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Question

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

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Solution

## Finding the value of $x$:$\begin{array}{l}{\left(\sqrt{\frac{3}{5}}\right)}^{1-2x}=\frac{125}{27}\\ ⇒{\left({\left(\frac{3}{5}\right)}^{\frac{1}{2}}\right)}^{1-2x}=\frac{{5}^{3}}{{3}^{3}}\left[\because \sqrt[n]{a}={a}^{1}{n}}\right]\\ ⇒{\left({\left(\frac{3}{5}\right)}^{\frac{1}{2}}\right)}^{1-2x}={\left(\frac{3}{5}\right)}^{-3}\left[\because {a}^{-n}=\frac{1}{{a}^{n}}\right]\\ ⇒\left({\left(\frac{3}{5}\right)}^{\frac{1}{2}-\frac{2x}{2}}\right)={\left(\frac{3}{5}\right)}^{-3}\text{}\left(\because {\left({a}^{m}\right)}^{n}={a}^{mn}\right)\\ ⇒\frac{1}{2}-\frac{2x}{2}=-3\text{}\left(\text{Basearesame,soequatingpowers}\right)\\ ⇒\frac{1-2x}{2}=-3\\ ⇒1-2x=-3×2\\ ⇒1-2x=-6\\ ⇒-2x=-6-1\\ ⇒2x=7\\ ⇒x=\frac{7}{2}\\ ⇒x=3\frac{1}{2}\end{array}$Hence,the value of $x=3\frac{1}{2}$

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