AMC 12 Mock Exam 1 Problem 9

Let $x=\log_2 a$ and $y=\log_2 b$. Since $\log_8 a=\frac x3$ and $\log_4 a=\frac x2$ (and similarly for $b$), the equations become $$\frac x3+\frac y2=5,\qquad \frac x2+\frac y3=5.$$ Adding gives $\frac56(x+y)=10$, so $x+y=12$. Subtracting gives $\frac{y-x}{6}=0$, so $x=y=6$. Hence $a=b=2^6=64$ and $a+b=\boxed{128}$.