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

Video 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

Video 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

Say true or false and justify your answer:

(i) \(10 × 10^{11} = 100^{11}\)

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

(iii) \(2^3 × 3^2 = 6^5\)

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

Solution

Video 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

Video 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

Video 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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