equation $\log_{2x}216=x$ Solve the equation $\log_{2x}216=x$ where $x$ is real. 
My attempt,
By trial and error $x=3$ as $6^3=216$
Is there any systematic way to solve it?

Noting that $$\log_{2x}(216) = \frac{\ln(216)}{\ln(2x)}$$ we can say \begin{align}\log_{2x}(216) &= x\\
\frac{\ln(216)}{\ln(2x)}&=x\\
\ln(216)&=x\ln(2x)\\
e^{\ln(216)}&=e^{x\ln(2x)}\\
e^{\ln(216)}&=e^{\ln((2x)^x)}\\
216&=(2x)^x
\end{align}
We can then note that $216=6^3$ and so \begin{align}6^3&=(2x)^x\\
(2\times 3)^3&=(2x)^x\end{align}
It is then trivial to see that $x=3$

We could also calculate this last step by noting that the exponent must be half the base.
We can then brute force this with integers:
Starting with an exponent of $1$, we have
\begin{align}2^1&=2\\
4^2&=16\\
6^3&=216\end{align}
And so our exponent, and thus $x$, must be $3$
