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 $3$.