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

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Question

Q2. 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) \(\begin{align} \frac{{{{\rm{a}}^5}}}{{{\rm{a}}^3}} \times \,{{\rm{a}}^8}\end{align} \)

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

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

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) \[\begin{align}\frac{{{2^3} \times {3^4} \times 4}}{{3 \times 32}} &= \frac{{{2^3} \times {3^4} \times {2^2}}}{{3 \times {2^5}}} &{{\rm{a}}^{\rm{m}}}{{\rm{ \times }}^{}}{{\rm{a}}^{\rm{n}}}= {\rm{ }}{{\rm{a}}^{{\rm{m + n}}}}\\&= \frac{{{2^{3 + 2}} \times {3^4}}}{{{3^1} \times {2^5}}}&{\rm{}}{{\rm{a}}^{\rm{m}}} \div {{\rm{a}}^{\rm{n}}} = {{\rm{a}}^{{\rm{m}} - {\rm{n}}}}\\&= \frac{{{2^5} \times {3^{4 - 1}}}}{{3 \times {2^5}}}\\&= {3^3}\end{align}\]

(ii) \[\begin{align}\left[ {{{\left( {{5^2}} \right)}^3} \times {5^4}} \right] \div {5^7} &= \left[ {{5^6} \times {5^4}} \right] + {5^7} &{{\rm{a}}^{\rm{m}}}{\rm{ \times }}\,\,{{\rm{a}}^{\rm{n}}}{\rm{ =  }}{{\rm{a}}^{{\rm{m + n}}}}\\&= \left[ {{5^{6 + 4}}} \right] \div {5^7} &{{\rm{a}}^{\rm{m}}}{\rm{ \div }}\,\,{{\rm{a}}^{\rm{n}}}{\rm{ =  }}{{\rm{a}}^{{\rm{m - n}}}}\\&= [{5^{10}}] \div {5^7}\\ &= \left[ {{5^{10 - 7}}} \right]\\&= {5^3}\end{align}\]

(iii) \[\begin{align}{25^4} \div {5^3} &= {\left( {{5^2}} \right)^4} \div {5^3}\\&= {5^8} \div {5^3}\\&= {5^{8 - 3}} = {5^5}\end{align}\]

(iv) \[\begin{align}\frac{{3{\rm{ }} \times {7^2} \times {\rm{ }}{{11}^8}}}{{21\,\, \times {\rm{ }}{{11}^3}}} &= \frac{{3{\rm{ }} \times {7^2} \times {\rm{ }}{{11}^8}}}{{3\, \times {\rm{ }}7{\rm{ }} \times {\rm{ }}{{11}^3}}}\\&= \frac{{{7^2} \times {\rm{ }}{{11}^8}}}{{7{\rm{ }} \times {\rm{ }}{{11}^3}}}\\&= {7^{2 - 1}}x{\rm{ }}{11^{8 - 3}} &  & {{\rm{a}}^{\rm{m}}}{\rm{ \div }}\,\,{{\rm{a}}^{\rm{n}}}{\rm{ =  }}{{\rm{a}}^{{\rm{m - n}}}}\\&= 7{\rm{ }}x{\rm{ }}{11^5}\end{align}\]

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

(vi) \[\begin{align}{2^0} + {3^0} + {4^0}\rm&= 1 + 1 + 1 &  & {{\rm{a}}^0} = 1\\&= 3\end{align}\]

(vii) \[\begin{align}{2^0} \times {3^0} \times {4^0} &= \rm{ }1 \times 1 \times 1 &  & {{\rm{a}}^0} = 1\\&= 1\end{align}\]

(viii) \[\begin{align}\left( {{3^0} + {2^0}} \right) \times {5^0} &= (1 + 1) \times 1 & & {{\rm{a}}^0} = {\rm{ }}1\\&= 2 \times 1 = 2\end{align}\]

(ix) \[\begin{align}\frac{{{2^8} \times {a^5}}}{{{4^3} \times {a^3}}} &= \frac{{{2^8} \times {a^5}}}{{{{\left( 2 \right)}^6} \times {a^3}}} &  & {{\rm{a}}^{\rm{m}}}{\rm{ \div }}{{\rm{a}}^{\rm{n}}}{\rm{ =  }}{{\rm{a}}^{{\rm{m - n}}}}\\&= {2^{8 - 6}} \times {a^{5 - 3}} = {2^2} \times {a^2}\\&= {(2a)^2}\end{align}\]

(x) \[\begin{align}\left( {\frac{{{a^5}}}{{{a^3}}}} \right) \times {a^8} &= \left( {{a^{5 - 3}}} \right) \times {a^8}& {{\rm{a}}^{\rm{m}}}{\rm{ \div }}\,\,{{\rm{a}}^{\rm{n}}}{\rm{ =  }}{{\rm{a}}^{{\rm{m - n}}}}\\&= {a^2} \times {a^8}\\&= {a^{2 + 8}}&{{\rm{a}}^{\rm{m}}}{\rm{ \times }}\,{{\rm{a}}^{\rm{n}}}{\rm{ =  }}{{\rm{a}}^{{\rm{m + n}}}}\\&= {a^{10}}\end{align}\]

(xi) \[\begin{align}\frac{{{4^5} \times {a^8}{b^3}}}{{{4^5} \times {a^5}{b^2}}} &= {4^{5 - 5}} \times {a^{8 - 5}} \times{b^3} &  & {{\rm{a}}^{\rm{m}}}{\rm{ \div }}\,{{\rm{a}}^{\rm{n}}}{\rm{ =  }}{{\rm{a}}^{{\rm{m - n}}}}\\&= {4^0} \times {a^3} \times {b^{3 - 2}}\, &  & \,\,\,\,\,{{\rm{a}}^{\rm{0}}}{\rm{ = 1}}\\&= 1 \times {a^3} \times b = {a^3}b\end{align}\]

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

  
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