Why $\log(W(e^2))=2-W(e^2)$?

Question

While trying to solve $x+e^x=2$ I found two solutions:

$$x=\log(W(e^2))=2-W(e^2)$$

It is not at all obvious to me why the two solutions would represent the same number. How are the two related?

Any help to make this more intuitive or insightful would be appreciated!

Answer 1

Let's start with the second equality:

\begin{align} x&=2-W(e^2)\\ \implies 2-x &= W(e^2)\\ \implies (2-x)e^{2-x} &= e^2\\ \implies 2e^{2}e^{-x} - xe^{2}e^{-x} &= e^{2}\\ \implies 2e^{-x} - xe^{-x} &= 1\\ \implies 2 - x &= e^{x} \end{align}

Now, with the first equality:

\begin{align} x &= \log(W(e^2))\\ \implies e^x &= W(e^2)\\ \implies e^xe^{e^x} &= e^{2}\\ \implies e^{x+e^{x}} &= e^{2}\\ \implies x+e^{x} &= 2\\ \implies 2 - x &= e^{x} \end{align}

and so we see that the two solutions are consistent.