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In a previous question, I learned that the equation $$x^{\frac{1}{x}} = c$$ have no solutions when $c = 0$. Below, I tried using Lambert's W function, and I found a solution at $x = 0$. Did I make a mistake?

If $x^{\frac{1}{x}} = c$ then $${\frac{1}{x}}\ln x = \ln c$$ Let $x = e^{-y}$. So $${\frac{1}{x}}\ln x = -y e^{y} = \ln c$$ The solution of $y e^y = -\ln c$ using Lambert's W is $$ y = W(-\ln c).$$ Hence $$ x = e^{-W(-\ln c)} $$ is a solution of $x^{\frac{1}{x}} = c$. Now, to solve $x^{\frac{1}{x}} = 0$, I tried $$ x = \lim_{c\to 0} e^{-W(-\ln c)} = 0,$$ since $\lim -\ln c \to \infty$ as $c\to 0$, and $\lim W(z) \to \infty$ as $z \to \infty$.

Hence, $x = 0$ is a solution to $x^{\frac{1}{x}} = 0$. But this contradicts the answers I got for my previous question. I must've made a mistake. Please point out where.

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Didn't anon point out in your previous question that (limitwise, from above) $x = 0$ is indeed a solution? So this does not contradict that. – TMM Mar 4 '12 at 1:14
@TMM $x \to 0^+$ is a solution, which not my case here. – user2468 Mar 4 '12 at 1:17
If $c=0$ then $\ln(c)$ is undefined. – Patrick Mar 4 '12 at 1:37
So, you claim $0^{1/0}$ is defined? – GEdgar Mar 4 '12 at 3:33
@GEdgar is your question for me or Patrick? – user2468 Mar 4 '12 at 4:54
up vote 3 down vote accepted

When you write, $x=e^{-y}$, you have ruled out $x=0$ as a solution, since there's no such $y$.

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