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.
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.