Question

Find the value of $x$, if ${\left(\sqrt{\frac{3}{5}}\right)}^{1-2x}=\frac{125}{27}$

Answer 1

Finding the value of $x$:

$\begin{array}{l}{\left(\sqrt{\frac{3}{5}}\right)}^{1-2x}=\frac{125}{27}\\ \Rightarrow {\left({\left(\frac{3}{5}\right)}^{\frac{1}{2}}\right)}^{1-2x}=\frac{5^3}{3^3}\left[\because \sqrt[n]{a}=a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}\right]\\ \Rightarrow {\left({\left(\frac{3}{5}\right)}^{\frac{1}{2}}\right)}^{1-2x}={\left(\frac{3}{5}\right)}^{-3}\left[\because a^{-n}=\frac{1}{a^n}\right]\\ \Rightarrow \left({\left(\frac{3}{5}\right)}^{\frac{1}{2}-\frac{2x}{2}}\right)={\left(\frac{3}{5}\right)}^{-3}\left(\because {\left(a^m\right)}^n=a^{mn}\right)\\ \Rightarrow \frac{1}{2}-\frac{2x}{2}=-3\left(\text{Base are same, so equating powers}\right)\\ \Rightarrow \frac{1-2x}{2}=-3\\ \Rightarrow 1-2x=-3\times 2\\ \Rightarrow 1-2x=-6\\ \Rightarrow -2x=-6-1\\ \Rightarrow 2x=7\\ \Rightarrow x=\frac{7}{2}\\ \Rightarrow x=3\frac{1}{2}\end{array}$

Hence,the value of $x=3\frac{1}{2}$