# Exercise 13.2 Exponents-and-Powers -NCERT Solutions Class 7

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## Chapter 13 Ex.13.2 Question 1

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

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

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

(iii) $${{\rm{a}}^3} \times {{\rm{a}}^2}$$

(iv) $${{7}^{x}}\times {{7}^{2}}$$

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

(vi) $${2^5} \times {\rm{ }}{5^5}$$

(vii) $${a^4} \times {b^4}$$

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

(ix) $$( 2^{20} \div 2^{15}) \times 2^3$$

(x) $${8^t} \div {\rm{ }}{8^2}$$

### Solution

What is known?

Exponential form of numbers.

What is unknown?

Exponential form by using laws of exponents.

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

(ii)  $${6^{15}} \div {6^{10}}$$

(iii)  $$a^3 \times a^2$$

(iv)  $${{7}^{x}}\times {{7}^{2}}$$

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

(vi) $${2^5} \times {\rm{ }}{5^5}$$

(vii)  $$a^4 \times b^4$$

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

(ix) $$( 2^{20} \div 2^{15}) \times 2^3$$

(x) $${8^t} \div {\rm{ }}{8^2}$$

Reasoning:

To solve this question, you must remember the laws of exponents. Here are some important laws of exponents:

1. $$a^m \times a^n = a^{m+n}$$

2. $$a^m \times b^m = (ab)^m$$

3.  $$a^m \div a^n = a^{m-n}$$

4. $$\left( a^m \right)^n=a^{mn}$$

5. $$a^o =1$$

Steps:

(i)

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

Using $$a^m \times a^n = a^{m+n}$$

(ii)

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

Using $$a^m \div a^n = a^{m-n}$$

(iii)

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

Using $$a^m \times a^n = a^{m+n}$$

(iv)

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

Using $$a^m \times a^n = a^{m+n}$$

(v)

\begin{align}(5^2)^3 \div 5^3 &= 5^6 \div 5^3\\&= 5^{6 - 3}\\&= 5^3\end{align}

Using $$\left( a^m \right)^n=a^{mn}$$  and  $$a^m \div a^n = a^{m-n}$$

(vi)

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

Using $$a^m \times b^m = (ab)^m$$

(vii)

$$a^4 \times b^4 = (ab)^4$$

Using $$\left( a^m \right)^n=a^{mn}$$

(viii)

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

Using $$\left( a^m \right)^n=a^{mn}$$

(ix)

\begin{align}\left( {{2}^{20}}\div {{2}^{15}} \right)\times {{2}^{3}} & ={{2}^{20-15}}\times{{2}^{3}} \\ & ={{2}^{5}}\times{{2}^{3}} \\& ={{2}^{8}} \\\end{align}

Using $$a^m \div a^n = a^{m-n}$$  and  $$a^m \times a^n = a^{m+n}$$

(x)

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

Using $$a^m \div a^n = a^{m-n}$$

## Chapter 13 Ex.13.2 Question 2

Simplify and express each of the following in exponential form:

(i) \begin{align}\frac{{{2^3} \times {3^3} \times 4}}{{3\,\times 32}}\end{align}

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

(iii) \begin{align} {25^4} \div\, {5^3}\end{align}

(iv) \begin{align} \frac{{3 \times {7^2} \times {{11}^8}}}{{21 \times {{11}^3}}}\end{align}

(v) \begin{align} \frac{{{3^{\bf{7}}}}}{{{3^4} \times {3^3}}}\end{align}

(vi) \begin{align} {2^0} + {3^0} + {4^0}\end{align}

(vii) \begin{align} {2^0} \times {3^0} \times {4^0}\end{align}

(viii) \begin{align} \left( {{3^0} + {2^0}} \right) \times {5^0}\end{align}

(ix) \begin{align} \frac {{{2^8} \times {a^5}}}{{4^3} \times {a^3}}\end{align}

(x) \begin{align} \frac{a^5}{a^3} \times a^8 \end{align}

(xi) \begin{align} \frac{{{4^5} \times {a^8}{b^3}}}{{4^5} \times \,{a^5}{b^2}}\end{align}

(xii) \begin{align} {\left( {{2^3} \times 2} \right)^2}\end{align}

### Solution

What is Known?

(i) \begin{align}\frac{{{2^3} \times {3^3} \times 4}}{{3 \times 32}}\end{align}

(ii) \begin{align} \left[ {\left( {{5^2}} \right) \times {5^4}} \right] \div{5^7}\end{align}

(iii) \begin{align} {25^4} \div {5^3}\end{align}

(iv) \begin{align} \frac{{3 \times {7^2} \times {{11}^8}}}{{21 \times {{11}^3}}}\end{align}

(v) \begin{align} \frac{{{3^{\bf{7}}}}}{{{3^4} \times {3^3}}}\end{align}

(vi) \begin{align} {2^0} + {3^0} + {4^0}\end{align}

(vii) \begin{align} {2^0} \times {3^0} \times 4\end{align}

(viii) \begin{align} \left( {{3^0} + {2^0}} \right) \times {5^0}\end{align}

(ix) \begin{align} \frac {{{2^8} \times {a^5}}}{{4^3} \times {a^3}}\end{align}

(x) \begin{align} \frac{{{{\rm{a}}^5}}}{{{\rm{a}}^3}} \times \,{{\rm{a}}^8}\end{align}

(xi) \begin{align} \frac{{{4^5} \times {a^8}{b^3}}}{{4^5} \times \,{a^5}{b^2}}\end{align}

(xii) \begin{align} {\left( {{2^3} \times 2} \right)^2}\end{align}

What is unknown?

Exponential form by using laws of exponents.

Reasoning:

To solve this question, you must remember the laws of exponents given in question $$1$$.

Steps:

(i) \begin{align}\frac{{{2^3} \times {3^3} \times 4}}{{3\,\times 32}}\end{align}

\begin{align}&= \frac{{{2^3} \! \times \! {3^4} \! \times \! {2^2}}}{{3 \! \times \! {2^5}}} &a^m \!\times\! a^n\!=\!a^{m + n}\\&= \frac{{{2^{3 + 2}} \! \times \! {3^4}}}{{{3^1} \! \times \! {2^5}}}&a^m\! \div\! a^n\!=\!a^{m -n}\\&= \frac{{{2^5} \! \times \! {3^{4 - 1}}}}{{3 \! \times \! {2^5}}}\\&= {3^3}\end{align}

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

\begin{align}&= \! \left[ {{5^6} \! \times \! {5^4}} \right] \! + \! {5^7} &a^m \! \times \! a^n \!=\! a^{m + n}\\&= \! \left[ {{5^{6 + 4}}} \right] \! \div \! {5^7} & a^m \! \div \! a^n \! =\! a^{m - n}\\&= \! [{5^{10}}] \! \div \! {5^7}\\ &= \! \left[ {{5^{10 - 7}}} \right]\\&= \! {5^3}\end{align}

(iii) \begin{align} {25^4} \div\, {5^3}\end{align}

\begin{align}&= {\left( {{5^2}} \right)^4} \div {5^3}\\&= {5^8} \div {5^3}\\&= {5^{8 - 3}} = {5^5}\end{align}

(iv) \begin{align}\frac{{3{\rm{ }} \times {7^2} \times {\rm{ }}{{11}^8}}}{{21\,\, \times {\rm{ }}{{11}^3}}}\end{align}

\begin{align}&= \!\frac{{3 \! \times \! {7^2} \! \times \! {{11}^8}}}{{3 \! \times \! 7 \! \times \! {{11}^3}}}\\&=\!\frac{{{7^2} \! \times \! {{11}^8}}}{{7 \! \times \! {{11}^3}}}\\&=\! {7^{2 - 1}} \! \times \! {\rm{ }}{11^{8 - 3}} \quad a^m \! \div \!a^n \!=\!a^{m - n}\\&=\!7 \! \times \! {11^5}\end{align}

(v) \begin{align}\frac{{{3^7}}}{{{3^4} \times {3^3}}}\end{align}

\begin{align}&=\!\!\frac{{{3^7}}}{{{3^{4 + 3}}}} \\&=\!\frac{{{3^7}}}{{{3^7}}} &\left[ a^m \! \times \! a^n \!= \! a^{m + n} \right]\\&=\! {3^{7 - 7}} &\left[ a^m \! \div \! a^n \!=\! a^{m - n} \right]\\&=\!{3^0}\\&=\!1\end{align}

(vi) \begin{align}{2^0} + {3^0} + {4^0}\end{align}

\begin{align}&= 1 + 1 + 1 \qquad a^0 = 1\\&= 3\end{align}

(vii) \begin{align}{2^0} \times {3^0} \times {4^0}\end{align}

\begin{align}&= 1 \times 1 \times 1 \qquad a^0 = 1\\&= 1\end{align}

(viii)  \begin{align}\left( {{3^0} + {2^0}} \right) \times {5^0} \end{align}

\begin{align}&= (1 + 1) \times 1 & & a^0 = 1\\&= 2 \times 1 \\&= 2\end{align}

(ix) \begin{align}\frac{{{2^8} \times {a^5}}}{{{4^3} \times {a^3}}}\end{align}

\begin{align}&= \! \frac{{{2^8} \! \times \! {a^5}}}{{{{\left( 2 \right)}^6} \! \times \! {a^3}}} & & a^m \div a^n \! = \! a^{m - n}\\&= \! {2^{8 - 6}} \! \times \! {a^{5 - 3}}\\ &= \! {2^2} \! \times \! {a^2}\\&= \! {(2a)^2}\end{align}

(x) \begin{align}\left( {\frac{{{a^5}}}{{{a^3}}}} \right) \times {a^8}\end{align}

\begin{align} &= \! \left( {{a^{5 - 3}}} \right) \! \times \! {a^8}& a^m \! \div \! a^n \! = \! a^{m - n}\\&= \! {a^2} \! \times \! {a^8}\\&= \! {a^{2 + 8}}& a^m \! \times a^n \! = \! a^{m + n}\\&= \! {a^{10}}\end{align}

(xi)  \begin{align}\frac{{{4^5} \times {a^8}{b^3}}}{{{4^5} \times {a^5}{b^2}}}\end{align}

\begin{align}&=\!{4^{5 - 5}}\!\times\!{a^{8 - 5}}\!\times\!{b^3} \quad a^m \! \div \!a^n\!=\!a^{m - n}\\&=\!{4^0}\!\times\!{a^3}\!\times\!{b^{3 - 2}}\, \quad\,\,\,\,\,a^0 = 1\\&=\!1\!\times\!{a^3} \!\times\! b \\&=\! {a^3}b\end{align}

(xii)\begin{align}{\left( {{2}^{3}}\times 2 \right)}^{2}\end{align}

\begin{align}& = \! {{\left( {{2}^{3+1}} \right)}^{2}}\\&=\!{{\left( {{2}^{4}} \right)}^{2}} \\&={{2}^{4\times 2}} \qquad \left( a^m \right)^n \!=\! a^{mn} \\ & = \! {{2}^{8}} \end{align}

## Chapter 13 Ex.13.2 Question 3

(i) $$10 × 10^{11} = 100^{11}$$

(ii) $$2^3 > 5^2$$

(iii) $$2^3 × 3^2 = 6^5$$

(iv) $$3^0 = 1000^0$$

### Solution

Steps:

(i)

\begin{align}10\,\,\times \,{{10}^{11}}={{100}^{11}} \end{align}

$$LHS$$ $$= 10 \times 10 ^{11} = 10^{11+1}=10^{12}$$

$$RHS$$ $$=$$ $$100^{11}=(10^2)^{11}=10^{22}$$

$$\therefore 10^{12} \ne 10^{22}$$

$$\therefore$$Thus,the statement is false.

(ii)

\begin{align}{{2}^{3}}>\text{ }{{5}^{2}}\end{align}

$$LHS \rm= 2^3=2 \times2\times2=8$$

$$RHS \rm = 5^2=5\times5=25$$

$$\therefore 2^3<5^2$$

Thus,the statement is false.

(iii)

\begin{align}{{2}^{5}}\rm x\text{ }{{3}^{5}}=\text{ }{{6}^{5}}\end{align}

$$LHS \rm\!=\!2^3\!\times\!3^2\!=\!2 \!\times\!2\!\times\!2\!\times\!3\!\times\!3\!=\!72$$

$$RHS \rm\!=\! 6^5\!=6\!\times\!6\!\times\!6\!\times\!6\!\times\!6\!=\!7776$$

$$\therefore 2^3 \times 3^2\ne 6^5$$

Thus,the statement is false.

(iv)

\begin{align}\,{3^0} = {\rm{ }}{1000^0} = {\rm{ }}1\end{align}

Thus ,the statement is true.

## Chapter 13 Ex.13.2 Question 4

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

(i) $$108 × 192$$

(ii) $$270$$

(iii) $$729 × 64$$

(iv) $$768$$

### Solution

What is Known?

(i) $$108 × 192$$

(ii) $$270$$

(iii) $$729 × 64$$

(iv) $$768$$

What is unknown?

Product of prime factors only in exponential form.

Reasoning:

To solve this question, you must remember the laws of exponents. Here are some important laws of exponents

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

2.  $${{{a}}^{{m}}}{{ \div }}\,\,{{{a}}^{{n}}} = {{{a}}^{{{m - n}}}}$$

3.  $${{\left( {{{{a}}^{{m}}}} \right)}^{{n}}} = {{{a}}^{{{mn}}}}$$

4.  $${{{a}}^{{0}}}= 1$$

Steps:

(i) $$108 \times192$$

\begin{align} \\&=2\!\!\times\!\!2\!\! \times\!\! 2\!\! \times\!\! 2\!\!\times\!\!2\!\!\times\!\!2\!\!\times\!\!\,2\!\! \times\!\!2\!\!\times\!\!3\!\!\times\!\!3\!\!\times\!\!3\!\!\times\!\!3\\&\qquad {a^m}\!\!\times\!\!{a^n}\!\!=\!\!{a^{m + n}}\\&={2^8}\!\!\times\!\!{3^4}\end{align}

(ii) $$270$$

\begin{align}\\&=2 \!\!\times\!\!3\!\!\times\!\!3\!\!\times\!\!3\!\!\times\!\!5 \quad {{{a}}^{{m}}}\!\!{{ \times }}{{{a}}^{{n}}}\,\!\!{{ = }}\,{{{a}}^{{{m + n}}}}\\&= 2 \times {3^3} \times 5 \\ &= 10 \times 3^3\end{align}

(iii) \begin{align}729 \times 64\end{align}

\begin{align}\\&\!\!\!\!=\!\!3\!\!\times\!\!3\!\!\times\!\!3\!\!\times\!\!3\!\!\times\!\!3\!\!\times\!\!3\!\!\times\!\!2\!\!\times\!\!2\!\!\times\!\!2\!\!\times\!\!2\!\!\times\!\!2\!\!\times\!\!2\\&\qquad{a^m}\!\!\times\!\!{a^n}\!\!=\!\!{a^{m + n}}\\&=\!\!{3^6}\!\!\times\!\!{2^6}\\&= {\left( {3 \times 2} \right)^6}\end{align}

(iv)$$768$$

\begin{align}\\&= \!2 \!\times\! 2 \!\times\! 2 \!\times\! 2 \!\times\! 2 \!\times\! 2 \!\times\! 2 \!\times\! 2 \!\times\! 3\\&\qquad{{{a}}^{{m}}}{{ \times }}{{{a}}^{{n}}}{{ = }}\,{{{a}}^{{{m + n}}}}\\&= {2^8} \times 3\end{align}

## Chapter 13 Ex.13.2 Question 5

Simplify:

(i) \begin{align}\frac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{8^3} \times 7}}\end{align}

(ii) \begin{align}\frac{{25 \times {5^2} \times {t^8}}}{{{{10}^3} \times {t^4}}} \end{align}

(iii) \begin{align}{\rm{ }}\,\frac{{{{\rm{3}}^{\rm{5}}} \times {\rm{ 1}}{{\rm{0}}^{\rm{5}}} \times {\rm{ 25 }}}}{{{{\rm{5}}^{\rm{7}}} \times {\rm{ }}{{\rm{6}}^{\rm{5}}}}}{\rm{ }} \end{align}

### Solution

Steps:

(i) \begin{align}\frac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{8^3} \times 7}}\end{align}

\begin{align}&=\frac {2^{5 \times 2\,}\times \;7^{3}}{(2^3)^3 \;\times \,7 } \\&=\frac {2^{10}\;\times\; 7^{3-1}}{2^9 } \quad[a^m\!\!\div\!\!a^n\!\!=\!\!a^{m - n}]\\&= {2^{10 - 9}}\!\!\times\!\!\,\,{7^{3 - 1}}\,\!\!=\!\!\,2^1 \times \,7^2\\&= 98\end{align}

(ii) \begin{align}\frac{{25 \times {5^2} \times {t^8}}}{{{{10}^3} \times {t^4}}} \end{align}

\begin{align}&=\frac{{{\left( 5 \right)}^{2}}\times {{5}^{2}}\times {{t}^{8}}}{{{(2\times 5)}^{3}}\times {{t}^{4}}} \\ &=\frac{{{\left( 5 \right)}^{2+2}}\times {{t}^{8}}}{{{2}^{3}}\times {{5}^{3}}\times {{t}^{4}}}\\& \text{Using }{a^m} \times {a^n} = {a^{m + n}} \\&=\frac{{{\left( 5 \right)}^{4}}\times {{t}^{8}}}{{{2}^{3}}\times {{5}^{3}}\times {{t}^{4}}} \\\\&=\frac{{{\left( 5 \right)}^{4-3}}\times {{t}^{8-4}}}{8}\\& \text{Using }{a^m} \div {a^n} = {a^{m - n}}\\&=\frac{5\times {{t}^{4}}}{8} \\\end{align}

(iii) \begin{align}{\rm{ }}\,\frac{{{{\rm{3}}^{\rm{5}}} \times {\rm{ 1}}{{\rm{0}}^{\rm{5}}} \times {\rm{ 25 }}}}{{{{\rm{5}}^{\rm{7}}} \times {\rm{ }}{{\rm{6}}^{\rm{5}}}}}{\rm{ }}\end{align}

\begin{align} &= \frac{{{3^5} \times {{\left( {2 \times 5} \right)}^5} \times \left( {5 \times 5} \right)}}{{{5^7} \times {{\left( {2 \times 3} \right)}^5}}}\\ &= \frac{{{3^5} \times {2^5} \times {5^5} \times 5^2}}{{{5^7} \times {2^5} \times {3^5}}}\\ &= \frac{{{3^5} \times {2^5} \times {5^7}}}{{\,\,\,{5^7} \times {2^5} \times {3^{5}}}} \\& \text{Using }{a^m} \times {a^n} = {a^{m + n}}\\\\&= {3^{5 - 5}} \times {2^{5 - 5}} \times {5^{7 - 7}} \\& \text{Using }{a^m} \div {a^n} = {a^{m - n}}\\\\& = {3^0} \times {2^0} \times {5^{0}} \\ & \text{Using }{a^{0}} = 1\\& = 1\end{align}

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