# Solving $\log_x\left(\frac{\log_4(x)}{\log_4(x)-3}\right)^{\log_3(x)}= 2$

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?

• Is $log_3(x)$ the exponent if the argument of the base $x$ logarithm?
– Sgg8
Jul 25 at 18:22
• Hey yes that's correct, the steps I've done so far aren't necessarily the right ways to go Jul 25 at 18:32
• 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. Jul 25 at 18:58
• Use $\log_a b\cdot \log_b c=\log_a c$
– Sil
Jul 25 at 19:42

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 $$x$$.

• 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? Jul 25 at 19:21
• @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 $x$
– Sgg8
Jul 25 at 19:29
• @mintteaplease I made an edit with full details in case it's still hard to grasp
– Sgg8
Jul 26 at 10:46

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,

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

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

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

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

$$x=4^{\frac{27}8}=2^{\frac{27}4}=64\sqrt8\;.$$