If $y=ax^be^{-cx}$ then $x=g(y)$, find $g$

Question

I have this function

$$y=0.384394\cdot x^{0.341429}\cdot e^{-0.004749 x}$$

Based on this function I would like to know how I can I get $x=g(y)$.

Answer 1

This is a job for the Lambert W function.

Look here: https://en.wikipedia.org/wiki/Lambert_W_function

This solves $y = ze^z$ for $z$ as a function of $y$.

The next step is to convert your function to this form.

Answer 2

Write it as $$ y = a x^b e^{-cx}$$ where we assume $a, b, c > 0$. Then $$ (y/a)^{1/b} = x e^{-cx/b}$$ so with $t = -cx/b$ we have $$ -\dfrac{c\; (y/a)^{1/b}}{b} = t e^{t} $$ Then $$t = W\left(- \frac{c\; (y/a)^{1/b}}{b}\right)$$ i.e. $$x = - \dfrac{b}{c}\; W\left(- \frac{c\; (y/a)^{1/b}}{b}\right)$$ where $W$ is a branch of the Lambert W function. If $0 < c (y/a)^{1/b}/b < 1/e$, there are two real branches (the $-1$ and $0$ branches), while if $ c (y/a)^{1/b}/b > 1/e$ neither branch is real. Here's a plot, with the $-1$ branch in blue and the $0$ branch in red.