Use Rouché's Theorem to show that, if $x e^x = t$, then $x = O(\log t)$ as $t \to \infty$

Question

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?