What if $x,y$ axis becomes log?

From the graph, when

$\begin{aligned} \log_4 x = 0 & \implies \log_8 y = 1 \\ x = 4^0 = 1 & \implies y = 8 \\ \implies 8 & = A (1^B) \\ \implies A & = 8 \end{aligned}$

When

$\begin{aligned} \log_8 y = 0 & \implies \log_4 x = 3 \\ y = 1 & \implies x = 64 \\ \implies 2 & = 8(64^B) \\ \implies 0 & = \log 8 + B \log 64 \\ B & = - \frac{\log 8}{\log 64} = - \frac{1}{2} \end{aligned}$

$\implies A + B = 8 - \dfrac{1}{2} = \boxed{7.5}$