How to solve this logarithmic equation through algebraic manipulation? I was working through an Art of Problem Solving workbook when I encountered a very frustrating problem.
Solve the equation $\log_{2x}216=x$, where $x\in \mathbb{R}$.
I understand how to find the answer by inspection (all you have to do is stare at it for a few seconds), but I'm trying to figure out how to solve it algebraically. Every time I try to manipulate the problem, I either find myself running in circles or I create some convoluted expression that's even harder to deal with than the original problem. Forgive me if this is a very easy question - I'm just completely lost here. Thanks!

$$ x = \log_{2x}{216} = \frac{\log{216}}{\log{2x}} = \frac{3\log{2}+3\log{3}}{\log{2}+\log{x}} = 3\frac{1+\log_2{3}}{1+\log_2{x}} $$
But then multiplying up gives
$$ x(1+\log_2{x}) = 3(1+\log_2{3}). $$
The left-hand side is an increasing function of $x$, so it only has one solution, which is obviously $x=3$.
