It is casually mentioned in de Bruijn that, for the equation
$$ x e^x = t, $$
"it is not very difficult" to show that $x = O(\log t)$ as $t \to \infty$ using Rouché's Theorem.
It's definitely not very difficult to show the result, but the obvious method doesn't need Rouché's Theorem at all. When $t > e$ we have $x > 1$, which immediately implies
$$ x = \log t - \log x < \log t. $$
So what's the meaning of the remark?