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I have the equation

$$\frac{b}{x}+\frac{ae^x}{1-e^x}=0$$

How do I prove this equation has no closed-form solution for $x$?

Edit: please note $a,b>0$

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  • $\begingroup$ Simplifies to $e^{-x}+\frac{a}{b}x-1=0$; therefore are you assuming that $\frac{a}{b}$ be a non-zero constant? $\endgroup$ Nov 16, 2019 at 15:42
  • $\begingroup$ Yeah sorry I should have mentioned, $a,b>0$ $\endgroup$ Nov 16, 2019 at 16:34
  • $\begingroup$ Mobeus Zoom, does your post then actually not explicitly and directly say that $\frac a b$ is a non-zero constant and hence $b=ka$ in answer below is actually relying on deduction made explicitly so far only in @JamesArathoon's comment? $\endgroup$
    – BCLC
    Jan 20, 2021 at 17:39

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I suppose that you mean no closed-form solution in terms of elementary functions.

Let $b=k a$ which makes the equation to be $$k+(x-k)e^x=0$$ Now, let $x=k+y$ to make $$y e^{k+y}+k=0 \implies y e^y=-ke^{-k}$$ and the solution is $$y=W_{-1}\left(-ke^{-k} \right)$$ where appears the lower branch of Lambert function. In the real domain, this function is defined for $0 < k \leq 1$.

Back to $(x,a,b)$ this gives $$x=\frac{b}{a}+W_{-1}\left(-\frac{b }{a}e^{-\frac{b}{a}}\right)$$

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  • $\begingroup$ So how do you show there is no closed-form solution in terms of elementary functions? $\endgroup$ Nov 17, 2019 at 7:24
  • $\begingroup$ @JairTaylor. This is the big question, I fully agree ! Cheers $\endgroup$ Nov 17, 2019 at 7:25
  • $\begingroup$ Thanks. I guess one could go from here and bring in the proof that Lambert W doesn't have a closed-form solution in elementary functions, even though this may not be a proof of the overall lemma. $\endgroup$ Nov 17, 2019 at 14:02
  • $\begingroup$ The other thing I'm wondering is, what if $b>a$ (so that $b/a<1$ is not true and the Lambert W isn't defined for our domain)? $\endgroup$ Nov 17, 2019 at 14:03

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