Equation system

Since $\log_b a=\dfrac{\log_a a}{\log a_b}=\dfrac{1}{\log_a b}$ (change of base formula), our equation becomes $\log_a b+\dfrac{1}{\log_a b}=\dfrac{10}{3}$ . Let $\log_a b=x$ . Therefore, $x+\dfrac{1}{x}=\dfrac{10}{3}$ . Solving for $x$ , we get $x=\dfrac{1}{3}$ or $x=3$ . Divide this into two cases:

Case 1 ( $x=\frac{1}{3}$ ): Since $\log_a b=\dfrac{1}{3}$ , $a^{\frac{1}{3}}=b$ . Therefore, $a=b^3$ . Since $a^b=64$ , $\left (b^3 \right )^b=64$ . Solving for $b$ , we get $\boxed{b=2}$ .

Case 2 ( $x=3$ ): Proceeding in the same manner as we did in Case 1, we eventually get $a^3=b$ . Therefore, $a=\sqrt[3]{b}$ , so $\sqrt[3]{b^b}=64=2^6$ . Cubing both sides, $b^b=2^{18}$ , which does not produce any integer solutions.

We can easily see that $b=2$ is the only solution for $b$ . From case 1, $a=b^3$ , so $a=8$ . Our final answer is $a+b=8+2=\boxed{10}$ . $\blacksquare$