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Ex.13.2 Q5 Exponents and Powers Solutions - NCERT Maths Class 7

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Question

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

 Video Solution
Exponents And Powers
Ex 13.2 | Question 5

Text 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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