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

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?

Travis Willse
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Abulkhair
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    Change of base is still a good idea, but you should choose carefully which base to change into! I suggest a base $2$ logarithm. You can also let $y = \log_2x$, and you'll get something that can be transformed into a quadratic in $y$. – Theo Bendit Apr 06 '22 at 04:15
  • @TheoBendit thank you this helped a lot – Abulkhair Apr 06 '22 at 04:29
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    @TheoBendit If one feels adventurous, changing everything to base $x$ could also work. – Arthur Apr 06 '22 at 06:58

2 Answers2

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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?

Átila Correia
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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 $x=2$

TreeGuy
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