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

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

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

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