2017 AMC 12A Problem 20

Let $u=\log_b a.$ Since $\log_b(a^{2017})=2017\log_b a,$ the equation is $u^{2017}=2017u,$ so $u=0$ or $u^{2016}=2017.$ If $u=0,$ then $a=1,$ valid for every one of the $199$ bases. If $u^{2016}=2017,$ then $u=\pm2017^{\frac{1}{2016}},$ giving $2$ values of $a$ for each base, i.e. $2\cdot199=398$ pairs. In total there are $199+398=597$ ordered pairs. Thus, the correct answer is E.