<s>[INST] ${\log _ {x}}8 - {\log _{4x}} 8 = {\log _{2x}} 16$ 
${\log _ {x}}8 - {\log _{4x}} 8 = {\log _{2x}} 16$

I tried solving this problem by change of base and by $\frac{1}{\log{x}}$, but I really cannot seem to solve it no matter how hard I try.
I could only answer it by substituting $x$ for mcq answers.
Where do I even begin to solve it?
 [/INST] HINT
As @TheoBendit has mentioned in the comments, I would recommend you to start with noticing
\begin{align*}
\log_{x}(8) - \log_{4x}(8) = \log_{2x}(16) & \Longleftrightarrow \frac{\log_{2}(8)}{\log_{2}(x)} - \frac{\log_{2}(8)}{\log_{2}(4x)} = \frac{\log_{2}(16)}{\log_{2}(2x)}
\end{align*}
where $x > 0$ and $x\not\in\{1,1/2,1/4\}$.
If we let that $y = \log_{2}(x)$, one arrives at equivalent equation:
\begin{align*}
\frac{3}{y} - \frac{3}{2 + y} = \frac{4}{1 + y} & \Longleftrightarrow \frac{6}{y(2 + y)} = \frac{4}{1 + y}\\\\
& \Longleftrightarrow
\begin{cases}
3(1 + y) = 2y(2 + y)\\\\
y(2 + y)(1 + y) \neq 0
\end{cases}\\\\
& \Longleftrightarrow
\begin{cases}
2y^{2} + y - 3 = 0\\\\
y(2 + y)(1 + y) \neq 0
\end{cases}
\end{align*}
Can you take it from here?
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