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If $ \left(\frac{-5}{7}\right)^{-4} \times\left(\frac{-5}{7}\right)^{12}=\left\{\left(\frac{-5}{7}\right)^{3}\right\}^{x} \times\left(\frac{-5}{7}\right)^{-1}, $ find the value of $ x $.
Given:
\( \left(\frac{-5}{7}\right)^{-4} \times\left(\frac{-5}{7}\right)^{12}=\left\{\left(\frac{-5}{7}\right)^{3}\right\}^{x} \times\left(\frac{-5}{7}\right)^{-1} \)
To do:
We have to find the value of \( x \).
Solution:
We know that,
$a^m \times b^m=(a\times b)^m$
$(a^m)^n=(a)^{mn}$
Therefore,
$(\frac{-5}{7})^{-4}\times(\frac{-5}{7})^{12}={(\frac{-5}{7})^{3}}^x\times(\frac{-5}{7})^{-1}$
$(\frac{-5}{7})^{-4+12}=(\frac{-5}{7})^{3x}\times(\frac{-5}{7})^{-1}$
$(\frac{-5}{7})^{8}=(\frac{-5}{7})^{3x-1}$
Comparing the powers on both sides, we get,
$8=3x-1$
$3x=8+1$
$3x=9$
$x=\frac{9}{3}$
$x=3$
The value of $x$ is $3$.
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