# Question #f12fd

Apr 26, 2015

Ok. The difference is 0 if

${3}^{3 x - 1} = {9}^{x + 4}$

Now we need to write both sides with the same base.
$9 = {3}^{2}$ so we'll use that.

${3}^{3 x - 1} = {\left({3}^{2}\right)}^{x + 4}$

${3}^{3 x - 1} = {3}^{2 \left(x + 4\right)}$

The last line can only be true if
$3 x - 1 = 2 \left(x + 4\right)$ and we can solve that.

$3 x - 1 = 2 x + 8$

$x = 9$

Check: ${3}^{26} = {9}^{13}$ looks good!