Question

Find $x$, if ${\left(\sqrt{\frac{3}{5}}\right)}^{-x+1}=\frac{125}{27}$

Answer 1

Given equation is
$\Rightarrow {\left(\sqrt{\frac{3}{5}}\right)}^{-x+1}=\frac{125}{27}$
$\Rightarrow {\left({\left(\frac{3}{5}\right)}^{\frac{1}{2}}\right)}^{-x+1}=\frac{5^3}{3^5}$
$\Rightarrow {\left(\frac{3}{5}\right)}^{-\frac{x}{2}+\frac{1}{2}}={\left(\frac{3}{5}\right)}^{-3}$
Both side base are same so by equating the powers
$\Rightarrow -\frac{x}{2}+\frac{1}{2}=-3$
$\Rightarrow \frac{-x+1}{2}=-3$
$\Rightarrow x=7$