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

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?

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For search engine purposes: the Lambert W function has the asymptotic relation $W(t)=O(\log\,t)$ – J. M. Aug 14 '11 at 0:49
Even more, $W(t) = \log t + \log \log t + o(1)$. – Antonio Vargas Aug 14 '11 at 21:55
@Antonio, log minus log log, actually. – Did Aug 15 '11 at 13:14
@Didier: Whoops, you're right! – Antonio Vargas Aug 22 '11 at 21:08
"It is casually mentioned in de Bruijn that..." Is this a book or a paper? Can you give its name? – Srivatsan Sep 17 '11 at 17:32