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I came across the equation $$x^{1/x}=2^{1/2}$$ where $x\in\mathbb R$. One can immediately see that $x=2$ is a solution, but it is easy to miss that $x=4$ satisfies the equation as well. Verfiying that $2,4$ are solutions is not hard, but how would one go about formally solving this equation, i.e. how could one solve for $x$? I am asking because if the equation was say $x^{1/x}=2^{1/3}$ then the two solutions would not be obvious.

Basically, how to fill in the dots: $$x^{1/x}=2^{1/2} \iff \ldots\iff x=2 \text{ or } x=4$$

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    $\begingroup$ for $0 < A < \frac{1}{e}$ there are two solutions to $\frac{\log x}{x} = A.$ For $A > \frac{1}{e}$ there are no solutions. Just draw the graph of $y = \frac{\log x}{x}$ for $x > 0.$ $\endgroup$ – Will Jagy Jan 17 '15 at 23:32
  • $\begingroup$ @WillJagy Thank you for this, but I am more interested in how to obtain the solutions, not so much in how to prove their existence. $\endgroup$ – Phil-ZXX Jan 17 '15 at 23:46
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    $\begingroup$ Tom, you asked about the well known example, $2^4 = 4^2.$ Someone came up with a reasonable conjecture on all rational solutions to $x^y = y^x.$ Having trouble finding that. I mean he asked his conjecture as an MSE question. $\endgroup$ – Will Jagy Jan 18 '15 at 1:26
  • $\begingroup$ a moderator found it for me, see answer at meta.math.stackexchange.com/questions/19299/… $\endgroup$ – Will Jagy Jan 18 '15 at 2:39
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Notice

$$x ^{1/x} = 2^{1/2} \Rightarrow x = 2^{x/2} \Rightarrow x^{2} = 2^x$$

then you may look want take a look here.

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  • $\begingroup$ There are two good answers there. Feel free to ask anything. $\endgroup$ – Aaron Maroja Jan 17 '15 at 23:32
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    $\begingroup$ Thank you. The accepted answer in your link does not actually provide any insights in how to obtain the solutions, whereas your solutions looks very nice (this was exactly what I was looking for). But I am still wondering, is there any way around the Lambert W function? $\endgroup$ – Phil-ZXX Jan 18 '15 at 0:08
  • $\begingroup$ @Tom Yes, see my answer. And no, you won't get explicit results except for the special cases like $2^4=4^2$. $\endgroup$ – user2345215 Jan 18 '15 at 0:08
  • $\begingroup$ @user2345215 I do agree with the reasoning of your answer, but it doesn't tell me how to obtain 2,4 (asssuming I did not know them). If the equation was $x^{1/x}=2^{1/3}$ we would have to actually derive the solutions, and I am looking for a way to do this (not a way to prove that 2,4 are the only solutions). $\endgroup$ – Phil-ZXX Jan 18 '15 at 0:12
  • $\begingroup$ @Tom I'm glad I could help you Tom, well there is Newton's approximation. $\endgroup$ – Aaron Maroja Jan 18 '15 at 0:23
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Note that $$(x^{1/x})'=(e^{(\ln x)/x})'=\tfrac{1-\ln x}{x^2}x^{1/x}$$ So the function is increasing on $(0,e)$ and decreasing on $(e,\infty)$. Therefore $2$ and $4$ are the only solutions.

From this follows that $x^{1/x}=c$ has two solutions for $c\in (1,e^{1/e})$, one for $c\in(0,1]\cup\{e^{1/e}\}$ and no for $c\in(e^{1/e},\infty)$.

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I suppose it is necessary to consider what happens when $x < 0.$ There is a question of interpretation of $x^{\frac{1}{x}}$ for $x \leq 0,$ ( for example, $(-1)^{r}$ could reasonable interpreted as a non-real complex number if $r$ is an irrational positive real number, if we consider $-1$ to be $e^{i \pi}$), so I assume the function only to be defined for $x >0$, where it is defined to be $e^{\frac{\log x}{x}}.$ As far as this question is concerned, it seems that the case $x >0$ is the most relevant case anyway.

Note that $x >0$ is a solution of $x^{\frac{1}{x}} = 2^{\frac{1}{2}}$ if and only if $\frac{\log x}{x} = \frac{\log 2}{2}.$ The derivative of $\frac{\log x}{x}$ is $\frac{ 1 - \log x}{x^{2}}$ which is positive when $x <e$ and negative when $x > e.$ Hence $\frac{\log x}{x}$ increases when $x <e$, then decreases when $x >e$, taking a maximum value when $x = e.$ The values of $\frac{ \log x}{x}$ are positive on $(1, \infty)$ and the function is differentiable on that interval.Now if $\frac{log{x}}{x} = \frac{\log{y}}{y}$ for $1 < x < y,$ then the derivative vanishes somewhere on $(x,y)$ by the Mean Value Theorem. Since the derivative vanishes only at $e$, we see that any value of $\frac{\log x}{x}$ can be attained at most twice. Since $\frac{\log{4}}{4} = \frac{2\log 2}{2 \times 2} = \frac{\log 2}{2},$ we see that $2$ and $4$ are the only values of $x$ for which $x^{\frac{1}{x}} = 2^{\frac{1}{2}}.$

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