given expression is:

$a^2 b^3 c^4$

(c) Given expression: $s \times s \times t \times t \times s \times s \times t$

Let's count the occurrences of each base:

The base $s$ appears 4 times ($s \times s \times s \times s$). Its exponential form is $s^4$.

The base $t$ appears 3 times ($t \times t \times t$). Its exponential form is $t^3$.

Combining these, the exponential form of the given expression is:

$s^4 t^3$

Let the number of times 30 must be added together be $x$.

Adding 30 together $x$ times is equivalent to multiplying 30 by $x$.

So, the problem can be expressed as the equation:

$x \times 30 = 30^7$

To find the value of $x$, we need to isolate it. Divide both sides of the equation by 30:

$x = \frac{30^7}{30}$

We can write $30$ as $30^1$. Using the property of exponents $\frac{a^m}{a^n} = a^{m-n}$, where $a=30$, $m=7$, and $n=1$:

$x = 30^{7-1}$

$x = 30^6$

Thus, 30 must be added together $30^6$ times to get a sum equal to $30^7$.

The number of times 30 must be added is $30^6$.

To express a number in exponential notation, we find its prime factorization and write it as a product of powers of its prime factors.

(a) Express 1024 in exponential notation:

We find the prime factorization of 1024:

So, $1024 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{10}$.

The exponential notation for 1024 is $2^{10}$.

(b) Express 1029 in exponential notation:

We find the prime factorization of 1029:

So, $1029 = 3 \times 7 \times 7 \times 7 = 3^1 \times 7^3$.

The exponential notation for 1029 is $3^1 \times 7^3$ or $3 \times 7^3$.

(c) Express $\frac{144}{875}$ in exponential notation:

We find the prime factorization of the numerator, 144:

So, $144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^4 \times 3^2$.

Next, we find the prime factorization of the denominator, 875:

So, $875 = 5 \times 5 \times 5 \times 7 = 5^3 \times 7^1$.

Now, we express the fraction using these exponential forms:

$\frac{144}{875} = \frac{2^4 \times 3^2}{5^3 \times 7^1}$

The exponential notation for $\frac{144}{875}$ is $\frac{2^4 \times 3^2}{5^3 \times 7^1}$ or $\frac{2^4 \times 3^2}{5^3 \times 7}$.

(a) To identify the greater number between $2^6$ and $6^2$, we calculate their values:

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

$6^2 = 6 \times 6 = 36$

Comparing the values, we have $64 > 36$.

Therefore, $2^6$ is the greater number.

(b) To identify the greater number between $2^9$ and $9^2$, we calculate their values:

$2^9 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 512$

$9^2 = 9 \times 9 = 81$

Comparing the values, we have $512 > 81$.

Therefore, $2^9$ is the greater number.

(c) To identify the greater number between $7.9 \times 10^4$ and $5.28 \times 10^5$, we compare their exponents of 10.

The first number is $7.9 \times 10^4$, where the exponent is 4.

The second number is $5.28 \times 10^5$, where the exponent is 5.

Since $5 > 4$, the number with the exponent 5 is greater.

Alternatively, we can express both numbers in usual form or with the same power of 10.

In usual form:

$7.9 \times 10^4 = 79,000$

$5.28 \times 10^5 = 528,000$

Comparing the values, we have $528,000 > 79,000$.

With the same power of 10 (e.g., $10^4$):

$7.9 \times 10^4$

$5.28 \times 10^5 = 5.28 \times 10^1 \times 10^4 = 52.8 \times 10^4$

Comparing $7.9 \times 10^4$ and $52.8 \times 10^4$, since $52.8 > 7.9$, the second number is greater.

Therefore, $5.28 \times 10^5$ is the greater number.

To express a number as a product of powers of its prime factors, we perform prime factorization.

(a) Express 9000 as a product of powers of its prime factors:

We find the prime factorization of 9000:

So, $9000 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 \times 5$.

In exponential notation, this is $2^3 \times 3^2 \times 5^3$.

The product of powers of prime factors for 9000 is $2^3 \times 3^2 \times 5^3$.

(b) Express 2025 as a product of powers of its prime factors:

We find the prime factorization of 2025:

So, $2025 = 3 \times 3 \times 3 \times 3 \times 5 \times 5$.

In exponential notation, this is $3^4 \times 5^2$.

The product of powers of prime factors for 2025 is $3^4 \times 5^2$.

(c) Express 800 as a product of powers of its prime factors:

We find the prime factorization of 800:

So, $800 = 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5$.

In exponential notation, this is $2^5 \times 5^2$.

The product of powers of prime factors for 800 is $2^5 \times 5^2$.

Solution:

To express these in single exponential form, we use the laws of exponents.

(a) $2^3 \times 3^3$

Using the law of exponents $a^m \times b^m = (ab)^m$:

$= 6^3$

(b) $2^4 \times 4^2$

First, we convert both terms to have the same base. Since $4 = 2^2$:

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

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

$= 2^4 \times 2^{2 \times 2}$

$= 2^4 \times 2^4$

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

Alternate approach:

We know $2^4 = (2^2)^2 = 4^2$.

$= 4^2 \times 4^2 = 4^{2+2} = 4^4$

Both $2^8$ and $4^4$ are single exponential forms.

(c) $5^2 \times 7^2$

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

$= 35^2$

(d) $(-5)^5 \times (-5)$

Note that $(-5)$ is the same as $(-5)^1$. Using the product law $a^m \times a^n = a^{m+n}$:

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

$= (-5)^6$

Since the exponent is an even number, $(-5)^6 = 5^6$.

(e) $(-3)^3 \times (-10)^3$

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

$= 30^3$

(f) $(-11)^2 \times (-2)^2$

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

$= 22^2$

To express a number in standard form (scientific notation), we write it as a number between 1 and 10 (inclusive of 1 but exclusive of 10), multiplied by a power of 10. The format is $a \times 10^n$, where $1 \leq |a| < 10$ and $n$ is an integer.

(a) Express 76,47,000 in standard form:

The number is 7,647,000.

The decimal point is initially at the end of the number.

We move the decimal point to the left until it is after the first non-zero digit (7).

The number of places moved is 6 (from the end to after 7).

Since the original number is large, the exponent of 10 is positive.

$7,647,000 = 7.647 \times 10^6$

The standard form is $7.647 \times 10^6$.

(b) Express 8,19,00,000 in standard form:

The number is 81,900,000.

The decimal point is initially at the end of the number.

We move the decimal point to the left until it is after the first non-zero digit (8).

The number of places moved is 7 (from the end to after 8).

Since the original number is large, the exponent of 10 is positive.

$81,900,000 = 8.19 \times 10^7$

The standard form is $8.19 \times 10^7$.

(c) Express 5, 83,00,00,00,000 in standard form:

The number is 583,000,000,000.

The decimal point is initially at the end of the number.

We move the decimal point to the left until it is after the first non-zero digit (5).

The number of places moved is 11 (from the end to after 5).

Since the original number is large, the exponent of 10 is positive.

$583,000,000,000 = 5.83 \times 10^{11}$

The standard form is $5.83 \times 10^{11}$.

(d) Express 24 billion in standard form:

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

So, 24 billion $= 24 \times 10^9$.

To express 24 in standard form, we write it as a number between 1 and 10 multiplied by a power of 10.

$24 = 2.4 \times 10^1$.

Substitute this back into the expression:

$24 \times 10^9 = (2.4 \times 10^1) \times 10^9$

Using the exponent property $10^m \times 10^n = 10^{m+n}$:

$= 2.4 \times 10^{1+9} = 2.4 \times 10^{10}$

The standard form is $2.4 \times 10^{10}$.

Given:

Speed of light in vacuum ($v$) = $3 \times 10^8$ m/s

Time taken for sunlight to reach Earth ($t$) = 8 minutes

To Find:

Distance of Sun from Earth in standard form.

Solution:

First, we need to convert the time from minutes to seconds, since the speed is given in meters per second.

1 minute = 60 seconds

Time in seconds ($t$) = 8 minutes $\times$ 60 seconds/minute

$t = 480$ seconds

The relationship between distance, speed, and time is given by:

Distance = Speed $\times$ Time

$d = v \times t$

Substitute the given values into the formula:

$d = (3 \times 10^8 \text{ m/s}) \times (480 \text{ s})$

$d = 3 \times 480 \times 10^8$ meters

$d = 1440 \times 10^8$ meters

Now, we need to express this distance in standard form ($a \times 10^n$, where $1 \leq |a| < 10$).

We need to write 1440 in standard form. The decimal point is after the last zero.

$1440 = 1.440 \times 10^3$

So, the distance is:

$d = (1.44 \times 10^3) \times 10^8$ meters

Using the property of exponents $10^m \times 10^n = 10^{m+n}$:

$d = 1.44 \times 10^{3+8}$ meters

$d = 1.44 \times 10^{11}$ meters

The distance of the Sun from Earth in standard form is $1.44 \times 10^{11}$ meters.

We use the properties of exponents to simplify each expression:

• $a^m \times a^n = a^{m+n}$ (Product of powers)
• $a^m \div a^n = a^{m-n}$ (Quotient of powers)
• $(a^m)^n = a^{mn}$ (Power of a power)
• $a^0 = 1$ (Zero exponent)

(a) Simplify $\left[ \left( \frac{3}{7} \right)^{4} \times \left( \frac{3}{7} \right)^{5} \right] \div \left( \frac{3}{7} \right)^{7}$

First, simplify the expression inside the square brackets using the product of powers rule:

$\left( \frac{3}{7} \right)^{4} \times \left( \frac{3}{7} \right)^{5} = \left( \frac{3}{7} \right)^{4+5} = \left( \frac{3}{7} \right)^{9}$

Now, substitute this back into the original expression:

$\left( \frac{3}{7} \right)^{9} \div \left( \frac{3}{7} \right)^{7}$

Using the quotient of powers rule:

$\left( \frac{3}{7} \right)^{9-7} = \left( \frac{3}{7} \right)^{2}$

The exponential form is $\left( \frac{3}{7} \right)^{2}$.

(b) Simplify $\left[ \left( \frac{7}{11} \right)^{5} \div \left( \frac{7}{11} \right)^{2} \right] \times \left( \frac{7}{11} \right)^{2}$

First, simplify the expression inside the square brackets using the quotient of powers rule:

$\left( \frac{7}{11} \right)^{5} \div \left( \frac{7}{11} \right)^{2} = \left( \frac{7}{11} \right)^{5-2} = \left( \frac{7}{11} \right)^{3}$

Now, substitute this back into the original expression:

$\left( \frac{7}{11} \right)^{3} \times \left( \frac{7}{11} \right)^{2}$

Using the product of powers rule:

$\left( \frac{7}{11} \right)^{3+2} = \left( \frac{7}{11} \right)^{5}$

The exponential form is $\left( \frac{7}{11} \right)^{5}$.

(c) Simplify $(3^7 \div 3^5)^4$

First, simplify the expression inside the parentheses using the quotient of powers rule:

$3^7 \div 3^5 = 3^{7-5} = 3^2$

Now, substitute this back into the original expression:

$(3^2)^4$

Using the power of a power rule:

$3^{2 \times 4} = 3^8$

The exponential form is $3^8$.

(d) Simplify $\left( \frac{a^{6}}{a^{4}} \right)$ × a⁵ × a⁰

First, simplify the expression inside the parentheses using the quotient of powers rule:

$\frac{a^6}{a^4} = a^{6-4} = a^2$

Now, substitute this back into the original expression:

$a^2 \times a^5 \times a^0$

We know that $a^0 = 1$ (assuming $a \neq 0$). Using the product of powers rule:

$a^2 \times a^5 \times 1 = a^{2+5} \times 1 = a^7 \times 1 = a^7$

The exponential form is $a^7$ (assuming $a \neq 0$).

(e) Simplify $\left[ \left( \frac{3}{5} \right)^{3} \times \left( \frac{3}{5} \right)^{8} \right] \div \left[ \left( \frac{3}{5} \right)^{2} \times \left( \frac{3}{5} \right)^{4} \right]$

Simplify the first part in the square brackets using the product of powers rule:

$\left( \frac{3}{5} \right)^{3} \times \left( \frac{3}{5} \right)^{8} = \left( \frac{3}{5} \right)^{3+8} = \left( \frac{3}{5} \right)^{11}$

Simplify the second part in the square brackets using the product of powers rule:

$\left( \frac{3}{5} \right)^{2} \times \left( \frac{3}{5} \right)^{4} = \left( \frac{3}{5} \right)^{2+4} = \left( \frac{3}{5} \right)^{6}$

Now, substitute these results back into the original expression:

$\left( \frac{3}{5} \right)^{11} \div \left( \frac{3}{5} \right)^{6}$

Using the quotient of powers rule:

$\left( \frac{3}{5} \right)^{11-6} = \left( \frac{3}{5} \right)^{5}$

The exponential form is $\left( \frac{3}{5} \right)^{5}$.

(f) Simplify $(5^{15} \div 5^{10}) \times 5^5$

First, simplify the expression inside the parentheses using the quotient of powers rule:

$5^{15} \div 5^{10} = 5^{15-10} = 5^5$

Now, substitute this back into the original expression:

$5^5 \times 5^5$

Using the product of powers rule:

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

The exponential form is $5^{10}$.

Solution:

(a) $\frac{7^{8} \;\times\; a^{10}b^{7}c^{12}}{7^{6} \;\times\; a^{8}b^{4}c^{12}}$

Using the law of exponents $\frac{x^m}{x^n} = x^{m-n}$ for each common base:

$= 7^{8-6} \times a^{10-8} \times b^{7-4} \times c^{12-12}$

$= 7^2 \times a^2 \times b^3 \times c^0$

$= 49 \times a^2 \times b^3 \times 1$

$= 49a^2b^3$

(b) $\frac{5^{4} \;\times\; 7^{4} \;\times\; 2^{7}}{8 \;\times\; 49 \;\times\; 5^{3}}$

First, express all numbers as powers of prime bases:

Substitute these values into the expression:

$= \frac{5^4 \times 7^4 \times 2^7}{2^3 \times 7^2 \times 5^3}$

Apply the law $\frac{x^m}{x^n} = x^{m-n}$:

$= 5^{4-3} \times 7^{4-2} \times 2^{7-3}$

$= 5^1 \times 7^2 \times 2^4$

$= 5 \times 49 \times 16$

$= 80 \times 49$

$= 3920$

(c) $\frac{125 \;\times\; 5^{2} \;\times\; a^{7}}{10^{3} \;\times\; a^{4}}$

Convert composite bases to prime bases:

Substituting into the expression:

$= \frac{5^3 \times 5^2 \times a^7}{2^3 \times 5^3 \times a^4}$

Cancelling common factor $5^3$ from numerator and denominator:

$= \frac{5^2 \times a^{7-4}}{2^3}$

$= \frac{25a^3}{8}$

(d) $\frac{3^{4} \;\times\; 12^{3} \;\times\; 36}{2^{5} \;\times\; 6^{3}}$

Prime factorise the bases:

$12 = 2^2 \times 3 \implies 12^3 = (2^2 \times 3)^3 = 2^6 \times 3^3$

$36 = 2^2 \times 3^2$

$6 = 2 \times 3 \implies 6^3 = (2 \times 3)^3 = 2^3 \times 3^3$

Now, rewrite the expression:

$= \frac{3^4 \times (2^6 \times 3^3) \times (2^2 \times 3^2)}{2^5 \times (2^3 \times 3^3)}$

Simplify numerator and denominator separately:

Numerator: $2^{6+2} \times 3^{4+3+2} = 2^8 \times 3^9$

Denominator: $2^{5+3} \times 3^3 = 2^8 \times 3^3$

Dividing:

$= \frac{2^8 \times 3^9}{2^8 \times 3^3} = 3^{9-3}$

$= 3^6 = 729$

$= 729$

(e) $\left( \frac{6 \;\times\; 10}{2^2 \;\times\; 5^3} \right)^{2} \times \frac{25}{27}$

Simplify the fraction inside the bracket first:

$\frac{2 \times 3 \times 2 \times 5}{2^2 \times 5^3} = \frac{2^2 \times 3 \times 5}{2^2 \times 5^3} = \frac{3}{5^2}$

Now apply the square:

$\left( \frac{3}{5^2} \right)^2 = \frac{3^2}{(5^2)^2} = \frac{3^2}{5^4}$

Multiply by $\frac{25}{27}$:

$= \frac{3^{2-3}}{5^{4-2}} = \frac{3^{-1}}{5^2}$

$= \frac{1}{3 \times 25} = \frac{1}{75}$

$= \frac{1}{75}$

(f) $\frac{15^{4} \;\times\; 18^{3}}{3^{3} \;\times\; 5^{2} \;\times\; 12^{2}}$

Convert to prime factors:

$15^4 = (3 \times 5)^4 = 3^4 \times 5^4$

$18^3 = (2 \times 3^2)^3 = 2^3 \times 3^6$

$12^2 = (2^2 \times 3)^2 = 2^4 \times 3^2$

Substituting:

$= \frac{3^4 \times 5^4 \times 2^3 \times 3^6}{3^3 \times 5^2 \times 2^4 \times 3^2}$

$= \frac{2^3 \times 3^{10} \times 5^4}{2^4 \times 3^5 \times 5^2}$

$= 2^{3-4} \times 3^{10-5} \times 5^{4-2}$

$= 2^{-1} \times 3^5 \times 5^2 = \frac{243 \times 25}{2}$

$= \frac{6075}{2}$ or $3037.5$

(g) $\frac{6^{4} \;\times\; 9^{2} \;\times\; 25^{3}}{3^{2} \;\times\; 4^{2} \;\times\; 15^{6}}$

Using prime bases $2, 3,$ and $5$:

Numerator: $(2 \times 3)^4 \times (3^2)^2 \times (5^2)^3 = 2^4 \times 3^4 \times 3^4 \times 5^6 $$ = 2^4 \times 3^8 \times 5^6$

Denominator: $3^2 \times (2^2)^2 \times (3 \times 5)^6 = 3^2 \times 2^4 \times 3^6 \times 5^6 $$ = 2^4 \times 3^8 \times 5^6$

Dividing numerator by denominator:

$= \frac{2^4 \times 3^8 \times 5^6}{2^4 \times 3^8 \times 5^6}$

$= 1$

Solution:

To convert a number into Scientific Notation (Standard Form), we express it in the form $k \times 10^n$, where $1 \leq k < 10$ and $n$ is an integer. In the Indian Perspective, we represent the areas with appropriate commas as per the Indian numbering system.

Step 1: Expressing Areas in Scientific Notation

Step 2: Arranging in Ascending Order

To arrange numbers in ascending order, we first compare the powers of $10$. If the powers are the same, we compare the decimal parts ($k$).

Comparing the values:

$1.554 \times 10^5$ (Gibson)

$1.9943 \times 10^5$ (Thar)

$6.475 \times 10^5$ (Great Victoria)

$9.324 \times 10^5$ (Kalahari)

$8.5988 \times 10^6$ (Sahara)

Final Ascending Order:

Gibson ($1.554 \times 10^5$) < Thar ($1.9943 \times 10^5$) < Great Victoria ($6.475 \times 10^5$) < Kalahari ($9.324 \times 10^5$) < Sahara ($8.5988 \times 10^6$)

Solution:

We convert the mass of each planet into scientific notation by moving the decimal point and counting the places.

Scientific Notation of Planetary Masses:

Descending Order:

To arrange in descending order (largest to smallest), we look at the exponent of $10$ first.

1. $1.9 \times 10^{27}$ (Jupiter)

2. $5.69 \times 10^{26}$ (Saturn)

3. $1.02 \times 10^{26}$ (Neptune)

4. $8.69 \times 10^{25}$ (Uranus)

5. $5.98 \times 10^{24}$ (Earth)

6. $4.87 \times 10^{24}$ (Venus)

7. $6.42 \times 10^{23}$ (Mars)

8. $3.3 \times 10^{23}$ (Mercury)

9. $1.31 \times 10^{22}$ (Pluto)

Final Sequence:

Jupiter > Saturn > Neptune > Uranus > Earth > Venus > Mars > Mercury > Pluto

Solution:

To express a number in Scientific Notation (also known as standard form), we write it as $k \times 10^n$, where $k$ is a decimal number such that $1 \leq k < 10$ and $n$ is an integer.

Below is the conversion of the given units of time into scientific notation based on the provided values:

Summary of Method:

For each value, the decimal point is moved to the left until only one non-zero digit remains to the left of the decimal. The number of places the decimal moves becomes the positive exponent of $10$.

For example, in the case of $86,400$:

The decimal point moves 4 places to the left to get $8.64$. Hence, it is written as $8.64 \times 10^4$.

Given:

To Find:

The approximate ratio of area covered with water to area covered by land.

Solution:

First, we convert the given areas into scientific notation:

1. Area of water $= 36,14,19,000 = 3.61419 \times 10^8 \text{ sq. km}$

2. Area of land $= 14,86,47,000 = 1.48647 \times 10^8 \text{ sq. km}$

Now, we find the ratio of water area to land area:

$\text{Ratio} = \frac{\text{Area of water}}{\text{Area of land}}$

$\text{Ratio} = \frac{3.61419 \times 10^8}{1.48647 \times 10^8}$

Cancelling the common term $10^8$ from the numerator and denominator:

$\text{Ratio} = \frac{3.61419}{1.48647}$

To find the approximate ratio, we can round off the decimals:

$\text{Ratio} \approx \frac{3.6}{1.5}$

$\text{Ratio} \approx \frac{\cancel{36}^{12}}{\cancel{15}_{5}} = \frac{12}{5}$

$\text{Ratio} \approx 2.4 : 1$

Therefore, the approximate ratio of area covered with water to area covered by land is $2.4 : 1$ (or approximately $12 : 5$).

Given:

$2^{n + 2} - 2^{n + 1} + 2^n = c \times 2^n$

Solution:

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

$2^n \cdot 2^2 - 2^n \cdot 2^1 + 2^n = c \times 2^n$

Taking $2^n$ as a common factor from all terms on the LHS:

$2^n (2^2 - 2^1 + 1) = c \times 2^n$

Simplifying the numerical expression inside the bracket:

$2^n (4 - 2 + 1) = c \times 2^n$

Dividing both sides by $2^n$ (where $2^n \neq 0$):

Thus, the value of $c$ is $3$.

Solution for (a):

Given, $1 \text{ light year} = 9,460,000,000,000 \text{ km}$.

To convert this to scientific notation, we move the decimal point 12 places to the left:

$1 \text{ light year} = 9.46 \times 10^{12} \text{ km}$

Solution for (b):

Given:

Distance between Earth and Sun $= 1.496 \times 10^8 \text{ km}$

Distance of one light year $= 9.46 \times 10^{12} \text{ km}$

Comparison:

To compare the two distances, we check the powers of 10:

Since $10^8 < 10^{12}$, it is evident that $1.496 \times 10^8$ is much smaller than $9.46 \times 10^{12}$.

Therefore, the distance between the Earth and the Sun is less than one light year.

Given:

Number of sides for a hexagon ($n$) = $6$

Formula for number of diagonals = $\frac{1}{2} (n^2 - 3n)$

Solution:

Substituting the value of $n = 6$ in the given formula:

$\text{Number of diagonals} = \frac{1}{2} (6^2 - 3 \times 6)$

$\text{Number of diagonals} = \frac{1}{2} (36 - 18)$

$\text{Number of diagonals} = \frac{1}{2} (18)$

Thus, a 6-sided figure (hexagon) has $9$ diagonals.

Given:

Initial number of bacteria = 1

Doubling time = 20 minutes

Total time duration = 6 hours

To Find:

Number of bacteria after 6 hours in exponential form and its evaluated value.

Solution:

First, we convert the total time from hours into minutes:

Total time = $6 \times 60 \text{ minutes} = 360 \text{ minutes}$

Now, we find the number of times the bacteria will divide (n):

$n = \frac{\text{Total Time}}{\text{Division Time}}$

$n = \frac{\cancel{360}^{18}}{\cancel{20}}$

$n = 18$ times

Since the number of bacteria doubles with every division, the population after $n$ divisions is given by $2^n$.

Exponential Form:

Number of bacteria = $2^{18}$

Given:

Weight of blubber = $2^2 \times 3^2 \times 5 \times 17 \text{ kg}$

Solution:

To evaluate the amount, we simplify the powers and then multiply:

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

$3^2 = 3 \times 3 = 9$

Substitute these values back into the expression:

Weight = $4 \times 9 \times 5 \times 17$

Weight = $36 \times 85$

Therefore, the average weight of the blubber is $3,060 \text{ kg}$.

Given:

Diameter of Red Blood Cell (RBC) = $7 \times 10^{-6} \text{ m}$

Diameter of Platelet = $2.33 \times 10^{-6} \text{ m}$

Solution:

To compare two numbers written in scientific notation, we first check the powers of $10$. If the powers are identical, we compare the decimal coefficients.

1. Power of 10 for RBC = $10^{-6}$

2. Power of 10 for Platelet = $10^{-6}$

Since the powers of $10$ are the same, we compare the coefficients $7$ and $2.33$.

Since $7 > 2.33$, it follows that $7 \times 10^{-6} > 2.33 \times 10^{-6}$.

Therefore, a red blood cell has a greater diameter than a platelet.

Solution:

(a) Writing a googol as a power:

A number represented by $1$ followed by $n$ zeros can be written as $10^n$.

Since a googol has 100 zeros:

Googol = $10^{100}$

(b) Googol times a googol:

We need to find the product: $\text{Googol} \times \text{Googol}$

$= 10^{100} \times 10^{100}$

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

$= 10^{100 + 100}$

$= 10^{200}$

Given Expression: $\frac{3^5}{9^5}$

Solution:

The student claimed that the value is $\frac{1}{3}$. Let us evaluate the expression correctly:

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

$\frac{3^5}{9^5} = \left( \frac{3}{9} \right)^5$

$\frac{3^5}{9^5} = \left( \frac{1}{3} \right)^5$

$\frac{3^5}{9^5} = \frac{1^5}{3^5} = \frac{1}{243}$

The Mistake:

The student likely divided the bases ($3$ and $9$) to get $\frac{1}{3}$ but forgot to apply the exponent ($5$) to the result. The student treated $\frac{3^5}{9^5}$ as if it were simply $\frac{3}{9}$.