In the solution of $x + e^x = 2$, how does $ln(W(e^2))$ become $2 - W(e^2)$?

Question

$x + e^x = 2$

1. Take $e$ of both sides

2. Apply $W(x)$ function then $\ln$ of both sides.

This gives me an answer of $\ln(W(e^2))$, but when I put the equation into Wolfram, it gives $2-W(e^2)$ as answer.

How does this happen?

Answer 1

$x=\ln( W(e^2))$ is correct solution, then you need just take

$$2-x=e^{\ln (W(e^2))}=W(e^2)$$

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

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

So, both are equal.

Answer 2

$W(z)e^{W(z)}=z$, therefore $$e^{2-W(e^2)}=\frac{e^2}{e^{W(e^2)}}=\frac{e^2}{e^2/W(e^2)}=W(e^2)$$

Which incidentally proves that, if $W(e^2)$ exists, then $W(e^2)>0$ and $2-W(e^2)=\ln W(e^2)$.

Answer 3

The $W$ function is defined so that $u=W(u)e^{W(u)}$ (for $u\ge0$). Letting $u=e^2$ and taking logs, we have

$$2=\ln(e^2)=\ln(W(e^2)e^{W(e^2)})=\ln(W(e^2))+W(e^2)$$

from which $\ln(W(e^2))=2-W(e^2)$ follows.