Find the all possible real solutions of $$x^y=y^x$$ $x,y$ both are real numbers.
My attempt:I observed the following solutions
$x=2,y=4$
$x=4,y=2$
$y=x$
Is there any other possible solutions?
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2Those are the only positive integer solutions. There will be infinitely many real solutions. – Mathmo123 Sep 24 '14 at 13:21
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Consider the equivalent $x^{1/x}=y^{1/y}$ – robjohn Sep 24 '14 at 13:21
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@ThomasAndrews: They ask: other than $x=y$. – robjohn Sep 24 '14 at 13:23
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The duplicate asks only for integers, but the accepted answer gives a general formula for all real solutions starting with $u\in\mathbb R$. Voting to close. – AlexR Sep 24 '14 at 13:25
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@robjohn Sure, just covering all the bases. Also, it's easy to see there are infinitely many solutions since $\log x/x$ has a local maximum where $x=e$. – Thomas Andrews Sep 24 '14 at 13:25
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@ThomasAndrews the OP also stated $y=x$ as a solution – Mathmo123 Sep 24 '14 at 13:26
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@DietrichBurde thank you so much. – Flip Sep 24 '14 at 13:32
1 Answers
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Applying the logarithm to both sides gives $$y \log x = x \log y,$$ and rearranging gives $$\frac{\log x}{x} = \frac{\log y}{y}$$, so we get a nontrivial solution (i.e., one for which $x \neq y$) for any value assumed twice by $f(x) := \frac{\log x}{x}$. More explicitly, if $x$ and $y$ are distinct values such that $f(x) = f(y)$, then by construction we have a nontrivial solution $x^y = y^x$.
The function $f$ is obviously continuous, and
- has a single root at $x = 1$,
- is strictly increasing on $(0, e)$ and strictly decreasing on $(e, \infty)$,
- has a unique maximum of $\frac{1}{e}$ at $x = e$, and
- has limit $\lim_{x \to \infty} f = 0$.
We can thus conclude that the values $f$ takes on twice are precisely those in $(0, \frac{1}{e})$. Given such a value, it is generally difficult to find its two preimages explicitly*, but nonetheless we can parameterize the solutions as $$(x, y) = \left(\left(1 + \frac{1}{u}\right)^u , \left(1 + \frac{1}{u}\right)^{u + 1}\right).$$
See $x^y = y^x$ for integers $x$ and $y$ for much more.
*In closed form, anyway, we can write down solutions in terms of the Lambert-W function, which appears on this site much more than it does in "real life"...
Travis Willse
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