NCERT Class 8 Maths Exponents and Powers
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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Exponents and Powers
Exercise 12.1
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:
\[\begin{align}&\frac{{{3}^{-5}}\times {{10}^{-5}}\times 125}{{{5}^{-7}}\times {{6}^{-5}}}\\&=\frac{{{3}^{-5}}\times {{\left( 2\times 5 \right)}^{-5}}\times {{5}^{3}}}{{{5}^{-7}}\times {{(2\times 3)}^{-5}}} \\& =\frac{{{3}^{-5}}\times {{2}^{-5}}\times {{5}^{-5}}\times {{5}^{3}}}{{{5}^{-7}}\times {{2}^{-5}}\times {{3}^{-5}}} \\& ={{3}^{-5-(-5)}}\times {{2}^{-5-(-5)}}\times {{5}^{-5-(-7)+3}} \\& ={{3}^{0}}\times {{2}^{0}}\times {{5}^{5}}\qquad\left[ \because {{a}^{0}}=1 \right] \\& =1\times 1\times {{5}^{5}}={{5}^{5}}\end{align}\]