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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 $4.96$)?

Vinícius
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    The Lambert-W function is multivalued on $(-1/e, 0)$. – player3236 Sep 29 '20 at 14:50
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    The solutions are $5 + W( - 5e^{ - 5} )$ where $W$ is either of the two real branches of the Lambert $W$-function. This because you can re-write your equation in the form $(x - 5)e^{(x - 5)} = - 5e^{ - 5} $. – Gary Sep 29 '20 at 14:54
  • Yeah, that's right @Gary, I messed with some minus signals in my calculations. I edited the post. – Vinícius Sep 29 '20 at 15:51

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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.

player3236
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  • This solves my doubts, I was only considering the principal branch of the Lambert-W function. Thank you! – Vinícius Sep 29 '20 at 15:52