I have an equation of the form
$$ x a + b = \exp(xc)$$
where I, in fact, know that $b =1$. Which implies that one solution to the equation is always at $x = 0$. I'm now searching for the other solution.
I learned from Wolfram Alpha about the existence of the Lambert W function, and tried to employ it here. I got
$$x_n = \frac{-a W_n(-c \frac{\exp(-bc/a)}{a}) - bc}{ac}$$
, where $W$ is the $n$ths branch of the Lambert W function.
However, I want to restrict my solution to the reals, and don't know how. Plugging in arbitrary numbers, I got that for $n=0$, I get $x=0$, and for $n=2$, I get another real solution, and anything else appears to be complex.
The only two branches of $W$ which can ever assume real values are the $n=0$ and $n=-1$ branches. For this to happen in your case, assuming that $a,b,c\in\mathbb{R}$, the argument to $W$ (from your solution, above) has to satisfy:
$$-\frac{1}{e}\le -\frac{c\cdot\exp\left(\frac{-cb}{a}\right)}{a},\,\,n=0$$ $$-\frac{1}{e}\le -\frac{c\cdot\exp\left(\frac{-cb}{a}\right)}{a}\lt 0,\,\,n=-1$$
If the first of the above conditions is satisfied you will get a real root for $n=0$. If the second is also satisfied you will get a second real root for $n=-1$. All other roots will be complex.