Solve $e^{-x} + \frac{x}{5} = 1$

Question

I know there are similar questions, but I want to discuss something I am missing to capture.

The obvious solution to this is $x = 0$ and the other one is given by $x = 5 + W(-5/e^5)$, where $W(x)$ is the W-Lambert function. But this second solution evaluates to zero too (or am I wrong?).

Ploting the graphs we can see two solutions to the equation; how can one obtain the second one (according to Wolfram, it is approximately $4.96$)?

Answer 1

If you are using WolframAlpha, you will obtain the solution:

$$x=W(-5e^{-5})+5$$

Since $0>-5e^{-5} > -e^{-1}$, the Lambert-W function has two values. On WolframAlpha you can input:

5+LambertW(0,-5e^(-5)) or 5+LambertW(-5e^(-5))

for the principal branch, where $W \ge -1$. This equates to $4.96\dots$;

5+LambertW(-1,-5e^(-5))

for the lower branch, where $W \le -1$. This equates to $0$ exactly.