Use Lambert W Function to Solve $\frac\alpha\beta\ln(x)-x+c=0$

Question

I need to solve this equation for x:

$\frac{α}{β}\ln(x)-x+c=0$

Apparently it has to be done with the Lambert W function. I think the answer is

$x=\frac{α\operatorname{W}(-\frac{βe^-\frac{Cβ}{α}}{α})}{β}$

But I don't know how to get to the answer.

Thanks in advance!

Answer 1

$ \frac{\alpha}{\beta}\ln{x}-x+c=0 $

$ \frac{\alpha}{\beta}\ln{x} = x-c $

$ \ln{x} = \frac{\beta}{\alpha}(x-c) $

$ x = e^{\frac{\beta}{\alpha}(x-c)} $

$ xe^{-\frac{\beta}{\alpha}x} = e^{-\frac{\beta}{\alpha}c} $

$ -\frac{\beta}{\alpha}xe^{-\frac{\beta}{\alpha}x} = -\frac{\beta}{\alpha}e^{-\frac{\beta}{\alpha}c} $

$ W(-\frac{\beta}{\alpha}xe^{-\frac{\beta}{\alpha}x}) = W(-\frac{\beta}{\alpha}e^{-\frac{\beta}{\alpha}c}) $

$ -\frac{\beta}{\alpha}x = W(-\frac{\beta}{\alpha}e^{-\frac{\beta}{\alpha}c}) $

$ x = -\frac{\alpha}{\beta}W(-\frac{\beta}{\alpha}e^{-\frac{\beta}{\alpha}c}) $