Is there a solution to this functional equation?

I was going through my old notebooks and I found a sheet of paper with this problem on it. I thought it would be a shame to let such an unreasonably difficult question go to waste, so I decided I would share it. The problem simply states:

Solve for $$f$$: $$f(x)=2\log(x)^2f\left(x^\frac{3}{8}\right)^2f\left(x^\frac{1}{4}\right)^2$$

No other information or context is given, but I'm assuming that $$f$$ is a complex valued function of a single real or complex variable (since evaluating the function for negative $$x$$ would require $$f(x)$$ to be complex), and that $$\log$$ is the natural logarithm (since no one would use it for log base-10 if they were talking about complex functions).

For curiosity's sake, I present it as a challenge to either solve for $$f$$ or prove that a solution does not exist there is one and only one solution (at least one solution exists, courtesy of Chrystomath). My own attempts at solving have been... unsuccessful.

Edit:

In their answer, Chrystomath provides a solution:

$$f(x)=a\log(x)^{-2/3}\quad:\quad a\in\left\{z\in\mathbb{C}\mid z^9=\frac{6^4}{8^9}\right\}$$

Which is quite possibly the most multivalued multivalued function I've ever seen.

I don't know whether or not the solution is unique, and I would still be interested in any other solutions.

• A better title would perhaps be more appreciated ;-) – tatan Mar 21 at 17:38
• @tatan Title changed, thanks for the advice. – R. Burton Mar 21 at 19:31

One possible solution. Try $$f(x)=a(\ln x)^k$$, then $$a(\ln x)^k=2(\ln x)^2a^2(\frac{3}{8})^{2k}(\ln x)^{2k}a^2(\frac{1}{4})^{2k}(\ln x)^{2k}=2a^4(\frac{3}{32})^{2k}(\ln x)^{2(1+2k)}$$ from which $$a$$ and $$k=-2/3$$ can be found.
• I can't believe I didn't think of that... Well, it's certainly a solution. It also appears that $a$ and $k$ are unique, with $k=-2/3$ and $a\in\left\{z\in\mathbb{C}\mid z^9=6^4/8^9\right\}$, the only real $a$ being $6^{4/9}/8$ (verified numerically by finding the zeros of $h(x,y)=y(\ln x)^k-2y^4(\frac{3}{32})^{2k}(\ln x)^{2(1+2k)}\bigg\vert_{k=-2/3}$). – R. Burton Mar 21 at 20:18