$$7^{2n-n-3}=1$$
$n=3$
$$7^{n-3}=1$$
$$\log(7^{n-3})=\log(1)$$
$$\left(n-3\right)\log(7)=\log(1)$$
$$n-3=\frac{\log(1)}{\log(7)}$$
$$n-3=\log_{7}\left(1\right)$$
$$n=-\left(-3\right)$$
Show Solution
Hide Solution
