(i) If $ x=\left(\frac{3}{2}\right)^{2} \times\left(\frac{2}{3}\right)^{-4} $, find the value of $ x^{-2} $
(ii) If $ x=\left(\frac{4}{5}\right)^{-2}\div\left(\frac{1}{4}\right)^{2} $, find the value of $ x^{-1} $.

To do:  

We have to find the value of

(i)  \( x^{-2} \) 

(ii)  \( x^{-1} \).

Solution:

We know that,

$a^m \times a^n=a^{m+n}$

(i) $x=(\frac{3}{2})^{2} \times(\frac{2}{3})^{-4}$

$x=(\frac{3}{2})^{2} \times(\frac{3}{2})^{4}$

$x=(\frac{3}{2})^{2+4}$

$x=(\frac{3}{2})^{6}$

Therefore,

$x^{-2}=[(\frac{3}{2})^{6}]^{-2}$

$=(\frac{3}{2})^{6\times(-2)}$

$=(\frac{3}{2})^{-12}$

$=(\frac{2}{3})^{12}$

(ii) $x=(\frac{4}{5})^{-2} \div (\frac{1}{4})^{2}$

$=(\frac{5}{4})^{2} \div (\frac{1}{4})^{2}$

$=(\frac{5}{4} \div \frac{1}{4})^{2}$

$=(\frac{5}{4} \times \frac{4}{1})^{2}$

$=(5)^{2}$

$=25$

Therefore,

$x^{-1}=(25)^{-1}$

$=\frac{1}{25}$

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Updated on: 10-Oct-2022

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