AMC 12 Mock Exam 4 Problem 23

Let $s=x+y>0$, $u=\ln x$, $v=\ln y$. The equations become $su=12v$ and $sv=3u$.

If $u=0$, then $v=0$, giving $(x,y)=(1,1)$, which works. If $v=0$, then $u=0$ likewise. Otherwise $u,v\ne0$, and multiplying gives $s^2uv=36uv$, so $s=6$. Then $6u=12v$, i.e., $x=y^2$, and $y^2+y=6$ gives $y=2$, $x=4$. Check: $4^6=2^{12}$ and $2^6=4^3$.

The solutions are $(1,1)$ and $(4,2)$, so the sum of the values of $x+y$ is $2+6=\boxed{8}$. (Overlooking $(1,1)$ gives $6$.)