smc

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 $\boxed{\text{A non-square, non-cube integer}}$.