${\log _ {x}}8 – {\log _{4x}} 8 = {\log _{2x}} 16$
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?
Best Answer
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?