NCERT Solutions for Class 7 Maths Chapter 13: Exponents and Powers - Exercise 13.2

Exercise 13.2

Refer for exercise 13.2 in the PDF

1. Using laws of exponents, simplify and write the answer in exponential form:

i) ${3^2} \times {3^4} \times {3^8}$

Ans: The given expression is: ${3^2} \times {3^4} \times {3^8}$

By laws of exponents, we have ${a^m} \times {a^n} = {a^{m + n}}$ .

So, we can write the given expression as,

${3^2} \times {3^4} \times {3^8} = {3^{\left( {2 + 4 + 8} \right)}}$

${3^2} \times {3^4} \times {3^8} = {3^{14}}$

Hence, the required answer is ${3^2} \times {3^4} \times {3^8} = {3^{14}}$.

ii) ${6^{15}} \div {6^{10}}$

Ans: The given expression is: ${6^{15}} \div {6^{10}}$

By laws of exponents, we have ${a^m} \div {a^n} = {a^{m - n}}$ .

So, we can write the given expression as,

${6^{15}} \div {6^{10}} = {6^{\left( {15 - 10} \right)}}$

${6^{15}} \div {6^{10}} = {6^5}$

Hence, the required answer is ${6^{15}} \div {6^{10}} = {6^5}$.

ii) ${a^3} \times {a^2}$

Ans: The given expression is: ${a^3} \times {a^2}$

By laws of exponents, we have ${a^m} \times {a^n} = {a^{m + n}}$ .

So, we can write the given expression as,

${a^3} \times {a^2} = {a^{\left( {3 + 2} \right)}}$

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

Hence, the required answer is ${a^3} \times {a^2} = {a^5}$.

iii) ${7^x} \times {7^2}$

Ans: The given expression is: ${7^x} \times {7^2}$

By laws of exponents, we have ${a^m} \times {a^n} = {a^{m + n}}$ .

So, we can write the given expression as,

${7^x} \times {7^2} = {7^{x + 2}}$

Hence, the required answer is ${7^x} \times {7^2} = {7^{x + 2}}$.

iv) ${\left( {{5^2}} \right)^3} \div {5^3}$

Ans: The given expression is: ${\left( {{5^2}} \right)^3} \div {5^3}$

By laws of exponents, we have ${\left( {{a^m}} \right)^n} = {a^{m \times n}}$

So, we can write the given expression as,

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

${\left( {{5^2}} \right)^3} \div {5^3} = {5^6} \div {5^3}$

By laws of exponents, we also have ${a^m} \div {a^n} = {a^{m - n}}$

${\left( {{5^2}} \right)^3} \div {5^3} = {5^{6 - 3}}$

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

Hence, the required answer is ${\left( {{5^2}} \right)^3} \div {5^3} = {5^3}$

v) ${2^5} \times {5^5}$

Ans: The given expression is: ${2^5} \times {5^5}$

By laws of exponents, we have ${a^m} \times {b^m} = {\left( {a \times b} \right)^m}$.

So, we can write the given expression as,

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

${2^5} \times {5^5} = {10^5}$

Hence, the required answer is ${2^5} \times {5^5} = {10^5}$.

vi) ${a^4} \times {b^4}$

Ans: The given expression is: ${a^4} \times {b^4}$

By laws of exponents, we have ${a^m} \times {b^m} = {\left( {a \times b} \right)^m}$ .

So, we can write the given expression as,

${a^4} \times {b^4} = {\left( {a \times b} \right)^4}$

Hence, the required answer is ${a^4} \times {b^4} = {\left( {a \times b} \right)^4}$ .

vii) ${\left( {{3^4}} \right)^3}$

Ans: The given expression is: ${\left( {{3^4}} \right)^3}$

By laws of exponents, we have ${\left( {{a^m}} \right)^n} = {a^{m \times n}}$

So, we can write the given expression as,

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

${\left( {{3^4}} \right)^3} = {3^{12}}$

Hence, the required answer is ${\left( {{3^4}} \right)^3} = {3^{12}}$.

viii) $\left( {{2^{20}} \div {2^{15}}} \right) \times {2^3}$

Ans: The given expression is: $\left( {{2^{20}} \div {2^{15}}} \right) \times {2^3}$

By laws of exponents, we also have ${a^m} \div {a^n} = {a^{m - n}}$

$\left( {{2^{20}} \div {2^{15}}} \right) \times {2^3} = {2^{\left( {20 - 15} \right)}} \times {2^3}$

$\left( {{2^{20}} \div {2^{15}}} \right) \times {2^3} = {2^5} \times {2^3}$

By laws of exponents, we have ${a^m} \times {a^n} = {a^{m + n}}$ .

So, we can write 

$\left( {{2^{20}} \div {2^{15}}} \right) \times {2^3} = {2^{5 + 3}}$

$\left( {{2^{20}} \div {2^{15}}} \right) \times {2^3} = {2^8}$

Hence, the required answer is $\left( {{2^{20}} \div {2^{15}}} \right) \times {2^3} = {2^8}$.

ix) ${8^t} \div {8^2}$

Ans: The given expression is: ${8^t} \div {8^2}$

By laws of exponents, we have ${a^m} \div {a^n} = {a^{m - n}}$ .

So, we can write the given expression as,

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

Hence, the required answer is ${8^t} \div {8^2} = {8^{t - 2}}$.

2. Simplify and express each of the following in exponential form:

i) $\dfrac{{{2^3} \times {3^4} \times 4}}{{3 \times 32}}$

Ans: The given expression is $\dfrac{{{2^3} \times {3^4} \times 4}}{{3 \times 32}}$

$32$can be written as,

$32 = {2^5}$

Also $4$ can be written as,

$4 = {2^2}$

Then the given expression becomes,

$\dfrac{{{2^3} \times {3^4} \times 4}}{{3 \times 32}} = \dfrac{{{2^3} \times {3^4} \times {2^2}}}{{3 \times {2^5}}}$

By laws of exponents, we have ${a^m} \times {a^n} = {a^{m + n}}$ .

So, we can write

$\dfrac{{{2^3} \times {3^4} \times {2^2}}}{{3 \times {2^5}}} = \dfrac{{{2^{3 + 2}} \times {3^4}}}{{3 \times {2^5}}}$

$\dfrac{{{2^3} \times {3^4} \times {2^2}}}{{3 \times {2^5}}} = \dfrac{{{2^5} \times {3^4}}}{{3 \times {2^5}}}$

By laws of exponents, we have  ${a^m} \div {a^n} = {a^{m - n}}$ .

So, we can write

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

$\dfrac{{{2^3} \times {3^4} \times {2^2}}}{{3 \times {2^5}}} = {2^0} \times {3^3}$

We have ${a^0} = 1$

$\dfrac{{{2^3} \times {3^4} \times {2^2}}}{{3 \times {2^5}}} = 1 \times {3^3}$

$\dfrac{{{2^3} \times {3^4} \times {2^2}}}{{3 \times {2^5}}} = {3^3}$

Hence, the required solution is $\dfrac{{{2^3} \times {3^4} \times {2^2}}}{{3 \times {2^5}}} = {3^3}$.

ii) $\left[ {{{\left( {{5^2}} \right)}^3} \times {5^4}} \right] \div {5^7}$

Ans: The given expression is $\left[ {{{\left( {{5^2}} \right)}^3} \times {5^4}} \right] \div {5^7}$

By laws of exponents, we have ${\left( {{a^m}} \right)^n} = {a^{m \times n}}$

So, we can write the given expression as

$\left[ {{{\left( {{5^2}} \right)}^3} \times {5^4}} \right] \div {5^7} = \left[ {{5^{2 \times 3}} \times {5^4}} \right] \div {5^7}$

$\left[ {{{\left( {{5^2}} \right)}^3} \times {5^4}} \right] \div {5^7} = \left[ {{5^6} \times {5^4}} \right] \div {5^7}$

By laws of exponents, we have ${a^m} \times {a^n} = {a^{m + n}}$ .

So, we can write

$\left[ {{{\left( {{5^2}} \right)}^3} \times {5^4}} \right] \div {5^7} = {5^{6 + 4}} \div {5^7}$

$\left[ {{{\left( {{5^2}} \right)}^3} \times {5^4}} \right] \div {5^7} = {5^{10}} \div {5^7}$

By laws of exponents, we have  ${a^m} \div {a^n} = {a^{m - n}}$ .

So, we can write

$\left[ {{{\left( {{5^2}} \right)}^3} \times {5^4}} \right] \div {5^7} = {5^{10 - 7}}$

$\left[ {{{\left( {{5^2}} \right)}^3} \times {5^4}} \right] \div {5^7} = {5^3}$

Hence, the required solution is $\left[ {{{\left( {{5^2}} \right)}^3} \times {5^4}} \right] \div {5^7} = {5^3}$.

iii) ${25^4} \div {5^3}$

Ans: The given expression is ${25^4} \div {5^3}$

$25$can be written as

$25 = {5^2}$

So, the expression will be

${25^4} \div {5^3} = {\left( {{5^2}} \right)^4} \div {5^3}$

By laws of exponents, we have ${\left( {{a^m}} \right)^n} = {a^{m \times n}}$

So, we can write 

${25^4} \div {5^3} = {5^{2 \times 4}} \div {5^3}$

${25^4} \div {5^3} = {5^8} \div {5^3}$

By laws of exponents, we have  ${a^m} \div {a^n} = {a^{m - n}}$ .

So, we can write

${25^4} \div {5^3} = {5^{8 - 3}}$

${25^4} \div {5^3} = {5^5}$

Hence, the required solution is ${25^4} \div {5^3} = {5^5}$.

iv) $\dfrac{{3 \times {7^2} \times {{11}^8}}}{{21 \times {{11}^3}}}$

Ans: The given expression is $\dfrac{{3 \times {7^2} \times {{11}^8}}}{{21 \times {{11}^3}}}$

$21$can be written as,

$21 = 7 \times 3$

Then the given expression becomes,

$\dfrac{{3 \times {7^2} \times {{11}^8}}}{{21 \times {{11}^3}}} = \dfrac{{3 \times {7^2} \times {{11}^8}}}{{7 \times 3 \times {{11}^3}}}$

By laws of exponents, we have  ${a^m} \div {a^n} = {a^{m - n}}$ .

So, we can write

$\dfrac{{3 \times {7^2} \times {{11}^8}}}{{21 \times {{11}^3}}} = {3^{1 - 1}} \times {7^{2 - 1}} \times {11^{8 - 3}}$

$\dfrac{{3 \times {7^2} \times {{11}^8}}}{{21 \times {{11}^3}}} = {3^0} \times {7^1} \times {11^5}$

We have ${a^0} = 1$

$\dfrac{{3 \times {7^2} \times {{11}^8}}}{{21 \times {{11}^3}}} = 1 \times 7 \times {11^5}$

$\dfrac{{3 \times {7^2} \times {{11}^8}}}{{21 \times {{11}^3}}} = 7 \times {11^5}$

Hence, the required solution is $\dfrac{{3 \times {7^2} \times {{11}^8}}}{{21 \times {{11}^3}}} = 7 \times {11^5}$.

v) $\dfrac{{{3^7}}}{{{3^4} \times {3^3}}}$

Ans: The given expression is $\dfrac{{{3^7}}}{{{3^4} \times {3^3}}}$

By laws of exponents, we have ${a^m} \times {a^n} = {a^{m + n}}$ .

So, we can write

$\dfrac{{{3^7}}}{{{3^4} \times {3^3}}} = \dfrac{{{3^7}}}{{{3^{4 + 3}}}}$

$\dfrac{{{3^7}}}{{{3^4} \times {3^3}}} = \dfrac{{{3^7}}}{{{3^7}}}$

By laws of exponents, we have  ${a^m} \div {a^n} = {a^{m - n}}$ .

So, we can write

$\dfrac{{{3^7}}}{{{3^4} \times {3^3}}} = {3^{7 - 7}}$

$\dfrac{{{3^7}}}{{{3^4} \times {3^3}}} = {3^0}$

We have ${a^0} = 1$

$\dfrac{{{3^7}}}{{{3^4} \times {3^3}}} = 1$

Hence, the required solution is $\dfrac{{{3^7}}}{{{3^4} \times {3^3}}} = 1$.

vi) ${2^0} + {3^0} + {4^0}$

Ans: The given expression is ${2^0} + {3^0} + {4^0}$ .

We have ${a^0} = 1$

${2^0} + {3^0} + {4^0} = 1 + 1 + 1$

By simplify,

${2^0} + {3^0} + {4^0} = 3$

Hence, the required solution is${2^0} + {3^0} + {4^0} = 3$.

vii) ${2^0} \times {3^0} \times {4^0}$

Ans: The given expression is ${2^0} \times {3^0} \times {4^0}$ .

We have ${a^0} = 1$

${2^0} \times {3^0} \times {4^0} = 1 \times 1 \times 1$

By simplify,

${2^0} \times {3^0} \times {4^0} = 1$

Hence, the required solution is${2^0} \times {3^0} \times {4^0} = 1$.

viii) $\left( {{3^0} + {2^0}} \right) \times {5^0}$

Ans: The given expression is $\left( {{3^0} + {2^0}} \right) \times {5^0}$ .

We have ${a^0} = 1$

$\left( {{3^0} + {2^0}} \right) \times {5^0} = \left( {1 + 1} \right) \times 1$

By simplify,

$\left( {{3^0} + {2^0}} \right) \times {5^0} = 2 \times 1$

$\left( {{3^0} + {2^0}} \right) \times {5^0} = 2$

Hence, the required solution is$\left( {{3^0} + {2^0}} \right) \times {5^0} = 2$.

ix) $\dfrac{{{2^8} \times {a^5}}}{{{4^3} \times {a^3}}}$

Ans: The given expression is $\dfrac{{{2^8} \times {a^5}}}{{{4^3} \times {a^3}}}$

$4$can be written as,

$4 = {2^2}$

Then the given expression becomes,

$\dfrac{{{2^8} \times {a^5}}}{{{4^3} \times {a^3}}} = \dfrac{{{2^8} \times {a^5}}}{{{{\left( {{2^2}} \right)}^3} \times {a^3}}}$

By laws of exponents, we have ${\left( {{a^m}} \right)^n} = {a^{m \times n}}$

So, we can write

$\dfrac{{{2^8} \times {a^5}}}{{{4^3} \times {a^3}}} = \dfrac{{{2^8} \times {a^5}}}{{{2^{\left( {2 \times 3} \right)}} \times {a^3}}}$

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

By laws of exponents, we have  ${a^m} \div {a^n} = {a^{m - n}}$ .

So, we can write

$\dfrac{{{2^8} \times {a^5}}}{{{4^3} \times {a^3}}} = {2^{8 - 6}} \times {a^{5 - 3}}$

$\dfrac{{{2^8} \times {a^5}}}{{{4^3} \times {a^3}}} = {2^2} \times {a^2}$

By laws of exponents, we have ${a^m} \times {b^m} = {\left( {a \times b} \right)^m}$ .

So, we can write

$\dfrac{{{2^8} \times {a^5}}}{{{4^3} \times {a^3}}} = {\left( {2a} \right)^2}$

Hence, the required solution is $\dfrac{{{2^8} \times {a^5}}}{{{4^3} \times {a^3}}} = {\left( {2a} \right)^2}$.

x) $\left( {\dfrac{{{a^5}}}{{{a^3}}}} \right) \times {a^8}$

Ans: The given expression is $\left( {\dfrac{{{a^5}}}{{{a^3}}}} \right) \times {a^8}$

By laws of exponents, we have  ${a^m} \div {a^n} = {a^{m - n}}$ .

So, we can write

$\left( {\dfrac{{{a^5}}}{{{a^3}}}} \right) \times {a^8} = {a^{5 - 3}} \times {a^8}$

$\left( {\dfrac{{{a^5}}}{{{a^3}}}} \right) \times {a^8} = {a^2} \times {a^8}$

By laws of exponents, we have ${a^m} \times {a^n} = {a^{m + n}}$ .

So, we can write

$\left( {\dfrac{{{a^5}}}{{{a^3}}}} \right) \times {a^8} = {a^{2 + 8}}$

$\left( {\dfrac{{{a^5}}}{{{a^3}}}} \right) \times {a^8} = {a^{10}}$

Hence, the required solution is $\left( {\dfrac{{{a^5}}}{{{a^3}}}} \right) \times {a^8} = {a^{10}}$.

xi) $\dfrac{{{4^5} \times {a^8}{b^3}}}{{{4^5} \times {a^5}{b^2}}}$

Ans: The given expression is $\dfrac{{{4^5} \times {a^8}{b^3}}}{{{4^5} \times {a^5}{b^2}}}$

By laws of exponents, we have  ${a^m} \div {a^n} = {a^{m - n}}$ .

So, we can write

$\dfrac{{{4^5} \times {a^8}{b^3}}}{{{4^5} \times {a^5}{b^2}}} = {4^{5 - 5}} \times {a^{8 - 5}} \times {b^{3 - 2}}$

$\dfrac{{{4^5} \times {a^8}{b^3}}}{{{4^5} \times {a^5}{b^2}}} = {4^0} \times {a^3} \times {b^1}$

We have ${a^0} = 1$

$\dfrac{{{4^5} \times {a^8}{b^3}}}{{{4^5} \times {a^5}{b^2}}} = 1 \times {a^3} \times b$

By simplifying, 

$\dfrac{{{4^5} \times {a^8}{b^3}}}{{{4^5} \times {a^5}{b^2}}} = {a^3}b$

Hence, the required solution is $\dfrac{{{4^5} \times {a^8}{b^3}}}{{{4^5} \times {a^5}{b^2}}} = {a^3}b$.

xii) ${\left( {{2^3} \times 2} \right)^2}$

Ans: The given expression is ${\left( {{2^3} \times 2} \right)^2}$

By laws of exponents, we have ${a^m} \times {a^n} = {a^{m + n}}$ .

So, we can write

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

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

By laws of exponents, we have ${\left( {{a^m}} \right)^n} = {a^{m \times n}}$

So, we can write

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

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

Hence, the required solution is ${\left( {{2^3} \times 2} \right)^2} = {2^8}$.

3. Say true or false and justify your answer:

i) $10 \times {10^{11}} = {100^{11}}$

Ans: The given expression is $10 \times {10^{11}} = {100^{11}}$

Taking left hand side,

By laws of exponents ${a^m} \times {a^n} = {a^{m + n}}$  we can write,

$10 \times {10^{11}} = {10^{1 + 11}}$

$10 \times {10^{11}} = {10^{12}}$

Taking right hand side,

${100^{11}} = {\left( {{{10}^2}} \right)^{11}}$

${100^{11}} = {10^{2 \times 11}}$

${100^{11}} = {10^{22}}$

So, ${\text{L}}{\text{.H}}{\text{.S}} \ne {\text{R}}{\text{.H}}{\text{.S}}$

Hence, the given statement is false.

ii) ${2^3} > {5^2}$

Ans: The given expression is ${2^3} > {5^2}$

Taking left hand side,

${2^3} = 8$

Taking right hand side,

${5^2} = 25$

Since, $25 > 8$

So, ${5^2} > {2^3}$

Hence, the given statement is false.

iii) ${2^3} \times {3^2} = {6^5}$

Ans: The given expression is ${2^3} \times {3^2} = {6^5}$

Taking left hand side,

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

${2^3} \times {3^2} = 72$

Taking right hand side,

${6^5} = 7776$

So,${2^3} \times {3^2} \ne {6^5}$

Hence, the given statement is false.

iv) ${3^0} = {\left( {1000} \right)^0}$

Ans: The given expression is ${3^0} = {\left( {1000} \right)^0}$

Taking left hand side,

Since, ${a^0} = 1$

${3^0} = 1$

Taking right hand side,

Since, ${a^0} = 1$

${\left( {1000} \right)^0} = 1$

So,${3^0} = {\left( {1000} \right)^0}$

Hence, the given statement is true.

4. Express each of the following as a product of prime factors only in exponential form:

i) $108 \times 192$

Ans: The given expression is $108 \times 192$

Finding prime factors of$108$,

So, $108$can be express as

$108 = 2 \times 2 \times 3 \times 3 \times 3$

$108 = {2^2} \times {3^3}$

Finding prime factors of$192$,

So, $192$can be express as

$192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3$

$192 = {2^6} \times 3$

So, $108 \times 192 = \left( {{2^2} \times {3^3}} \right) \times \left( {{2^6} \times 3} \right)$

$108 \times 192 = \left( {{2^2} \times {3^3}} \right) \times \left( {{2^6} \times 3} \right)$

Since, ${a^m} \times {a^n} = {a^{m + n}}$

$108 \times 192 = {2^{2 + 6}} \times {3^{3 + 1}}$

$108 \times 192 = {2^8} \times {3^4}$

Hence, the required solution is $108 \times 192 = {2^8} \times {3^4}$.

ii) $270$

Ans: The given expression is $270$

Finding prime factors of$270$,

So, $270$can be express as

$270 = 2 \times 3 \times 3 \times 3 \times 5$

$270 = 2 \times {3^3} \times 5$

Hence, the required solution is $270 = 2 \times {3^3} \times 5$.

iii) $729 \times 64$

Ans: The given expression is $729 \times 64$

Finding prime factors of$729$,

So, $729$can be express as

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

$729 = {3^6}$

Finding prime factors of$64$ ,

So, $64$can be express as

$64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2$

$64 = {2^6}$

So, $729 \times 64 = {3^6} \times {2^6}$

Hence, the required solution is $729 \times 64 = {3^6} \times {2^6}$ .

iii) $768$

Ans: The given expression is $768$

Finding prime factors of$768$,

So, $768$can be express as

$768 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3$

$768 = {2^8} \times 3$

Hence, the required solution is $768 = {2^8} \times 3$.

5. Simplify:

i) $\dfrac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{8^3} \times 7}}$

Ans: The given expression is $\dfrac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{8^3} \times 7}}$

$8$can be written as,

$8 = {2^3}$

Then the given expression becomes,

$\dfrac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{8^3} \times 7}} = \dfrac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{{\left( {{2^3}} \right)}^3} \times 7}}$

By laws of exponents, we have ${\left( {{a^m}} \right)^n} = {a^{m \times n}}$

So, we can write

$\dfrac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{8^3} \times 7}} = \dfrac{{{2^{\left( {5 \times 2} \right)}} \times {7^3}}}{{{2^{\left( {3 \times 3} \right)}} \times 7}}$

$\dfrac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{8^3} \times 7}} = \dfrac{{{2^{10}} \times {7^3}}}{{{2^9} \times 7}}$

By laws of exponents, we have  ${a^m} \div {a^n} = {a^{m - n}}$ .

So, we can write

$\dfrac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{8^3} \times 7}} = {2^{10 - 9}} \times {7^{3 - 1}}$

$\dfrac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{8^3} \times 7}} = {2^1} \times {7^2}$

Since, ${7^2} = 49$

$\dfrac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{8^3} \times 7}} = 2 \times 49$

$\dfrac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{8^3} \times 7}} = 98$

Hence, the required solution is $\dfrac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{8^3} \times 7}} = 98$.

ii) $\dfrac{{25 \times {5^2} \times {t^8}}}{{{{10}^3} \times {t^4}}}$

Ans:  The given expression is $\dfrac{{25 \times {5^2} \times {t^8}}}{{{{10}^3} \times {t^4}}}$

$25$can be written as,

$25 = {5^2}$

$10$can be express as,

$10 = 2 \times 5$

Then the given expression becomes,

$\dfrac{{25 \times {5^2} \times {t^8}}}{{{{10}^3} \times {t^4}}} = \dfrac{{{5^2} \times {5^2} \times {t^8}}}{{{{\left( {2 \times 5} \right)}^3} \times {t^4}}}$

By laws of exponents, we have ${\left( {a \times b} \right)^m} = {a^m} \times {b^m}$

So, we can write

$\dfrac{{25 \times {5^2} \times {t^8}}}{{{{10}^3} \times {t^4}}} = \dfrac{{{5^2} \times {5^2} \times {t^8}}}{{{2^3} \times {5^3} \times {t^4}}}$

By laws of exponents, we have  ${a^m} \times {a^n} = {a^{m + n}}$ .

So, we can write

$\dfrac{{25 \times {5^2} \times {t^8}}}{{{{10}^3} \times {t^4}}} = \dfrac{{{5^{2 + 2}} \times {t^8}}}{{{2^3} \times {5^3} \times {t^4}}}$

$\dfrac{{25 \times {5^2} \times {t^8}}}{{{{10}^3} \times {t^4}}} = \dfrac{{{5^4} \times {t^8}}}{{{2^3} \times {5^3} \times {t^4}}}$

By laws of exponents, we have  ${a^m} \div {a^n} = {a^{m - n}}$ .

So, we can write

$\dfrac{{25 \times {5^2} \times {t^8}}}{{{{10}^3} \times {t^4}}} = \dfrac{{{5^{4 - 3}} \times {t^{8 - 4}}}}{{{2^3}}}$

$\dfrac{{25 \times {5^2} \times {t^8}}}{{{{10}^3} \times {t^4}}} = \dfrac{{{5^1} \times {t^4}}}{{{2^3}}}$

Since, ${2^3} = 8$

$\dfrac{{25 \times {5^2} \times {t^8}}}{{{{10}^3} \times {t^4}}} = \dfrac{{5{t^4}}}{8}$

Hence, the required solution is $\dfrac{{25 \times {5^2} \times {t^8}}}{{{{10}^3} \times {t^4}}} = \dfrac{{5{t^4}}}{8}$.

iii) $\dfrac{{{3^5} \times {{10}^5} \times 25}}{{{5^7} \times {6^5}}}$

Ans:  The given expression is $\dfrac{{{3^5} \times {{10}^5} \times 25}}{{{5^7} \times {6^5}}}$

$25$can be written as,

$25 = {5^2}$

$10$can be express as,

$10 = 2 \times 5$

$6$can be express as,

$6 = 2 \times 3$

Then the given expression becomes,

$\dfrac{{{3^5} \times {{10}^5} \times 25}}{{{5^7} \times {6^5}}} = \dfrac{{{3^5} \times {{\left( {2 \times 5} \right)}^5} \times {5^2}}}{{{5^7} \times {{\left( {2 \times 3} \right)}^5}}}$

By laws of exponents, we have ${\left( {a \times b} \right)^m} = {a^m} \times {b^m}$

So, we can write

$\dfrac{{{3^5} \times {{10}^5} \times 25}}{{{5^7} \times {6^5}}} = \dfrac{{{3^5} \times {2^5} \times {5^5} \times {5^2}}}{{{5^7} \times {2^5} \times {3^5}}}$

By laws of exponents, we have  ${a^m} \times {a^n} = {a^{m + n}}$ .

So, we can write

$\dfrac{{{3^5} \times {{10}^5} \times 25}}{{{5^7} \times {6^5}}} = \dfrac{{{3^5} \times {2^5} \times {5^{5 + 2}}}}{{{5^7} \times {2^5} \times {3^5}}}$

$\dfrac{{{3^5} \times {{10}^5} \times 25}}{{{5^7} \times {6^5}}} = \dfrac{{{3^5} \times {2^5} \times {5^7}}}{{{5^7} \times {2^5} \times {3^5}}}$

By laws of exponents, we have  ${a^m} \div {a^n} = {a^{m - n}}$ .

So, we can write

$\dfrac{{{3^5} \times {{10}^5} \times 25}}{{{5^7} \times {6^5}}} = {3^{5 - 5}} \times {2^{5 - 5}} \times {5^{7 - 7}}$

$\dfrac{{{3^5} \times {{10}^5} \times 25}}{{{5^7} \times {6^5}}} = {3^0} \times {2^0} \times {5^0}$

Since, ${a^0} = 1$

$\dfrac{{{3^5} \times {{10}^5} \times 25}}{{{5^7} \times {6^5}}} = 1 \times 1 \times 1$

$\dfrac{{{3^5} \times {{10}^5} \times 25}}{{{5^7} \times {6^5}}} = 1$

Hence, the required solution is $\dfrac{{{3^5} \times {{10}^5} \times 25}}{{{5^7} \times {6^5}}} = 1$.

NCERT Solutions For Class 7 Maths Chapter 13 Exponents and Powers Exercise 13.2

Opting for the NCERT solutions for Ex 13.2 Class 7 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 13.2 Class 7 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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