${\log _ {x}}8 - {\log _{4x}} 8 = {\log _{2x}} 16$ 
${\log _ {x}}8 - {\log _{4x}} 8 = {\log _{2x}} 16$

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?
 A: 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?
A: First, you can use the change of base formula on both sides of the equation.  You'll notice that the problem is easiest when put into log base-2.
$\frac{\log_2(8)}{\log_2(x)}-\frac{\log_2(8)}{\log_2(4x)}=\frac{\log_2(16)}{\log_2(2x)}$
Next, simplify everything.
$\frac{3}{\log_2(x)}-\frac{3}{2+\log_2(x)}=\frac{4}{1+\log_2(x)}$
Next, get rid of all of the denominators by multiplying each term by all of the denominators.
$3(2+\log_2(x))(1+\log_2(x))-3\log_2(x)(1+\log_2(x))=4\log_2(x)(2+\log_2(x))$
Next, combine all like terms and move everything to one side.
$4\log_2(x)^2+2\log_2(x)-6=0$
Notice that the equation is in the form of a quadratic that can be factored.
$2(2\log_2(x)+3)(\log_2(x)-1)=0$
You'll find that:
$\log_2(x)=-\frac{3}{2}$ and $\log_2(x)=1$
Now get rid of the logarithms to get $x$ alone.
$x=2^{-\frac{3}{2}}$ and $x=2$
