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When answering this question, I contemplated across the following problem:

Find $x$ such that $x^{\frac{1}{x}}=0.$

I thought RHS $ = 0$ is a special case, so I attempted solving $$\frac{1}{x} \ln x = \ln c$$ from some constant $c > 0,$ but I do not know how. I heard about Lambert W, but I am not familiar with it.

BTW, Maple solve(x^(1/x) = 0) gives $x=0$, and WolframAlpha gives (no solution exist)!

How can I solve this?

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To solve $\ln x = cx$, you need to reach for far more advanced tools than basic ones; you need Lambert's W function. –  Arturo Magidin Mar 3 '12 at 23:43
    
So if I understood correctly, the solution to $$\frac{1}{x} \ln x = \ln c$$ using Lambert's W is $$x = e^{-W(-\ln c)}.$$ –  user2468 Mar 3 '12 at 23:56

2 Answers 2

up vote 7 down vote accepted

The only way you can ever have $a^b=0$ is to have $a=0$, but in the current case this leads to $0^{1/0}$, which is undefined. So, there's no solution.

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However, $\lim_{x\to0}x^\frac1x=0$. –  Ilmari Karonen Mar 3 '12 at 23:44

The functin $f(x)=x^{1/x}=\exp(\frac{1}{x}\log x)$ never equals zero, however the limit as $x\to 0^+$ evaluates to zero, so if we consider the domain $[0,\infty)$ we could remove the discontinuity at $x=0$ if we desired.

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