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

Question

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?

Answer 1

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 $x$.

Answer 2

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,

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

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

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

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

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