# Ch. - 12 Exponents and Powers

## 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( {\frac{5}{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[ {\frac{5}{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}

The chapter 12 begins with an introduction to Exponents and Powers with an example of the mass of the earth. Then, the powers with negative exponents are discussed with relevant examples.Following this,  the Laws of Exponents involving negative exponents are explained in the next section.In this section, the standard identities are also mentioned.The use of exponents to express small numbers in standard form is described in detail in the next section.Finally, comparing very large and very small numbers are briefly explained in the last part of the chapter.

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