Exercise 10.1 (Revised) - Chapter 12 - Exponents & Powers - Ncert Solutions class 8 - Maths

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Chapter 10: Exponents & Powers - NCERT Solutions for Class 8 Maths

Ex 10.1 Question 1.

Evaluate:
(i) $3^{-2}$
(ii) $(-4)^{-2}$
(iii) $\left(\frac{1}{2}\right)^{-5}$

Answer.

(i) $3^{-2}=\frac{1}{3^2}$
$
\begin{aligned}
& {\left[\because a^{-m}=\frac{1}{a^m}\right]} \\
& =\frac{1}{9}
\end{aligned}
$
$
\begin{aligned}
& \text { (ii) }(-4)^{-2}=\frac{1}{(-4)} \\
& {\left[\because a^{-m}=\frac{1}{a^m}\right]} \\
& =\frac{1}{16}
\end{aligned}
$
(iii) $\left(\frac{1}{2}\right)^{-5}=\left(\frac{2}{1}\right)^5$

$\begin{aligned}
& {\left[\because a^{-m}=\frac{1}{a^m}\right]} \\
& =(2)^5=32
\end{aligned}$

Ex 10.1 Question 2.

Simplify and express the result in power notation with positive exponent:
(i) $(-4)^5 \div(-4)^8$
(ii) $\left(\frac{1}{2^3}\right)^2$
(iii) $(-3)^4 \times\left(\frac{5}{3}\right)^4$
(iv) $\left(3^{-7} \div 3^{-10}\right) \times 3^{-5}$
(v) $2^{-3} \times(-7)^{-3}$

Answer.

(i) $(-4)^5 \div(-4)^8=(-4)^{5-8}\left[\because a^m \div a^n=a^{m-n}\right]$
$
=(-4)^{-3}=\frac{1}{(-4)^3}\left[\because a^{-m}=\frac{1}{a^m}\right]
$
(ii) $\left(\frac{1}{2^3}\right)^2=\frac{1^2}{\left(2^3\right)^2}$

$
\begin{aligned}
& {\left[\because\left(\frac{a}{b}\right)^m=\frac{a^m}{a^n}\right]} \\
& =\frac{1}{2^{3 \times 2}}=\frac{1}{2^6}\left[\because\left(a^m\right)^n=a^{m \times n}\right]
\end{aligned}
$
(iii) $(-3)^4 \times\left(\frac{5}{3}\right)^4=(-3)^4 \times \frac{5^4}{3^4}\left[\because\left(\frac{a}{b}\right)^m=\frac{a^m}{a^n}\right]$
$
=\left\{(-1)^4 \times 3^4\right\} \times \frac{5^4}{3^4}
$

$
\begin{aligned}
& {\left[\because(a b)^m=a^m b^m\right]} \\
& =3^{4-4} \times 5^4\left[\because a^m \div a^n=a^{m-n}\right] \\
& =3^0 \times 5^4=5^4\left[\because a^0=1\right]
\end{aligned}
$
$
\begin{aligned}
& \text { (iv) }\left(3^{-7} \div 3^{-10}\right) \times 3^{-5}=3^{-7-(-10)} \times 3^{-5} \quad\left[\because a^m \div a^n=a^{m-n}\right] \\
& =3^{-7+10} \times 3^{-5}=3^3 \times 3^{-5}=3^{3+(-5)}\left[\because a^m \times a^n=a^{m+n}\right] \\
& =3^{-2}=\frac{1}{3^2}\left[\because a^{-m}=\frac{1}{a^m}\right]
\end{aligned}
$
(v) $2^{-3} \times(-7)^{-3}=\frac{1}{2^3} \times \frac{1}{(-7)^3}\left[\because a^{-m}=\frac{1}{a^m}\right]$
$
=\frac{1}{\{2 \times(-7)\}^3}=\frac{1}{(-14)^3}\left[\because(a b)^m=a^m b^m\right]
$

Ex 10.1 Question 3.

Find the value of:
(i) $\left(3^0+4^{-1}\right) \times 2^2$
(ii) $\left(2^{-1} \times 4^{-1}\right) \div 2^{-2}$
(iii) $\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}$
(iv) $\left(3^{-1}+4^{-1}+5^{-1}\right)^0$
(v) $\left\{\left(\frac{-2}{3}\right)^{-2}\right\}^2$

Answer.
$
\begin{aligned}
& \text { (i) }\left(3^0+4^{-1}\right) \times 2^2=\left(1+\frac{1}{4}\right) \times 2^2\left[\because a^{-m}=\frac{1}{a^m}\right] \\
& =\left(\frac{4+1}{4}\right) \times 2^2=\frac{5}{4} \times 2^2=\frac{5}{2^2} \times 2^2=5 \times 2^{2-2} \quad\left[\because a^m \div a^n=a^{m-n}\right] \\
& =5 \times 2^0=5 \times 1=5 \quad\left[\because a^0=1\right]
\end{aligned}
$
(ii) $\left(2^{-1} \times 4^{-1}\right) \div 2^{-2}=\left(\frac{1}{2^1} \times \frac{1}{4^1}\right) \div 2^{-2}\left[\because a^{-m}=\frac{1}{a^m}\right]$
$
\begin{aligned}
& =\left(\frac{1}{2} \times \frac{1}{2^2}\right) \div 2^{-2}=\frac{1}{2^3} \div 2^{-2}\left[\because a^m \times a^n=a^{m+n}\right] \\
& =2^{-3} \div 2^{-2}=2^{-3-(-2)}=2^{-3+2}=2^{-1}\left[\because a^m \div a^n=a^{m-n}\right] \\
& =\frac{1}{2}\left[\because a^{-m}=\frac{1}{a^m}\right]
\end{aligned}
$
(iii) $\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}$

$
\begin{aligned}
& =\left(2^{-1}\right)^{-2}+\left(3^{-1}\right)^{-2}+\left(4^{-1}\right)^{-2} \\
& {\left[\because a^{-m}=\frac{1}{a^m}\right]} \\
& =2^{-1 \times(-2)}+3^{-\operatorname{lx}(-2)}+4^{-1 \times(-2)}\left[\because\left(a^m\right)^n=a^{m \times n}\right] \\
& =2^2+3^2+4^2=4+9+16=29
\end{aligned}
$
(iv) $\left(3^{-1}+4^{-1}+5^{-1}\right)^0=\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)^0\left[\because a^{-m}=\frac{1}{a^m}\right]$

$
\begin{aligned}
& =\left(\frac{20+15+12}{60}\right)^0=\left(\frac{47}{60}\right)^0=1 \\
& {\left[\because a^0=1\right]}
\end{aligned}
$
(v) $\left\{\left(\frac{-2}{3}\right)^{-2}\right\}^2=\left(\frac{-2}{3}\right)^{-2 \times 2}\left[\because\left(a^m\right)^n=a^{m \times n}\right]$
$
\begin{aligned}
& =\left(\frac{-2}{3}\right)^{-4}=\left(\frac{-3}{2}\right)^4\left[\because a^{-m}=\frac{1}{a^m}\right] \\
& =\frac{81}{16}
\end{aligned}
$
Ex 10.1 Question 4.

Evaluate:
(i) $\frac{8^{-1} \times 5^3}{2^{-4}}$
(ii) $\left(5^{-1} \times 2^{-1}\right) \times 6^{-1}$

Answer.

(i) $\frac{8^{-1} \times 5^3}{2^{-4}}=\frac{\left(2^3\right)^{-1} \times 5^3}{2^{-4}}=\frac{2^{-3} \times 5^3}{2^{-4}}\left[\because\left(a^m\right)^n=a^{m \times n}\right]$ $=2^{-3-(-4)} \times 5^3=2^{-3+4} \times 5^3 \quad\left[\because \quad a^m \div a^n=a^{m-n}\right]$

$
=2 \times 125=250
$
(ii) $\left(5^{-1} \times 2^{-1}\right) \times 6^{-1}=\left(\frac{1}{5} \times \frac{1}{2}\right) \times \frac{1}{6}\left[\because a^{-m}=\frac{1}{a^m}\right]$
$
=\frac{1}{10} \times \frac{1}{6}=\frac{1}{60}
$
Ex 10.1 Question 5.

Find the value of $m$ for which $5^m \div 5^{-3}=5^5$.

Answer.

$5^m \div 5^{-3}=5^5$

$
\begin{aligned}
& \Rightarrow 5^{m-(-3)}=5^5 \\
& {\left[\because a^m \div a^n=a^{m-n}\right]} \\
& \Rightarrow 5^{m+3}=5^5
\end{aligned}
$

Comparing exponents both sides, we get
$
\begin{aligned}
& \Rightarrow m+3=5 \\
& \Rightarrow m=5-3 \\
& \Rightarrow m=2
\end{aligned}
$
Ex 10.1 Question 6.

Evaluate:
(i) $\left\{\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}$ (ii) $\left(\frac{5}{8}\right)^{-7} \times\left(\frac{8}{5}\right)^{-4}$

Answer.
$
\begin{aligned}
& \text { (i) }\left\{\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}=\left\{\left(\frac{3}{1}\right)^1-\left(\frac{4}{1}\right)^1\right\}\left[\because a^{-m}=\frac{1}{a^m}\right] \\
& =\{3-4\}=-1
\end{aligned}
$

$\begin{aligned}
& \text { (ii) }\left(\frac{5}{8}\right)^{-7} \times\left(\frac{8}{5}\right)^{-4}=\frac{5^{-7}}{8^{-7}} \times \frac{8^{-4}}{5^{-4}}\left[\because\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\right] \\
& =5^{-7-(-4)} \times 8^{-4-(-7)} \quad\left[\because a^m \div a^n=a^{m-n}\right] \\
& =5^{-7+4} \times 8^{-4-7}=5^{-3} \times 8^3=\frac{8^3}{5^3}\left[\because a^{-m}=\frac{1}{a^m}\right] \\
& =\frac{512}{125}
\end{aligned}$

Ex 10.1 Question 7. Simplify:
(i) $\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}} \quad(t \neq 0)$
(ii) $\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}$

Answer.

(i) $\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}}$
$
\begin{aligned}
& =\frac{5^2 \times t^{-4}}{5^{-3} \times 5 \times 2 \times t^{-8}} \\
& =\frac{5^{2-(-3)-1} \times t^{-4-(-8)}}{2} \\
& {\left[\because a^m \div a^n=a^{m-n}\right]} \\
& =\frac{5^{2+3-1} \times t^{-4+8}}{2}=\frac{5^4 \times t^4}{2}=\frac{625}{2} t^4
\end{aligned}
$
(ii) $\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}$

$\begin{aligned}
& =\frac{3^{-5} \times(2 \times 5)^{-5} \times 5^3}{5^{-7} \times(2 \times 3)^{-5}} \\
& =\frac{3^{-5} \times 2^{-5} \times 5^{-5} \times 5^3}{5^{-7} \times 2^{-5} \times 3^{-5}} \\
& {\left[\because(a b)^m=a^m b^m\right]} \\
& =\frac{3^{-5} \times 2^{-5} \times 5^{-5+3}}{5^{-7} \times 2^{-5} \times 3^{-5}}=\frac{3^{-5} \times 2^{-5} \times 5^{-2}}{5^{-7} \times 2^{-5} \times 3^{-5}}\left[\because a^m \times a^n=a^{m-n}\right]
\end{aligned}$

$\begin{aligned}
& =3^{-5-(-5)} \times 2^{-5-(-5)} \times 5^{-2-(-7)}\left[\because a^m \div a^n=a^{m-n}\right] \\
& =3^{-5+5} \times 2^{-5+5} \times 5^{-2+7}=3^0 \times 2^0 \times 5^5 \\
& =1 \times 1 \times 3125\left[\because a^0=1\right] \\
& =3125
\end{aligned}$