NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers

1.Evaluate:
(i) \(3^{-2}\)

(ii) \((-4)^{-2}\)

(iii) \((\frac { 1 }{ 2 })^{-5}\)

Answer :

(i) We have:
\(3^{-2}=\frac{1}{3^2}=\frac{1}9\quad\)[∵\(a^{-n}=\frac{1}{a^n}\)]

(ii)We have:
\((-4)^{-2}=\frac{1}{(-4)^2}=\frac{1}{16}\quad\)[∵\(a^{-n}=\frac{1}{a^n}\)]

(iii)We have:
\((\frac { 1 }{ 2 })^{-5}=2^5=32\quad\)[∵\((\frac{a}b)^{-n}=(\frac{b}a)^n\)]

2.Simplify and express the result in power notation with a positive exponent. (i)\((-4)^5÷(-4)^8\)
(ii)\((\frac{1}{2^3})^2\)

(iii)\((-3)^4\times (\frac{5}3)^4\)

(iv)\((3^{-7}÷3^{-10})\times 3^{-5}\)

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

Answer :

(i)We have:\((-4)^5÷(-4)^8\)
\(=(-4)^{5-8}=(-4)^{-3}=\frac{1}{(-4)^3}\quad\)[∵\(a^{m}÷a^{n}=a^{m-n}\)]

\(=(-\frac{1}4)^3\)

(ii)We have:\((\frac{1}{2^3})^2\)

\((\frac{1}{2^3})^2=\frac{(1)^2}{(2^3)^2}=\frac{1}{2^6}=(\frac{1}2)^6\)

(iii)We have:\((-3)^4\times (\frac{5}3)^4\)

\(=(-3)^4\times (\frac{5}3)^4=(-3)^4\times \frac{(5)^4}{(3)^4}\)

\(=\frac{(3)^4\times(5)^4}{(3)^4}=(5)^4\)

(iv)We have:\((3^{-7}÷3^{-10})\times 3^{-5}\)

\(=(3^{-7}÷3^{-10})\times 3^{-5}=3^{-7-(-10)}\times 3^{-5}\)

\(=3^{-7+10}\times 3^{-5}=3^3\times 3^{-5}=3^{(3-5)}\)

\(=3^{-2}=\frac{1}{3^2}=(\frac{1}3)^2\)

(v)We have:\(2^{-3}\times(-7)^{-3}\)

\(=2^{-3}\times(-7)^{-3}=[2\times(-7)]^{-3}=(-14)^{-3}\)

\(=-(14)^{-3}=-\frac{1}{14^3}=(-\frac{1}{14})^3\)

3.Find the value of
(i)\(3^0+4^{-1}\times 2^2\)

(ii)\(2^{-1}\times4^{-1}÷ 2^{-2}\)

(iii)\((\frac{1}2)^{-2}+(\frac{1}3)^{-2}+(\frac{1}4)^{-2}\)

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

(v)\([(\frac{-2}3)^{-2}]^2\)

Answer :

(i)We have:\(3^0+4^{-1}\times 2^2\)

\(=3^0+4^{-1}\times 2^2=(1+\frac{1}4)\times4\)

\(=(\frac{4+1}4)\times4=\frac{5}4\times 4=5\)

(ii)We have:\(2^{-1}\times4^{-1}÷ 2^{-2}\)

\(=2^{-1}\times4^{-1}÷ 2^{-2}=(\frac{1}2\times\frac{1}4)÷\frac{1}{2^2}\)

\(=\frac{1}8÷\frac{1}4=\frac{1}8\times\frac{4}1\)

\(=\frac{1}2\)

(iii)We have:\((\frac{1}2)^{-2}+(\frac{1}3)^{-2}+(\frac{1}4)^{-2}\)

\(=(\frac{1}2)^{-2}+(\frac{1}3)^{-2}+(\frac{1}4)^{-2}=2^2+3^2+4^2\)

\(=4+9+16=29\)

(iv)We have:\((3^{-1}+4^{-1}+5^{-1})^0\)

\(=(3^{-1}+4^{-1}+5^{-1})^0=1\quad\)[∵\(a^0=1\)]

(v)We have:\([(\frac{-2}3)^{-2}]^2\)

\(=[(\frac{-2}3)^{-2}]^2=[(-\frac{3}2)^2]^2\)

\(=(\frac{9}4)^2=\frac{81}{16}\)

4.Evaluate
(i) \(\frac { { 8 }^{ -1 }\times { 5 }^{ 3 } }{ { 2 }^{ -4 } } \)

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

Answer :

(i)We have:\(\frac { { 8 }^{ -1 }\times { 5 }^{ 3 } }{ { 2 }^{ -4 } } \)

\(=\frac{8^{-1}\times 5^3}{2^{-4}}=\frac{1}8\times 5^3\times 2^4\)

\(=\frac{1}8\times125\times 16=125\times 2=250\)

(ii)We have:\((5^{-1} \times 2^{-1}) \times 6^{-1}\)

\(=(\frac{1}5\times\frac{1}2)\times\frac{1}6\)

\(=\frac{1}{10}\times\frac{1}6=\frac{1}{60}\)

5.Find the value of m for which \(5^m ÷ 5^{-3} = 5^5\).

Answer :

We have given: \(5^m ÷ 5^{-3} = 5^5\).

\(\Rightarrow 5^{m-(-3)} = 5^5\quad\) [∵\( a^m ÷ a^n = a^{m-n}\)]

\(\Rightarrow 5^{m+3} = 5^5\)

Now, Comparing the powers of equal bases, we have m + 3 = 5

\(\Rightarrow m = 5 – 3 = 2\), i.e., m = 2

6.Evaluate:
(i)\([(\frac{1}3)^{-1}-(\frac{1}4)^{-1}]^{-1}\)

(ii)\((\frac{5}8)^{-7}\times (\frac{8}5)^{-4}\)

Answer :

(i)We have:\([(\frac{1}3)^{-1}-(\frac{1}4)^{-1}]^{-1}\)

\(=(3^1-4^1)^{-1}=(-1)^{-1}=\frac{1}{(-1)^1}=\frac{1}{-1}=-1\)

(ii)We have: \((\frac{5}8)^{-7}\times (\frac{8}5)^{-4}\)
\(=(\frac{8}5)^7\times(\frac{5}8)^4=\frac{8^7}{5^7}\times\frac{5^4}{8^4}\)

\(=\frac{8^{7-4}}{5^{7-4}}=\frac{8^3}{5^3}=\frac{512}{125}\)

7.Simplify :
(i)\(\frac{25\times t^{-4}}{5^{-3}\times10\times t^{-8}}(t\neq 0)\)

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

Answer :

(i)We have:\(\frac{25\times t^{-4}}{5^{-3}\times10\times t^{-8}}(t\neq 0)\)

\(=\frac{25\times 5^3}{10}\times t^{(-4+8)}=\frac{5\times 5^3}2\times t^4\)

\(=\frac{625}2t^4\)
(ii)We have:\(\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times 6^{-5}}\)

\(=\frac{5^{7}\times 6^{5}\times 125}{3^{5}\times 10^{5}}\)

\(=\frac{5^{7}\times (3\times 2)^{5}\times 5^3}{3^{5}\times (2\times5)^{5}}=\frac{5^{7}\times 3^{5}\times 2^5\times 5^3}{3^{5}\times 2^5\times5^{5}}=5^{7+3-5}=5^5\)