The value of $ x $ for which $ \left(\frac{7}{8}\right)^{-4} \times\left(\frac{7}{8}\right)^{3 x}=\left(\frac{7}{8}\right)^{8} $, is equal to

Given:

$(\frac{7}{8})^{-4} \times(\frac{7}{8})^{3 x}=(\frac{7}{8})^{8}$.

To do:

We have to find the value of $x$.

Solution:

We know that,

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

Therefore,

LHS

$(\frac{7}{8})^{-4} \times(\frac{7}{8})^{3 x}=(\frac{7}{8})^{-4+3x}$

$=(\frac{7}{8})^{3x-4}$

RHS $=(\frac{7}{8})^{8}$

On comparing LHS and RHS,

$3x-4=8$

$3x=8+4$

$x=\frac{12}{3}$

$x=4$

The value of $x$ is $4$. 

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Updated on: 10-Oct-2022

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