# Exercise 12.1 Exponents and Powers- NCERT Solutions Class 8

## Chapter 12 Ex.12.1 Question 1

Evaluate

(i) \begin{align}{3^{ - 2}}\end{align}

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

(iii) \begin{align}{\left( {\frac{1}{2}} \right)^{ - 5}}\end{align}

### Solution

(i) Evaluate $$\,{( - 4)^{ - 2}}$$

What is known?

Exponential form

What is unknown?

Value in the number

Reasoning:

We know \begin{align}{a^{ - m}} = \frac{1}{{{a^m}}}\end{align} for any non-zero integer

Steps:

\begin{align}{3^{ - 2}}&= \frac{1}{{{3^2}}}\\& = \frac{1}{9}\end{align}

Related Problems:

\begin{align}{( - 2)^{ - 4}}\end{align} & \begin{align}{(4)^{ - 3}}\end{align}

(ii) Evaluate \begin{align}{(-4)^{ - 2}}\end{align}

What is known?

Exponential form

What is unknown?

Value in the number

Reasoning:

We know \begin{align}{a^{ - m}} = \frac{1}{{{a^m}}}\end{align} for any non-zero integer

Steps:

\begin{align}{(-4)^{ - 2}}&= \frac{1}{{{{(-4)}^2}}} = \frac{1}{{(-4) \times (-4)}}\\&= \frac{1}{{16}}\end{align}

(iii) Evaluate \begin{align}{\left( {\frac{1}{2}} \right)^{ - 5}}\end{align}

What is known?

Exponential form

What is unknown?

Value in the number

Reasoning:

As per the fact used in previous question

Steps:

\begin{align}{\left( {\frac{1}{2}} \right)^{ - 5}} &= {(2)^5}\\&= 2 \times 2 \times 2 \times 2 \times 2\\&= 32\end{align}

## Chapter 12 Ex.12.1 Question 2

Simplify and express the result in power notation with positive exponent.

(i) \begin{align} {( - 4)^5} \div {( - 4)^8}\end{align}

(ii) \begin{align} {\left( {\frac{1}{{{2^3}}}} \right)^2}\end{align}

(iii) \begin{align} {( - 3)^4} \times {\left( {\fracabc{3}} \right)^4}\end{align}

(iv) \begin{align}\,\,({{3}^{-7}}\div {{3}^{-10}})\times {{3}^{-5}}\end{align}

(v) \begin{align} {2^{ - 3}} \times {( - 7)^{ - 3}}\end{align}

### Solution

(i) Evaluate \begin{align}{( - 4)^5} \div {( - 4)^8}\end{align}

What is known?

Expression in exponential form

What is unknown?

Result in power notation with positive exponent

Reasoning:

As we know \begin{align}\frac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\end{align} where \begin{align}m\end{align} & \begin{align}n\end{align} are integers.

Steps:

\begin{align}{{(-4)}^{5}}\div {{(-4)}^{8}}& =\frac{{{(-4)}^{5}}}{{{(-4)}^{8}}}={{(-4)}^{5-8}} \\ {{(-4)}^{-3}}&={{\left( \frac{1}{-4} \right)}^{3}}\end{align}

(ii) Evaluate \begin{align}\quad {\left( {\frac{1}{{{2^3}}}} \right)^2}\end{align}

What is known?

Expression in exponential form

What is unknown?

Result in power notation with positive exponent

Reasoning:

As we know for any non-zero integer a,\begin{align}{({a^{{m}}})^{{n}}} = {a^{{{mn}}}}\end{align}

Steps:

\begin{align}{\left( {\frac{1}{{{2^3}}}} \right)^2}&= \frac{1}{{{2^6}}}\end{align}

(iii) Evaluate \begin{align} {( - 3)^4} \times {\left( {\frac{5}{3}} \right)^4}\end{align}

What is known?

Expression in exponential form

What is unknown?

Result in power notation with positive exponent

Reasoning:

We know that \begin{align}{a^{{m}}} \times {b^{{m}}} = {(ab)^{{m}}}\end{align} where $$a$$ & $$b$$ are non-zero integers and $$m$$ is any integer

Steps:

\begin{align}& {{(-3)}^{4}}\times {{\left( \frac{5}{3} \right)}^{4}} \\& {{(-1\times 3)}^{4}}\times \frac{{{5}^{4}}}{{{3}^{4}}} \\& {{(-1)}^{4}}\times {{{\not\!{3}}}^{4}}\times \frac{{{5}^{4}}}{{{{\not\!{3}}}^{4}}} \\& {{(-1)}^{4}}\times {{5}^{4}}={{5}^{4}}\qquad[\because {{(-1)}^{4}}=1] \\\end{align}

(iv) Evaluate \begin{align}\,\,({{3}^{-7}}\div {{3}^{-10}})\times {{3}^{-5}}\end{align}

What is known?

Expression in exponential form

What is unknown?

Result in power notation with positive exponent

Reasoning:

We know \begin{align}\frac{{{{{a}}^{{m}}}}}{{{{{a}}^{{n}}}}}{{ = }}{{{a}}^{{{m - n}}}}\end{align} & \begin{align}{{{a}}^{{m}}}{{ \times }}{{{a}}^{{n}}}{{ = }}{{{a}}^{{{m + n}}}}\end{align}

Steps:

\begin{align}({3^{ - 7}} \! \div \! {3^{ - 10}}) \! \times \! {3^{ - 5}} & \! = \! ({3^{ - 7 - ( - 10)}}) \! \times \! {3^{ - 5}}\\& \! = \! ({3^{ - 7 + 10}}) \! \times \! {3^{ - 5}}\\& \! = \! {3^3} \! + \! ({3^{ - 5}})\\& \! = \! {3^{3 + ( - 5)}}\\& \! = \! {3^{ - 2}}\\& \! = \! \frac{1}{{{3}^{2}}}\end{align}

(v) Evaluate $$\quad {2^{ - 3}} \times {( - 7)^{ - 3}}$$

What is known?

Expression in exponential form

What is unknown?

Result in power notation with positive exponent

Reasoning:

As we know \begin{align} {{{a}}^{{m}}}\,{{ \times }}\,{{{b}}^{{m}}}{{ = }}{\left( {{{ab}}} \right)^{{m}}}\end{align}

Steps:

\begin{align}{2^{ - 3}} \! \times \! {( - 7)^{ - 3}} & \! = \! {[2 \! \times \! ( - 7)]^{ - 3}}\\& \! = \! [ - 14]^{ - 3} \\ \text{Since }&[{a^{ - m}} \! = \! \frac{1}{{{a^m}}}]\\& \! = \! {\left( {\frac{{ - 1}}{{14}}} \right)^3}\\\end{align}

## Chapter 12 Ex.12.1 Question 3

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}

### 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}

## Chapter 12 Ex.12.1 Question 4

Evaluate

(i)\begin{align}\, \frac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}}\end{align}

(ii)$$\, ({5^{ - 1}} \times {2^{ - 1}}) \times {6^{ - 1}}$$

### Solution

(i) Evaluate\begin{align}\, \frac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}}\end{align}

What is known?

Expression in exponential form

What is unknown?

Value of the expression

Reasoning:

\begin{align}{a^{ - m}}& = \frac{1}{{{a^m}}}\\{a^m} \div {a^n} &= {a^{m - n}}\end{align}

Steps:

\begin{align}&\frac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}}\\&= \frac{{{2^4} \times {5^3}}}{{{8^1}}} \quad \left[ {{a^{ - {{m}}}} = \frac{1}{{{a^{{m}}}}}} \right]\\&= \frac{{{2^4} \times {5^3}}}{{{2^3}}}\\&= {2^{4 - 3}} \times {5^3} \quad \left[ {{a^{{m}}} \div {a^{{n}}} = {a^{{{m - n}}}}} \right]\\&= 2 \times 125\\&= 250\end{align}

(ii) Evaluate $$\quad ({5^{ - 1}} \times {2^{ - 1}}) \times {6^{ - 1}}$$

What is known?

Expression in exponential form

What is unknown?

Value of the expression

Reasoning:

${a^{{m}}} \times {b^{{m}}} = {(ab)^{{m}}}$

Steps:

\begin{align}&({5^{ - 1}} \times {2^{ - 1}}) \times {6^{ - 1}} \\&= {10^{ - 1}} \times {6^{ - 1}} \\&= {(10 \times 6)^{ - 1}} \\&\qquad \left[ {\because \,{a^m} \times {b^m} = {{\left( {ab} \right)}^m}} \right] \\&= {(60)^{ - 1}} \\&= \frac{1}{{60}} \\\end{align}

Related problems:

(i) $$({7^{ - 2}} \times {14^{ - 2}}) \times {3^{ - 1}}$$

(ii) \begin{align}\frac{{{2^{ - 2}} \times {5^2}}}{{{8^{ - 3}}}}\end{align}

## Chapter 12 Ex.12.1 Question 5

Find the value of m for which \begin{align}({5^m} \div {5^{ - 3}}) = {5^5}\end{align}

### Solution

What is known?

Expression in exponential form

What is unknown?

Value of $$m$$

Reasoning:

${a^m} \div {a^n} = {a^{m - n}}$

Steps:

\begin{align}({5^m} \div {5^{ - 3}})&= {5^5}\\{5^{m - ( - 3)}} &= {5^5}\\{5^{m + 3}} &= {5^5}\end{align}

On both the side powers have the same base, so their exponents must be equal.

$$\therefore + 3 = 5 \to m = 5 - 3 \to m = 2$$

## Chapter 12 Ex.12.1 Question 6

Evaluate

(i) $\left[ {\left( {\frac{{\rm{1}}}{{\rm{3}}}} \right)^{{\rm{ - 1}}} - \left( {\frac{{\rm{1}}}{{\rm{4}}}} \right)^{{\rm{ - 1}}} } \right]^{ - 1}$

(ii) \begin{align}{\left[ {\fracabc{8}} \right]^{ - 7}} \! \times \! {\left[ {\frac{8}{5}} \right]^{ - 4}}\end{align}

### Solution

(i) Evaluate $\left[ {\left( {\frac{{\rm{1}}}{{\rm{3}}}} \right)^{{\rm{ - 1}}} - \left( {\frac{{\rm{1}}}{{\rm{4}}}} \right)^{{\rm{ - 1}}} } \right]^{ - 1}$

What is known?

Expression in exponential form

What is unknown?

Value of the expression

Reasoning:

\begin{align}{\left[ {\frac{a}{b}} \right]^{ - m}} \! \times \! {\left[ {\frac{b}{a}} \right]^{ m}}\end{align}

Steps:

\begin{align} \left[ {\left( {\frac{1}{3}} \right)^{ - 1} \! - \! \left( {\frac{1}{4}} \right)^{ - 1} } \right]^{ - 1} & \! = \! \left[ {3^1 \! - \! 4^1 } \right]^{ - 1} \\ & \! = \! \left[ {3 \! - \! 4} \right]^{ - 1} \\ & \! = \! \left[ { - 1} \right]^{ - 1} \\ & \! = \! \left[ {\frac{{ - 1}}{1}} \right]^1 \\ & \! = \! - 1 \\ \end{align}

(ii) Evaluate \begin{align}{\left[ {\frac{5}{8}} \right]^{ - 7}} \! \times \! {\left[ {\frac{8}{5}} \right]^{ - 4}}\end{align}

What is known?

Expression in exponential form

What is unknown?

Value of the expression

Reasoning:

\begin{align}{\left[ {\frac{a}{b}} \right]^{ - m}} \! \times \! {\left[ {\frac{b}{a}} \right]^{ m}}\end{align}

Steps:

\begin{align}{\left[ {\frac{5}{8}} \right]^{ - 7}} \! \times \! {\left[ {\frac{8}{5}} \right]^{ - 4}}& \! = \! {\left[ {\frac{8}{5}} \right]^7} \! \times \! {\left[ {\frac{8}{5}} \right]^{ - 4}}\\& \! = \! {\left[ {\frac{8}{5}} \right]^3}\\& \! = \! \frac{{512}}{{125}}\end{align}

## Chapter 12 Ex.12.1 Question 7

Simplify.

(i) \begin{align}\frac{{25 \times {t^{ - 4}}}}{{{5^{ - 3}} \times 10 \times {t^{ - 8}}}} \end{align}

(ii) \begin{align} \frac{{{3^{ - 5}} \times {{10}^{ - 5}} \times 125}}{{{5^{ - 7}} \times {6^{ - 5}}}}\end{align}

### Solution

(i) Evaluate \begin{align}\frac{{25 \times {t^{ - 4}}}}{{{5^{ - 3}} \times 10 \times {t^{ - 8}}}} \end{align}

What is known?

Expression in exponential form

What is unknown?

Value of the expression

Reasoning:

\begin{align}{a^{{m}}}{{ \times}}{a^{{n}}}{{ =}}{a^{{{m + n}}}} \quad \text{and} \quad \frac{{{a^{{m}}}}}{{{a^{{n}}}}}{{=}}{a^{{{m - n}}}} \end{align}

Steps:

\begin{align}\frac{{25 \times {t^{ - 4}}}}{{{5^{ - 3}} \times 10 \times {t^{ - 8}}}} &= \frac{{{5^2} \times {t^{ - 4}}}}{{{5^{ - 3}} \times 5 \times 2 \times {t^{ - 8}}}} \\\\& \left[ {{a^m} \times {a^n} = {a^{m + n}}} \right]\\\\ &= \frac{{{5^2} \times {t^{ - 4}}}}{{{5^{ - 3 + 1}} \times 2 \times {t^{ - 8}}}}\\&= \frac{{{5^2} \times {t^{ - 4}}}}{{{5^{ - 2}} \times 2 \times {t^{ - 8}}}}\\&= \frac{{{5^{2 - ( - 2)}} \times {t^{ - 4 - ( - 8)}}}}{2} \\\\& \left[ {\frac{{{a^m}}}{{{a^n}}} = {a^{m - n}}} \right]\\\\&= \frac{{{5^4} \times {t^{ - 4 + 8}}}}{2}\\&= \frac{{625{t^4}}}{2}\end{align}

(ii) Evaluate \begin{align} \frac{{{3^{ - 5}} \times {{10}^{ - 5}} \times 125}}{{{5^{ - 7}} \times {6^{ - 5}}}}\end{align}

What is known?

Expression in exponential form

What is unknown?

Value of the expression

Reasoning:

$\frac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\text{ and }{a^0} = 1$

Steps:

\begin{align}&\frac{{{3}^{-5}}\times {{10}^{-5}}\times 125}{{{5}^{-7}}\times {{6}^{-5}}}\\&=\frac{{{3}^{-5}}\times {{\left( 2\times 5 \right)}^{-5}}\times {{5}^{3}}}{{{5}^{-7}}\times {{(2\times 3)}^{-5}}} \\& =\frac{{{3}^{-5}}\times {{2}^{-5}}\times {{5}^{-5}}\times {{5}^{3}}}{{{5}^{-7}}\times {{2}^{-5}}\times {{3}^{-5}}} \\& ={{3}^{-5-(-5)}}\times {{2}^{-5-(-5)}}\times {{5}^{-5-(-7)+3}} \\& ={{3}^{0}}\times {{2}^{0}}\times {{5}^{5}}\qquad\left[ \because {{a}^{0}}=1 \right] \\& =1\times 1\times {{5}^{5}}={{5}^{5}}\end{align}

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