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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 $a,b>0$

Mobeus Zoom
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  • Simplifies to $e^{-x}+\frac{a}{b}x-1=0$; therefore are you assuming that $\frac{a}{b}$ be a non-zero constant? – James Arathoon Nov 16 '19 at 15:42
  • Yeah sorry I should have mentioned, $a,b>0$ – Mobeus Zoom Nov 16 '19 at 16:34
  • 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? – BCLC Jan 20 '21 at 17:39

2 Answers2

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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 $$x=\frac{b}{a}+W_{-1}\left(-\frac{b }{a}e^{-\frac{b}{a}}\right)$$

Claude Leibovici
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My answer is for solutions in terms of elementary functions.

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

The left-hand side of equation (1) is an expression of both $x$ and $e^x$, algebraic over $\mathbb{C}$. It can be shown that such expressions don't have elementary inverses:
How can we show that $A(z,e^z)$ and $A(\ln (z),z)$ have no elementary inverse?

For $a,b\in\overline{\mathbb{Q}}$, equation (1) is an algebraic equation of both $x$ and $e^x$. Lin (1983) proved, assuming Schanuel's conjecture ist true, that such equations, if irreducible, don't have solutions in the elementary numbers.

IV_
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