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Is there a closed-form solution for $x$ in the following equation? If so, how does it look like? If not, can you explain why?

$$ A\log\frac{jx-a}{jx+a}=\log\frac{j-ax}{j+ax} $$

where $A,x\in\mathbb{R}$ and $j$ is the imaginary unit? The equation stems from some trigonometric equation which I formulated into this way. I have tried different ways, but the $A\neq 1$ makes it very difficult, as taking the exp on both sides, gives me $(jx-a)^A$ on the left-hand side, and then I can't proceed.

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This is not an answer. But as a possible approach, take the exp of both sides, and let $jx=y$ $$\left(\frac{y -a}{y+a}\right)^A=\frac{1+ay}{1-ay}$$ I think this form is easier to proceed.

Edit- As a further attempt, if $a$ is real then you can take another change of variable, $ajx=t$ to get: $$\left(\frac{t -a^2}{t+a^2}\right)^A=\frac{t+1}{t-1}$$ Now if you plot the two functions: $$f_1(t)=\frac{t -a^2}{t+a^2},\quad f_2(t)=\frac{t+1}{t-1}$$ you can see that for non-integer $A$, there is no real solutions.

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    $\begingroup$ Clearly not solvable in general (I think), so this is the best one can likely do. $\endgroup$ – Simply Beautiful Art Feb 18 '17 at 12:29
  • $\begingroup$ What is amazing is that, assuming that $A$ is an integer, the solutions exist (with nasty radicals) up to $A=8$ solving, for $y$ equation $(1-a y) (y-a)^A-(1+a y) (y+a)^A=0$ $\endgroup$ – Claude Leibovici Feb 18 '17 at 12:53
  • $\begingroup$ Thanks for your attempt. Though it looks simpler, still the problem with A in the exponent remains. $\endgroup$ – Maximilian Matthé Feb 18 '17 at 13:03
  • $\begingroup$ @SimplyBeautifulArt can you explain, why it is clearly not solvable? Would it become solvable if we allow more advanced functions like lambertz W-function or similar? $\endgroup$ – Maximilian Matthé Feb 18 '17 at 13:04
  • $\begingroup$ @ClaudeLeibovici Oh, nice :D $\endgroup$ – Simply Beautiful Art Feb 18 '17 at 13:10

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