Answer
$x=\frac{7+log{\frac{3}{8}}}{2}$
Work Step by Step
We know that by definition if $y=a^x$, then $log_a {y}=x$ , also $log_e {x}=ln {x}$, hence if $y=e^x$, then $ln{e^x}=log_e {e^x}=x$ and vice versa and that $ln{\frac{1}{x}}=-ln{x}$.
Hence if $8\cdot10^{2x-7}=3$, then $10^{2x-7}=\frac{3}{8}$. Solve the equation above to obtain (after taking $log$ of both sides): \begin{align*}2x-7=log{\frac{3}{8}}\end{align*} \begin{align*}2x=7+log{\frac{3}{8}}\end{align*} \begin{align*}x=\frac{7+log{\frac{3}{8}}}{2}\end{align*}
