Answer
$x=\displaystyle \{1,\quad \frac{\log 5}{\log 3}\}$
Work Step by Step
Substituting $\quad t=3^{x}\qquad (t\gt 0)$
the equation becomes
$t^{2}-8t+15=0$
Factor by finding two factors of 15 with sum $-8$... they are $-3$ and $-5$
$(t-3)(t-5)=0$
$t=3, t=5$
$t=3\Rightarrow\quad 3^{x}=3\Rightarrow\quad x=1.$
$ t=5\Rightarrow\quad 3^{x}=5\quad$apply log(...) in both sides
$x\log 3=\log 5$
$x=\displaystyle \frac{\log 5}{\log 3}$
$x=\displaystyle \{1,\quad \frac{\log 5}{\log 3}\}$
