Question

If ${\left(\sqrt{\frac{3}{5}}\right)}^{x+1}=\frac{125}{27}$ find the value of $x$

Answer 1

Finding the value of $x$:

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

Hence,the value of $x=-7$