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

Go back to  'Exponents and Powers'

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

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

  
Download Cuemath NCERT App
Related Sections
Related Sections

Learn from the best math teachers and top your exams

Learn from the best

math teachers and top

your exams


Personalized Curriculum
Instant Doubts clarification
Cover latest CBSE Syllabus
Unlimited Mock & Practice tests
Covers CBSE, ICSE, IB curriculum

Instant doubt clearing with Cuemath Advanced Math Program
0