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I have the following equation $e^{-x/b}(a+x) = e^{x/b}(a-x)$

where $b > 0$, and $a > 0$

I need to solve for $x$. I can do it numerically, but would prefer if there was a closed form solution.

It seems to me that there likely is no closed form solution, but thought I'd ask the experts here, just in case.

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    $\begingroup$ Notice that $e^{-x/b}(a+x)=e^{x/b}(a-x)\iff e^{2x/b}=(a+x)/(a-x)$. Subsitute $y=2x/b$, and obtain $e^y=(2a/b+y)/(2a/b-y)$, so you only have to solve $e^y=(c+y)/(c-y)$, $c>0$. $\endgroup$ – Luiz Cordeiro Jul 2 '14 at 21:38
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    $\begingroup$ Letting $u = \frac{2x}{b}$ in the equation $\exp\left(\frac{2x}{b}\right) = \frac{a+x}{a-x}$ gives $e^u = \frac{2a+bu}{2a-bu},$ and I am nearly certain that this can't be solved explicitly (at least without using something like the Lambert function) even when $a=b=1.$ $\endgroup$ – Dave L. Renfro Jul 2 '14 at 21:38
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    $\begingroup$ Mathematica can't solve the original equation or the one supplied by @DaveL.Renfro not very surprising though. I highly doubt that this can be solved in closed form. $\endgroup$ – Alice Ryhl Jul 2 '14 at 21:52
  • $\begingroup$ Another tidy presentation of the equation is $\frac xa = \tanh(\frac xb)$. $\endgroup$ – user21467 Jul 2 '14 at 23:58
  • $\begingroup$ $x=0$ is one solution $\endgroup$ – Claude Leibovici Jul 3 '14 at 8:03
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Is there a closed form solution to this equation ?

No. Not even one in terms of Lambert's W function.

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As Steven Taschuk commented, the equation can write $$\frac xa = \tanh(\frac xb)$$ Changing variable $x=by$, this write $$\frac {b}{a} y = \tanh(y)$$ Taking into account the shape of $\tanh(y)$ and essentially the fact that, at $y=0$, $\Big(tanh(y)\Big)'=1$, there is a solution only if $a\gt b$. Otherwise, the only possible solution is $y=0$ which is valid for any non zero values of $a$ and $b$. If $y_*$ is a solution, $-y_*$ is the other one. So, we have a maximum of three solutions.

Unfortunately, it seems that there is no closed form for the solution and that numerical methods should be required.

Using Taylor expansion to the third order, a first approximation is given by $$y_*= \sqrt{3\frac{a-b}{a}}$$

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