Find the values of $x$ in each of the following:$ \left(\sqrt{\frac{3}{5}}\right)^{x+1}=\frac{125}{27} $

Given:

\( \left(\sqrt{\frac{3}{5}}\right)^{x+1}=\frac{125}{27} \)

To do: 

We have to find the value of $x$.

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$

Therefore,

$(\sqrt{\frac{3}{5}})^{x+1}=\frac{125}{27}$

$\Rightarrow [(\frac{3}{5})^{\frac{1}{2}}]^{x+1}=(\frac{5}{3})^{3}$

$\Rightarrow (\frac{3}{5})^{\frac{x+1}{2}}=(\frac{3}{5})^{-3}$

Comparing both sides, we get,

$\frac{x+1}{2}=-3$

$\Rightarrow x+1=-3\times2$

$\Rightarrow x+1=-6$

$\Rightarrow x=-6-1$

$\Rightarrow x=-7$

The value of $x$ is $-7$.        

Tutorialspoint

Updated on: 10-Oct-2022

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