# How do you solve 2 Log (x - 4) = 4 log 2?

We must have $x - 4 > 0 \implies x > 4$ hence
$2 \log \left(x - 4\right) = 4 \log 2 \implies \log {\left(x - 4\right)}^{2} = \log {2}^{4} \implies {\left(x - 4\right)}^{2} = {2}^{4} \implies x - 4 = \pm 4 \implies x = 8 \mathmr{and} x = 0$
Finally $x = 8$