# How do you solve 4^(x+4) = 8^x?

Jul 23, 2015

I found: $x = 8$

#### Explanation:

You can write it as:
${\left({2}^{2}\right)}^{x + 4} = {\left({2}^{3}\right)}^{x}$
${2}^{2 \left(x + 4\right)} = {2}^{3 x}$ take the ${\log}_{2}$ on both sides to get rid of the $2$ (where you have $\cancel{{\log}_{2}} \left({\cancel{2}}^{x}\right) = x$) and get:
$2 \left(x + 4\right) = 3 x$
$2 x + 8 = 3 x$
$x = 8$