Solving $ax=e^{bx}$

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I've been asked by a friend to help him solve this equation, but since we couldn't find the right answer, I thought about posting it here. Firstly, I thought about derivating both sides and get:

$a=be^{bx}$

and from here we could find the answer pretty quickly, but then I noticed that what I did was incorrect because I can't derivate both sides - the functions aren't equal.

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By the Lambert W function, we have,

$$ax=e^{bx}$$

$$xe^{-bx}=\frac1a$$

$$-bxe^{-bx}=-\frac ba$$

$$-bx=W\left(-\frac ba\right)$$

$$x=-\frac1bW\left(-\frac ba\right)$$

If $0<b/a<e^{-1}$, then there will be two real solutions, denoted by the two branches of the Lambert W function.

If $b/a<0$, then there is only one real solution.

If $b=0$, then the solution is $x=\frac1a$.

If $b/a=e^{-1}$, then $x=\frac1b$.

If $b/a>e^{-1}$, then there are no real solutions.


See here and here for some basic approximations.

I also made a quick approximation calculator-ish thing for the Lambert W function:

Desmos.com

$f(x)$ works best for large inputs and $g(x)$ works best for inputs close to zero.