Answer
$\{0\}$
Work Step by Step
... Substitute $t=3^{x}$
... the equation becomes
$t^{2}+t-2=0 \quad $... factors of $-2$ whose sum is $+1$...
$(t+2)(t-1)=0$
t is either $-2$ or $1$.
$t=3^{x}=-2$ cannot be a solution as $3^{x}$ is never negative.
$t=3^{x}=1$
$3^{x}=3^{0}\quad $...Apply$: \quad $If $a^{u}=a^{v},$ then $u=v$
$x=0$
The solution set is $\{0\}$
