Solve the following
Question:

(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 $x^{-1}$.

Solution:

(i) First, we have to find x.

$x=\left(\frac{3}{2}\right)^{2} \times\left(\frac{2}{3}\right)^{-4} \quad \ldots>\left(a^{-1}=1 / a\right)$

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

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

Hence, x−2 is:

$x^{-2}=\left(\left(\frac{3}{2}\right)^{6}\right)^{-2} \quad \cdots>\left(a^{-1}=1 / a\right)$

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

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

(ii) First, we have to find x.

$x=\left(\frac{4}{5}\right)^{-2} \div\left(\frac{1}{4}\right)^{2} \quad \cdots\left((a / b)^{n}=\left(a^{n}\right) /\left(b^{n}\right)\right)$

$=\left(\frac{4^{-2}}{5^{-2}}\right) \times 4^{2}$

$=\frac{4^{0}}{5^{-2}}$

$=\frac{1}{5^{-2}} \quad \ldots\left(a^{0}=1\right)$

Hence, the value of x−1 is:

$x^{-1}=\left(\frac{1}{5^{-2}}\right)^{-1}$

$=\left(5^{2}\right)^{-1} \quad \cdots>\left(a^{-1}=1 / a\right)$

$=\frac{1}{5^{2}} \quad \cdots\left(a^{-1}=1 / a\right)$