If $\log_{2x}216 = x$, where $x$ is real, then $x$ is:
$\textbf{(A)}\ \text{A non-square, non-cube integer}\qquad$
$\textbf{(B)}\ \text{A non-square, non-cube, non-integral rational number}\qquad$
$\textbf{(C)}\ \text{An irrational number}\qquad$
$\textbf{(D)}\ \text{A perfect square}\qquad$
$\textbf{(E)}\ \text{A perfect cube}$
Rewrite the equation to $(2x)^x = 216$.  Since $2x$ is the base of the logarithm, we only need to check x-values that are positive.
With trial and error, $3$ is a solution because $(2 \cdot 3)^3 = 6^3 = 216$.  Since $(2x)^x$ gets larger as x gets larger, $3$ is the only solution, so the answer is $\boxed{\textbf{(A)}}$.