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Ex.13.2 Q2 Exponents and Powers Solutions - NCERT Maths Class 7

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Question

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

 Video Solution
Exponents And Powers
Ex 13.2 | Question 2

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

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

  
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