## College Algebra (10th Edition)

$\{0\}$
...$3^{x+1}=3^{x}\cdot 3$ ... 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 $\{0\}$