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I am trying to solve this equation: $$\log_x\left(\frac{\log_4(x)}{\log_4(x)-3}\right)^{\log_3(x)}= 2$$

I'd like some advice on what to go about it, so far I have made it into:

$$\begin{aligned}\log_3(x)\cdot\log_x\left(\frac{\log_4(x)}{\log_4(x)-3}\right)=2 \\\log_3(x)\cdot\log_x(\log_4(x))-\log_x(\log_4(x)-3)=2\end{aligned}$$

And I'm unsure how to proceed from here?

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  • $\begingroup$ Is $log_3(x)$ the exponent if the argument of the base $x$ logarithm? $\endgroup$
    – Sgg8
    Jul 25 at 18:22
  • $\begingroup$ Hey yes that's correct, the steps I've done so far aren't necessarily the right ways to go $\endgroup$
    – mintteaplease
    Jul 25 at 18:32
  • $\begingroup$ I would use change of base on all the logarithms to natural log(or log 2 if you want) and then just combine them all and repeatedly take e^(or 2^) of both sides. Also, WA claims $x=64 \cdot 2^{3/4}$ is the solution so looks like stuff should simplify or at least be nice. $\endgroup$
    – dgeyfman
    Jul 25 at 18:58
  • $\begingroup$ Use $\log_a b\cdot \log_b c=\log_a c$ $\endgroup$
    – Sil
    Jul 25 at 19:42
  • $\begingroup$ @mintteaplease, please read my answer and tell me if you like it. $\endgroup$
    – Angelo
    Jul 25 at 20:24

2 Answers 2

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Using change of base rule:

$\log_3(x)= \frac{\log(x)}{\log(3)};$ $\log_4(x)= \frac{\log(x)}{\log(4)};$ $\log_x\left(\frac{\log_4(x)}{\log_4(x)-3}\right)=\frac{\log\left(\frac{\log_4(x)}{\log_4(x)-3}\right)}{\log(x)}$

$2= \log_x\left(\frac{\log_4(x)}{\log_4(x)-3}\right)^{\log_3(x)}= \log_3(x)\cdot\log_x\left(\frac{\log_4(x)}{\log_4(x)-3}\right)= $ $\begin{aligned}= \frac{\log\left(\frac{\log(x)}{\log(4)\left(\frac{\log(x)}{\log(4)}-3\right)}\right)}{\log(x)} \frac{\log(x)}{\log(3)}&=2\\ \implies \log\left(\frac{\log(x)}{\log(4)\left(\frac{\log(x)}{\log(4)}-3\right)}\right)&=2 \log(3)=\log(9)\end{aligned}$

Then $\frac{\log(x)}{\log(4)\left(\frac{\log(x)}{\log(4)}-3\right)}=9.$

Now, isolate $\log(x)$ and solve for $x$.

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  • $\begingroup$ Sorry I'm a bit unsure how you converted my function to the first expression, or how the change of base rule should be applied here, do you mind breaking it down a bit? $\endgroup$
    – mintteaplease
    Jul 25 at 19:21
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    $\begingroup$ @mintteaplease the change of base rule suggests that $log_{3}(x) = \frac{log(x)}{log(3)}$. Same for that other huge scarry logarithm with base $x$ $\endgroup$
    – Sgg8
    Jul 25 at 19:29
  • $\begingroup$ @mintteaplease I made an edit with full details in case it's still hard to grasp $\endgroup$
    – Sgg8
    Jul 26 at 10:46
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In order to solve your equation, it is sufficient to use the definition of logarithm and the property $\,\log_b(a)=\dfrac1{\log_a(b)}\,.$

Since $\,\log_x\!\left(\!\dfrac{\log_4(x)}{\log_4(x)-3}\!\right)^{\!\log_3(x)}\!\!\!\!\!\!\!=2\;,\;$ by definition of logarithm

$\log_x(\cdot)\,,\,$ it follows that

$x^2=\left(\dfrac{\log_4(x)}{\log_4(x)-3}\right)^{\log_3(x)}\;.$

Moreover,

$x^{2\log_x(3)}=\left(\dfrac{\log_4(x)}{\log_4(x)-3}\right)^{\log_3(x)\cdot\log_x(3)}\;,$

$\left(x^{\log_x(3)}\right)^2=\left(\dfrac{\log_4(x)}{\log_4(x)-3}\right)^{\log_3(x)\cdot\frac1{\log_3(x)}}\;,$

$3^2=\dfrac{\log_4(x)}{\log_4(x)-3}\;\;,$

$\log_4(x)=\dfrac{27}8\;\;,$

$x=4^{\frac{27}8}=2^{\frac{27}4}=64\sqrt[4]8\;.$

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