# How do you solve log _ 2 (4x-8)=1?

$x = \frac{5}{2}$
using : ${\log}_{b} a = n \Leftrightarrow a = {b}^{n}$
then  log_2 (4x - 8 ) = 1 → 4x - 8 = 2^1 = 2
so 4x - 8 = 2  → 4x = 10 → x = 10/4 = 5/2