Ex.13.2 Q1 Exponents and Powers Solution- NCERT Maths Class 7

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Question

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) \({{\rm{a}}^{\rm{4}}}{\rm{ \times }}{{\rm{b}}^{\rm{4}}}\) 

(viii) \({\left( {{3^4}} \right)^3}\)

(ix) \(\left( {{2^{20}} \div {\rm{ }}{2^{15}}} \right){\rm{ }}\,x{\rm{ }}\,{2^3}\)

 (x) \({8^t} \div {\rm{ }}{8^2}\)

Text 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)  \({{\rm{a}}^3} \times {{\rm{a}}^2}\)

(iv)  \({{7}^{x}}\times {{7}^{2}}\)

(v) \({\left( {{5^2}} \right)^3} \div {5^3}\)

(vi) \({2^5} \times {\rm{ }}{5^5}\)

(vii)  \({{\rm{a}}^{\rm{4}}}{\rm{ \times }}{{\rm{b}}^{\rm{4}}}\)

(viii) \({\left( {{3^4}} \right)^3}\)

(ix) \(\left( {{2^{20}} \div {\rm{ }}{2^{15}}} \right){\rm{ }}\,x{\rm{ }}\,{2^3}\)

 (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. \({{\rm{a}}^{\rm{m}}} \times \,{{\rm{a}}^{\rm{n}}} = {{\rm{a}}^{{\rm{m}} + {\rm{n}}}}\)

2. \({{\rm{a}}^{\rm{m}}}{\rm{ \times }}\,{{\rm{b}}^{\rm{m}}}{\rm{ = }}\,{{\rm{(ab)}}^{\rm{m}}}\)

3.  \({{\rm{a}}^{\rm{m}}}{{\rm{ \div }}^{}}{{\rm{a}}^{\rm{n}}}\,{\rm{ = }}\,{{\rm{a}}^{{\rm{m - n}}}}\)

4. \({\left( {{{\rm{a}}^{\rm{m}}}} \right)^{\rm{n}}}\,{\rm{ = }}\,{{\rm{a}}^{{\rm{mn}}}}\)

5. \({{\rm{a}}^{\rm{o}}}\,{\rm{ = 1}}\)

Steps:

(i) \({3^2} \times {3^4} \times {3^8} = {3^{2 + 4 + 8}} = {3^{14}}\)

\(\left[ {{{\rm{a}}^{\rm{m}}}\,{\rm{ \times }}\,{{\rm{a}}^{\rm{n}}}{\rm{ = }}\,{{\rm{a}}^{{\rm{m + n}}}}} \right]\)

(ii) \({6^{15}} \div {6^{10}} = {6^{15 - 10}} = {6^5}\)

\(\left[ {{{\rm{a}}^{\rm{m}}}{\rm{ \div }}\,\,{{\rm{a}}^{\rm{n}}}{\rm{ = }}{{\rm{a}}^{{\rm{m - n}}}}} \right]\)

(iii) \({{\rm{a}}^3} \times {{\rm{a}}^2} = {{\rm{a}}^{3 + 2}} = {{\rm{a}}^5}\)

\(\left[ {{{\rm{a}}^{\rm{m}}}{\rm{ \times }}{{\rm{a}}^{\rm{n}}}\,{\rm{ = }}\,{{\rm{a}}^{{\rm{m + n}}}}} \right]\)

(iv) \(7x \times {7^2} = 7x + 2\)

\(\left[ {{{\rm{a}}^{\rm{m}}}{\rm{ \times }}\,\,{{\rm{a}}^{\rm{n}}}\,{\rm{ = }}\,{{\rm{a}}^{{\rm{m + n}}}}} \right]\)

\(\begin{align}{\left( {\rm{v}} \right)\,\,{{\left( {{5^2}} \right)}^3} \div }{5^3} &= {5^6} \div {5^3}\\{}&{ = {5^{6 - 3}}}\\{}&{ = {5^3}}\end{align}\)

\[\begin{align}\text{ }\!\!~\!\!\text{ }&\left[ {{\left( {{\text{a}}^{\text{m}}} \right)}^{\text{n}}}\text{= }{{\text{a}}^{\text{mn}}} \right]  \\\text{ }\!\!~\!\!\text{ }&\left[ {{\text{a}}^{\text{m}}}\text{ }\!\!\div\!\!\text{ }\,\,\,{{\text{a}}^{\text{n}}}\text{= }{{\text{a}}^{\text{m - n}}} \right]  \\\end{align}\]

(vi) \({2^5} \times {5^5} = {(2\rm x 5)^5} = {(10)^5}\)

\(\left[ {{{\rm{a}}^{\rm{m}}} \times \,\,{{\rm{b}}^{\rm{m}}}\,{\rm{ = }}\,{{\left( {{\rm{ab}}} \right)}^{\rm{m}}}} \right]\)

(vii) \({{\rm{a}}^4} \times {{\rm{b}}^4} = {({\rm{ab}})^4}\)

\[\begin{align}\left[ {{\text{a}}^{\text{m}}}\text{ }\!\!\times\!\!\text{ }\,\,{{\text{b}}^{\text{m}}}\text{= }{{\left( \text{ab} \right)}^{\text{m}}} \right] \\\text{ }\!\!~\!\!\text{ } \\\end{align}\]

(viii) \({\left( {{3^4}} \right)^3} = {(3)^{12}}\)

\[\left[ {{{\left( {{{\rm{a}}^{\rm{m}}}} \right)}^{\rm{n}}}{\rm{ = }}{{\rm{a}}^{{\rm{mn}}}}} \right]\]

(ix) 

\(\begin{align}\left( {{2}^{20}}\div {{2}^{15}} \right)\times {{2}^{3}} & ={{2}^{20-15}}\text{x}{{2}^{3}}  \\
& ={{2}^{5}}\text{x}{{2}^{3}}  \\& ={{2}^{8}}  \\\end{align}\)

 

 

\[\begin{align} \left[ {{\text{a}}^{\text{m}}}\text{ }\!\!\div\!\!\text{ }\,\,{{\text{a}}^{\text{n}}}\text{= }{{\text{a}}^{\text{m - n)}}} \right]  \\\left[ {{\text{a}}^{\text{m}}}\text{ }\!\!\times\!\!\text{ }\,\,{{\text{a}}^{\text{n}}}\text{= }{{\text{a}}^{\text{m + n}}} \right]  \\\end{align}\]

 

(x) \({8^t} \div {8^2} = {8^{t - 2}}\)

\(\left[ {{{\rm{a}}^{\rm{m}}}{\rm{ \div }}{{\rm{a}}^{\rm{n}}}\,{\rm{ = }}\,{{\rm{a}}^{{\rm{m - n}}}}} \right]\)

  
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