Answer
$\{0\}$
Work Step by Step
...$3^{x+1}=3^{x}\cdot 3$
... Substitute $t=3^{x}$
... the equation becomes
$t^{2}+3t-4=0 \quad $... factors of $-4$ whose sum is $+3$...
$(t-1)(t+4)=0$
t is either $-4$ or $1$.
$t=3^{x}=-4$ 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 $\{0\}$
