Find $n$ so that
$(\frac{4}{5})^3)\times(\frac{4}{5})^{-6}=(\frac{4}{5})^{2n-1}$.

Given:

$(\frac{4}{5})^3)\times(\frac{4}{5})^{-6}=(\frac{4}{5})^{2n-1}$.

To do:

We have to find the value of $n$.
Solution:

 We know that,

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

Therefore,

LHS

$(\frac{4}{5})^3)\times(\frac{4}{5})^{-6}=(\frac{4}{5})^{3+(-6)})$

$=(\frac{4}{5})^{(-3)})$

RHS$=(\frac{4}{5})^{2n-1}$

Equating the powers on LHS and RHS, we get,

$-3=2n-1$

$2n=-3+1$

$2n=-2$

$n=\frac{-2}{2}$

$n=-1$

The value of $n$ is $-1$.

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Simply Easy Learning

Updated on: 10-Oct-2022

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