# Ex.12.1 Q3 Exponents and Powers Solution - NCERT Maths Class 8

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## Question

Find the value of

(i) \begin{align} ({3^0} \times {4^{ - 1}}) \times {2^{ - 2}} \end{align}

(ii) \begin{align} ({2^{ - 1}} \times {4^{ - 1}}) \div {2^{ - 2}} \end{align}

(iii) \begin{align} {\left( {\frac{1}{2}} \right)^{ - 2}} + {\left( {\frac{1}{3}} \right)^{ - 2}} + {\left( {\frac{1}{4}} \right)^{ - 2}} \end{align}

(iv) \begin{align} {({3^{ - 1}} + {4^{ - 1}} + {5^{ - 1}})^0}\end{align}

(v) \begin{align} {\left\{ {{{\left( {\frac{{ - 2}}{3}} \right)}^{ - 2}}} \right\}^2}\end{align}

Video Solution
Exponents And Powers
Ex 12.1 | Question 3

## Text Solution

(i)\begin{align} \,({3^{\circ}} \times {4^{ - 1}}) \times {2^{ - 2}} \end{align}

What is known?

Expression in exponential form

What is unknown?

Value of the expression

Reasoning:

$${a^0} = 1$$ and \begin{align}{a^{ - {m}}} = \frac{1}{{{a^{m}}}} \end{align}

Steps:

\begin{align}&({3^0} \times {4^{ - 1}}) \times {2^2}\\&= (1 + \frac{1}{4}) \times {2^2}\\&= \left( {\frac{{4 + 1}}{4}} \right) \times {2^2}\\&= \left( {\frac{5}{4}} \right) \times {2^2}\\&= \frac{5}{{{2^2}}} \times {2^2}\,\,\,\,\,[4 = 2 \times 2 = {2^2}]\\&= \,\,\,5\end{align}

(ii) \begin{align}\,({2^{ - 1}} \times {4^{ - 1}}) \div {2^{ - 2}} \end{align}

What is known?

Expression in exponential form

What is unknown?

Value of the expression

Reasoning:

\begin{align}&{\left( {a^m} \right)^n} \!=\! {a^{mn}},{a^m} \!\times \!{a^n} \!=\! {a^{m + n}},\\&{a^{ - m}}\! = \!\frac{1}{a^m} \end{align}

Steps:

\begin{align}&\left( {{2^{ - 1}} \times {4^{ - 1}}} \right) \div {2^{ - 2}}\\& = \left[ {{2^{ - 1}} \times {{\left\{ {{{\left( 2 \right)}^2}} \right\}}^{ - 1}}} \right] \div {2^{ - 2}} \\&= \left( {{2^{ - 1}} \!\times \!{2^{ - 2}}} \right) \!\div\! {2^{ - 2}} \\& \qquad \left[ {\because\;{a^m}\! \times \!{a^n} \!= \!{a^{m + n}}} \right] \\&= {2^{ - 3}} \div {2^{ - 2}} \\&= {2^{ - 3 - \left( { - 2} \right)}} \\ & \qquad \left[ {\because \;{a^m} \div {a^n} = {a^{m - n}}} \right] \\&= {2^{ - 3 + 2}} \\&= {2^{ - 1}} \\&= \frac{1}{2} \\ & \qquad \left[ {\because \;{a^{-m}} = \frac{1}{{{a^m}}}} \right] \\\end{align}

(iii) \begin{align}\, {\left( {\frac{1}{2}} \right)^{ - 2}} + {\left( {\frac{1}{3}} \right)^{ - 2}} + {\left( {\frac{1}{4}} \right)^{ - 2}} \end{align}

What is known?

Expression in exponential form

What is unknown?

Value of the expression

Reasoning:

\begin{align}{\left( {\frac{a}{b}} \right)^{ - {\rm{m}}}} = {\left( {\frac{b}{a}} \right)^{\rm{m}}}\end{align}

Steps:

\begin{align}&\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2} \\&=\left(\frac{2}{1}\right)^{2}+\left(\frac{3}{1}\right)^{2}+\left(\frac{4}{1}\right)^{2} \\ &=(2)^{2}+(3)^{2}+(4)^{2} \\ &=4+9+16 \\ &=29 \end{align}

(iv) \begin{align} \quad {({3^{ - 1}} + {4^{ - 1}} + {5^{ - 1}})^0}\end{align}

What is known?

Expression in exponential form

What is unknown?

Value of the expression

Reasoning:

\begin{align}{a^0} = 1\end{align} and $${a^{ - {\rm{m}}}} = \frac{1}{{{a^{\rm{m}}}}}$$

Steps:

\begin{align}&{\left( {{3^{ - 1}} + {4^{ - 1}} + {5^{ - 1}}} \right)^0}\\ &= {\left[ {\frac{1}{3} + \frac{1}{4} + \frac{1}{5}} \right]^0}\quad \left[ {{a^{ - m}} = \frac{1}{{{a^m}}}} \right]\\&= 1 \qquad \qquad \left[ {{a^0} = 1} \right]\end{align}

(v) \begin{align} \quad {\left\{ {{{\left( {\frac{{ - 2}}{3}} \right)}^{ - 2}}} \right\}^2}\end{align}

What is known?

Expression in exponential form

What is unknown?

Value of the expression

Reasoning:

\begin{align}{a^{ - {\rm{m}}}} = \frac{1}{{{a^{\rm{m}}}}}\end{align} and \begin{align}{\left( {\frac{a}{b}} \right)^{\rm{m}}} = \frac{{{a^{\rm{m}}}}}{{{b^{\rm{m}}}}} \end{align}

Steps:

\begin{align}&{\left\{ {{{\left( {\frac{{ - 2}}{3}} \right)}^{ - 2}}} \right\}^2}\\& = {\left\{ {{{\left( {\frac{3}{{ - 2}}} \right)}^2}} \right\}^2} \qquad \left[ {{a^{ - {\rm{m}}}} = \frac{1}{{{a^m}}}} \right]\\&= {\left\{ {\frac{{{3^2}}}{{{{\left( { - 2} \right)}^2}}}} \right\}^2} \qquad \left[ {{{\left( {\frac{a}{b}} \right)}^{\rm{m}}} = \frac{{{a^{\rm{m}}}}}{{{b^{\rm{m}}}}}} \right]\\&= {\left( {\frac{9}{4}} \right)^2}\\&= \frac{{81}}{{16}}\end{align}

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