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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 $-7$.
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