# Prove there is no closed-form solution to equations with e^x and x

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 $$a,b>0$$

• Simplifies to $e^{-x}+\frac{a}{b}x-1=0$; therefore are you assuming that $\frac{a}{b}$ be a non-zero constant? Nov 16, 2019 at 15:42
• Yeah sorry I should have mentioned, $a,b>0$ Nov 16, 2019 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, 2021 at 17:39

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 $$x=\frac{b}{a}+W_{-1}\left(-\frac{b }{a}e^{-\frac{b}{a}}\right)$$
• 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)? Nov 17, 2019 at 14:03