So I've encountered this exercise where it asks you to find given $a, m, n \in \mathbb{R}$ an $x \in \mathbb{R^+}: mx + n + e^x = a$. I was told it can be algebraically solved using Lambert's W function. I'm not sure it can be done without $m, n$ and $a$ being concrete numbers and not parameters.
So, that's pretty much it: Can $mx + n + e^x = a$ be solved with Lambert W function?
Anything you can do with concrete numbers you can usually do with parameters: $$\begin{align*}& e^x = a-n-mx \\ \implies & 1 = (a-n-mx)e^{-x} \\ \implies & \frac{1}{m} = \left(\frac{a-n}{m}-x\right)e^{-x} \\ \implies & \frac{e^{s}}{m} = \left(s-x\right)e^{s-x} \qquad\qquad s:= (a-n)/m \\ \implies & W\left(\frac{e^s}{m}\right) = s-x \\ \implies & x = s - W\left(\frac{e^s}{m}\right) \end{align*} $$
Here I've written $W$ to mean the proper branch (i.e. either $W_0$ or $W_{-1}$), depending on the value of $\frac{e^s}{m}$.