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# NCERT Solutions for Class 7 Maths Exercise 13.2

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NCERT solutions for Maths Exponents and Powers

## NCERT Solutions for Class 7 Maths Exponents and Powers

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

(i){tex}{3^2} \times {3^4} \times {3^8}\;\;\;\;{/tex}

(ii) {tex}{6^{15}}{/tex}{tex} \div {/tex}{tex}\;{6^{10}}{/tex}

(iii) {tex}{a^3} \times {a^2}{/tex}

(iv) {tex}{7^x} \times {7^2}{/tex}

(v) {tex}{\left( {{5^2}} \right)^2}{/tex}{tex} \div {/tex}{tex}{5^3}{/tex}

(vi) {tex}{2^5} \times {5^5}{/tex}

(vii) {tex}{a^4} \times {b^4}{/tex}

(viii) {tex}{\left( {{3^4}} \right)^3}{/tex}

(ix) ({tex}{2^{20}}{/tex}{tex} \div {/tex}{tex}{2^{15}}{/tex}) x{tex}{2^3}{/tex}

(x) {tex}{8^t} \div {8^2}{/tex}

(i) {tex}{3^2} \times {3^4} \times {3^8} = {3^{\left( {2 + 4 + 8} \right)}} = {3^{14}}{/tex} {tex}\left[ {\because {\text{ }}{a^m} \times {a^n} = {a^{m + n}}} \right]{/tex}

(ii) {tex}{6^{15}} \div {6^{10}} = {6^{15 – 10}} = {6^5}{/tex} {tex}\left[ {\because {\text{ }}{a^m} \div {a^n} = {a^{m – n}}} \right]{/tex}

(iii) {tex}{a^3} \times {a^2} = {a^{3 + 2}} = {a^5}{/tex} {tex}\left[ {\because {\text{ }}{a^m} \times {a^n} = {a^{m + n}}} \right]{/tex}

(iv) {tex}{7^x} \times {7^2} = {7^{x + 2}}{/tex} {tex}\left[ {\because {\text{ }}{a^m} \times {a^n} = {a^{m + n}}} \right]{/tex}

(v) {tex}{\left( {{5^2}} \right)^3} \div {5^3} = {5^{2 \times 3}} \div {5^3} = {5^6} \div {5^3}{/tex} {tex}\left[ {\because {\text{ }}{{\left( {{a^m}} \right)}^n} = {a^{m \times n}}} \right]{/tex}

=

(vi) {tex}{2^5} \times {5^5} = {\left( {2 \times 5} \right)^5} = {10^5}{/tex} {tex}\left[ {\because {\text{ }}{a^m} \times {b^m} = {{\left( {a \times b} \right)}^m}} \right]{/tex}

(vii) {tex}{a^4} \times {b^4} = {\left( {a \times b} \right)^4}{/tex} {tex}\left[ {\because {\text{ }}{a^m} \times {b^m} = {{\left( {a \times b} \right)}^m}} \right]{/tex}

(viii) {tex}{\left( {{3^4}} \right)^3}{/tex} = {tex}{3^{4 \times 3}} = {3^{12}}{/tex} {tex}\left[ {\because {\text{ }}{{\left( {{a^m}} \right)}^n} = {a^{m \times n}}} \right]{/tex}

(ix) {tex}\left( {{2^{20}} \div {2^{15}}} \right) \times {2^3}{/tex} = {tex}\left( {{2^{20 – 15}}} \right) \times {2^3}{/tex} {tex}\left[ {\because {\text{ }}{a^m} \div {a^n} = {a^{m – n}}} \right]{/tex}

=

(x) {tex}{8^t} \div {8^2} = {8^{t – 2}}{/tex} {tex}\left[ {\because {\text{ }}{a^m} \div {a^n} = {a^{m – n}}} \right]{/tex}

NCERT Solutions for Class 7 Maths Exercise 13.2

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

(i) {tex}\frac{{{2^3} \times {3^4} \times 4}}{{3 \times 32}}{/tex}

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

(iii) {tex}{25^4} \div {5^3}{/tex}

(iv) {tex}\frac{{3 \times {7^2} \times {{11}^8}}}{{21 \times 11}}{/tex}

(v) {tex}\frac{{{3^7}}}{{{3^4} \times {3^3}}}{/tex}

(vi) {tex}{2^0} + {3^0} + {4^0}{/tex}

(vii) {tex}{2^0} \times {3^0} \times {4^0}{/tex}

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

(ix) {tex}\frac{{{2^8} \times {a^5}}}{{{4^3} \times {a^3}}}{/tex}

(x) {tex}\left( {\frac{{{a^5}}}{{{a^3}}}} \right) \times {a^8}{/tex}

(xi) {tex}\frac{{{4^5} \times {a^8}{b^3}}}{{{4^5} \times {a^5}{b^2}}}{/tex}

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

(i) {tex}\frac{{{2^3} \times {3^4} \times 4}}{{3 \times 32}} = \frac{{{2^3} \times {3^4} \times {2^2}}}{{3 \times {2^5}}} = \frac{{{2^{3 + 2}} \times {3^4}}}{{3 \times {2^5}}}{/tex} {tex}\left[ {\because {\text{ }}{a^m} \times {a^n} = {a^{m + n}}} \right]{/tex}

= {tex}\frac{{{2^5} \times {3^4}}}{{3 \times {2^5}}} = {2^{5 – 5}} \times {3^{4 – 3}}{/tex} {tex}\left[ {\because {\text{ }}{a^m} \div {a^n} = {a^{m – n}}} \right]{/tex}

= {tex}{2^0} \times {3^3}{/tex} = {tex}1 \times {3^3} = {3^3}{/tex}

(ii) {tex}\left[ {{{\left( {{5^2}} \right)}^3} \times {5^4}} \right] \div {5^7} = \left[ {{5^6} \times {5^4}} \right] \div {5^7}{/tex} {tex}\left[ {\because {\text{ }}{{\left( {{a^m}} \right)}^n} = {a^{m \times n}}} \right]{/tex}

= {tex}\left[ {{5^{6 + 4}}} \right] \div {5^7} = {5^{10}} \div {5^7}{/tex} {tex}\left[ {\because {\text{ }}{a^m} \times {a^n} = {a^{m + n}}} \right]{/tex}

= {tex}{5^{10 – 7}} = {5^3}{/tex} {tex}\left[ {\because {\text{ }}{a^m} \div {a^n} = {a^{m – n}}} \right]{/tex}

(iii) {tex}{25^4} \div {5^3} = {\left( {{5^2}} \right)^4} \div {5^3} = {5^8} \div {5^3}{/tex} {tex}\left[ {\because {\text{ }}{{\left( {{a^m}} \right)}^n} = {a^{m \times n}}} \right]{/tex}

=

(iv) {tex}\frac{{3 \times {7^2} \times {{11}^8}}}{{21 \times {{11}^3}}} = \frac{{3 \times {7^2} \times {{11}^8}}}{{3 \times 7 \times {{11}^3}}} = {3^{1 – 1}} \times {7^{2 – 1}} \times {11^{8 – 3}}{/tex} {tex}\left[ {\because {\text{ }}{a^m} \div {a^n} = {a^{m – n}}} \right]{/tex}

= {tex}{3^0} \times {7^1} \times {11^5}{/tex} = {tex}7 \times {11^5}{/tex}

(v) {tex}\frac{{{3^7}}}{{{3^4} \times {3^3}}} = \frac{{{3^7}}}{{{3^{4 + 3}}}} = \frac{{{3^7}}}{{{3^7}}}{/tex} {tex}\left[ {\because {\text{ }}{a^m} \times {a^n} = {a^{m + n}}} \right]{/tex}

= {tex}{3^{7 – 7}} = {3^0} = 1{/tex} {tex}\left[ {\because {\text{ }}{a^m} \div {a^n} = {a^{m – n}}} \right]{/tex}

(vi) {tex}{2^0} + {3^0} + {4^0} = 1 + 1 + 1 = 3{/tex} {tex}\left[ {\because {\text{ }}{a^0} = 1} \right]{/tex}

(vii) {tex}{2^0} \times {3^0} \times {4^0} = 1 \times 1 \times 1 = 1{/tex} {tex}\left[ {\because {\text{ }}{a^0} = 1} \right]{/tex}

(viii) {tex}\left( {{3^0} + {2^0}} \right) \times {5^0} = \left( {1 + 1} \right) \times 1 = 2 \times 1 = 2{/tex} {tex}\left[ {\because {\text{ }}{a^0} = 1} \right]{/tex}

(ix) {tex}\frac{{{2^8} \times {a^5}}}{{{4^3} \times {a^3}}} = \frac{{{2^8} \times {a^5}}}{{{{\left( {{2^2}} \right)}^3} \times {a^3}}} = \frac{{{2^8} \times {a^5}}}{{{2^6} \times {a^3}}}{/tex} {tex}\left[ {\because {\text{ }}{{\left( {{a^m}} \right)}^n} = {a^{m \times n}}} \right]{/tex}

=

=

(x) {tex}\left( {\frac{{{a^5}}}{{{a^3}}}} \right) \times {a^8} = \left( {{a^{5 – 3}}} \right) \times {a^8} = {a^2} \times {a^8}{/tex} {tex}\left[ {\because {\text{ }}{a^m} \div {a^n} = {a^{m – n}}} \right]{/tex}

=

(xi) {tex}\frac{{{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}} = {4^0} \times {a^3} \times b{/tex} {tex}\left[ {\because {\text{ }}{a^m} \div {a^n} = {a^{m – n}}} \right]{/tex}

=

(xii) {tex}{\left( {{2^3} \times 2} \right)^2} = {\left( {{2^{3 + 1}}} \right)^2} = {\left( {{2^4}} \right)^2}{/tex} {tex}\left[ {\because {\text{ }}{a^m} \times {a^n} = {a^{m + n}}} \right]{/tex}

= {tex}{2^{4 \times 2}} = {2^8}{/tex} {tex}\left[ {\because {\text{ }}{{\left( {{a^m}} \right)}^n} = {a^{m \times n}}} \right]{/tex}

NCERT Solutions for Class 7 Maths Exercise 13.2

(i) {tex}10 \times {10^{11}}{/tex}={tex}{100^{11}}{/tex}

(ii) {tex}{2^3} > {\text{ }}{5^2}{/tex}

(iii){tex}{2^3} \times {3^2} = {6^5}{/tex}

(iv) {tex}{3^0} = {\text{ }}{\left( {1000} \right)^0}{/tex}

(i) {tex}10 \times {10^{11}} = {100^{11}}{/tex}

L.H.S. {tex}{10^{1 + 11}}{/tex} = {tex}{10^{12}}{/tex} and R.H.S. {tex}{\left( {{{10}^2}} \right)^{11}} = {10^{22}}{/tex}

Since, L.H.S. {tex} \ne {/tex} R.H.S.

Therefore, it is false.

(ii) {tex}{2^3} > {5^2}{/tex}

L.H.S. {tex}{2^3} = 8{/tex}andR.H.S.{tex}{5^2} = 25{/tex}

Since, L.H.S. is not greater than R.H.S.

Therefore, it is false.

(iii) {tex}{2^3} \times {3^2} = {6^5}{/tex}

L.H.S. {tex}{2^3} \times {3^2} = 8 \times 9 = 72{/tex} and R.H.S. {tex}{6^5} = 7,776{/tex}

Since, L.H.S. {tex} \ne {/tex} R.H.S.

Therefore, it is false.

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

L.H.S. {tex}{3^0} = 1{/tex} and R.H.S. {tex}{\left( {1000} \right)^0}{/tex} = 1

Since, L.H.S. = R.H.S.

Therefore, it is true.

NCERT Solutions for Class 7 Maths Exercise 13.2

###### Question 4.Express each of the following as a product of prime factors only i^n exponential form:

(i) 108 x 192

(ii) 270

(iii) 729 x 64

(iv) 768

(i) 108 x 192

= {tex}\left( {{2^2} \times {3^3}} \right) \times \left( {{2^6} \times 3} \right){/tex}

= {tex}{2^{2 + 6}} \times {3^{3 + 1}}{/tex}

= {tex}{2^8} \times {3^4}{/tex}

(ii) 270

= {tex}2 \times {3^5} \times 5{/tex}

(iii) 729 x 64

= {tex}{3^6} \times {2^6}{/tex}

(iv) 768

= {tex}{2^8} \times 3{/tex}

NCERT Solutions for Class 7 Maths Exercise 13.2

###### Question 5.Simplify:

(i) {tex}\frac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{8^3} \times 7}}{/tex}

(ii) {tex}\frac{{25 \times {5^2} \times {t^8}}}{{{{10}^3} \times {t^4}}}{/tex}

(iii) {tex}\frac{{{3^5} \times {{10}^5} \times 25}}{{{5^7} \times {6^5}}}{/tex}

(i) {tex}\frac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{8^3} \times 7}} = \frac{{{2^{5 \times 2}} \times {7^3}}}{{{{\left( {{2^3}} \right)}^3} \times 7}}{/tex}

= {tex}\frac{{{2^{10}} \times {7^3}}}{{{2^9} \times 7}}{/tex}

= {tex}{2^{10 – 9}} \times {7^{3 – 1}} = 2 \times {7^2}{/tex}

= 2 x 49

= 98

(ii) {tex}\frac{{25 \times {5^2} \times {t^8}}}{{{{10}^3} \times {t^4}}} = \frac{{{5^2} \times {5^2} \times {t^8}}}{{{{\left( {5 \times 2} \right)}^3} \times {t^4}}}{/tex}

= {tex}\frac{{{5^{2 + 2}} \times {t^{8 – 4}}}}{{{2^3} \times {3^3}}}{/tex}

= {tex}\frac{{{5^4} \times {t^4}}}{{{2^3} \times {5^3}}}{/tex}

= {tex}\frac{{{5^{4 – 3}} \times {t^4}}}{{{2^3}}}{/tex}

= {tex}\frac{{5{t^4}}}{8}{/tex}

(iii) {tex}\frac{{{3^5} \times {{10}^5} \times 25}}{{{5^7} \times {6^5}}}{/tex} = {tex}\frac{{{3^5} \times {{\left( {2 \times 5} \right)}^5} \times {5^2}}}{{{5^7} \times {{\left( {2 \times 3} \right)}^5}}}{/tex}

= {tex}\frac{{{3^5} \times {2^5} \times {5^5} \times {5^2}}}{{{5^7} \times {2^5} \times {3^5}}}{/tex}

= {tex}\frac{{{3^5} \times {2^5} \times {5^{5 + 2}}}}{{{5^7} \times {2^5} \times {3^5}}}{/tex}

= {tex}\frac{{{3^5} \times {2^5} \times {5^7}}}{{{5^7} \times {2^5} \times {3^5}}}{/tex}

= {tex}{2^{5 – 5}} \times {3^{5 – 5}} \times {5^{5 – 5}}{/tex}

= {tex}{2^0} \times {3^0} \times {5^0}{/tex}

= 1 x 1 x 1

= 1

## NCERT Solutions for Class 7 Maths Exercise 13.2

NCERT Solutions Class 7 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 7 Maths includes text book solutions from Class 7 Maths Book . NCERT Solutions for CBSE Class 7 Maths have total 15 chapters. 7 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 7 solutions PDF and Maths ncert class 7 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.

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