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.


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.

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

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  • $\begingroup$ 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 $h(x,y)=y(\ln x)^k-2y^4(\frac{3}{32})^{2k}(\ln x)^{2(1+2k)}\bigg\vert_{k=-2/3}$). $\endgroup$ – R. Burton Mar 21 '19 at 20:18

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