What is $x$ given that $x^{log_4(x)- 2} = 2^{3log_4(x)-1}$?

Question

My math teacher had given me four problems on a sheet of paper, and I am unable to solve this last one:

Given that $x^{\log_4(x - 2)} = 2^{3\log_4(x-1)}$, solve for $x$.

I've tried numerous things such as simplifying the right hand side to $(x-1)^\frac{3}{2}$ and changing the left hand side to $(x-2)^{\log_4x}$ to no avail.

From the way the question has been written, the equality could also be read as: $x^{(\log_4x)-2} = 2^{3(\log_4x)-1}$

Edit: It seems that the equation is supposed to be $x^{(\log_4x)-2} = 2^{3((\log_4x)-1)}$

From WolframAlpha, I know the solutions are $x=2$ and $x=64$, but I don't know how to derive it.

Answer 1

$$x^{(\log_4x)-2} = 2^{3(\log_4x-1)}\\\implies (\log_4(x)-2)\log_4(x) = (3\log_4(x)-3)\log_4(2)\stackrel{u=\log_4(x)}{\implies}(u-2)u = \frac 12(3u - 3)$$ Now you have a quadratic in $u$. Can you solve it from there?

Answer 2

Notice that $2 = 4^{1/2}$. Express your solution in base 4 as $x = 4^t$ and try to solve this again. It is also worth mentioning that $\log$ doesn't "distribute" as you have done. That is, $\log(x-1) \neq \log(x) - 1$.

Answer 3

Notice that $$x^{\log_4(x - 2)} = 2^{3\log_4(x-1)}=8^{\log_4(x-1)}\Rightarrow x\gt 8$$

On the other hand $$x^{\log_4(x - 2)} = 2^{3\log_4(x-1)}=4^{\frac32\log_4(x-1)}=4^{\log_4(x-1)^{\frac 32}}$$ It follows the trascendental equation $$x^{\log_4(x - 2)}=(x-1)^{\frac 32}$$

An approximation gives (we have seen above that $x\gt 8$) $$\color{red}{x\approx 9.1799}$$

Answer 4

Let $A=\log_4 x.$ Then $4^A=x.$ So $$2^{-1+3\log_4 x}=4^{-1/2+(3/2)\log_4x}=4^{-1/2+3A/2}$$ and $$x^{-2+\log_4x}=(4^A)^{(-2+\log_4x)}=(4^A)^{(-2+A)}=4^{-2A+A^2}.$$ So $x$ is a solution iff $-1/2+3A/2=-2A+A^2.$