Question

If $x={\left(\frac{3}{2}\right)}^2\times {\left(\frac{2}{3}\right)}^{-4}$, then $x^{-2}$ is $(\frac{2}{m}{)}^{12}$. The value of $m$ is

Answer 1

$x={\left(\frac{3}{2}\right)}^2\times {\left(\frac{2}{3}\right)}^{-4}$
$\Rightarrow x={\left(\frac{3}{2}\right)}^2\times {\left(\frac{3}{2}\right)}^4$
$\Rightarrow x={\left(\frac{3}{2}\right)}^{2+4}$
$\Rightarrow x={\left(\frac{3}{2}\right)}^6$
$\Rightarrow x^{-2}={\left\{{\left(\frac{3}{2}\right)}^6\right\}}^{-2}$
$\Rightarrow x^{-2}={\left(\frac{3}{2}\right)}^{-12}$
$\Rightarrow x^{-2}={\left(\frac{2}{3}\right)}^{12}$