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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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