How many ordered pairs $(a,b)$ such that $a$ is a positive real number and $b$ is an integer between $2$ and $200$, inclusive, satisfy the equation $(\log_b a)^{2017}=\log_b(a^{2017})?$
$\textbf{(A)}\ 198\qquad\textbf{(B)}\ 199\qquad\textbf{(C)}\ 398\qquad\textbf{(D)}\ 399\qquad\textbf{(E)}\ 597$
By the properties of logarithms, we can rearrange the equation to read $x^{2017}=2017x$ with $x=\log_b a$. If $x\neq 0$, we may divide by it and get $x^{2016}=2017$, which implies $x=\pm \root{2016}\of{2017}$. Hence, we have $3$ possible values $x$, namely 
\[x=0,\qquad x=2017^{\frac1{2016}},\, \text{and}\quad  x=-2017^{\frac1{2016}}.\]
Since $\log_b a=x$ is equivalent to $a=b^x$, each possible value $x$ yields exactly $199$ solutions $(b,a)$, as we can assign $a=b^x$ to each $b=2,3,\dots,200$. In total, we have $3\cdot 199=\boxed{\textbf{(E) } 597}$ solutions.