The following equation has been of interest for a long time. I'm bringing this to your attention again:
$$x^y = y^x$$
How do you attack to find all real and imaginary solutions for this equation? or
What is the nicest solution that you have seen?
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3Start by taking logs and graphing the function $f(x)=\dfrac{\log x}x$. – Ted Shifrin Aug 25 '13 at 19:54
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x=y. but its trivial. x^y < y^x to wolfram alpha shows how it really works. – tp1 Aug 25 '13 at 20:07
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if your $x,y$ are complex and not real, there is no principal value for either expression and you have a mess. If you demand at least one of them real and positive you get something sensible. – Will Jagy Aug 25 '13 at 20:56
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As far as I know, the only non complex pair not on y=x is (2,4). I don't have a proof... – imranfat Aug 25 '13 at 22:48
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Make titles more informative, please. – Pedro Sep 10 '13 at 02:01
3 Answers
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The standard way I have seen to solve $x^y = y^x$ is to define $\frac{y}{x} = r$.
Then $y = xr$, so $x^{xr} = (xr)^x$ or $x^r = xr$ or $x^{r-1} = r$ or $x = r^{1/(r-1)}$.
From this, $y =xr = r^{1+1/(r-1)} = r^{r/(r-1)} $.
Any value you put in for $r$ (except $r=1$) gives you $x$ and $y$ that satisfy $x^y = y^x$, and conversely.
For example, if you start with $r = i$, $\frac{1}{r-1} =\frac{1}{i-1} =\frac{-i-1}{(-i-1)(i-1)} =\frac{-i-1}{2} $ and $\frac{r}{r-1} =\frac{i}{i-1} =\frac{i(-i-1)}{(-i-1)(i-1)} =\frac{1-i}{2} $. Then $x = i^{(-i-1)/2} $ and $y =i^{(1-i)/2} $. (I'll let you work the standard form for $x$ and $y$.)
marty cohen
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Allowing the Lambert W function,
$$y=e^{W_k\left(\frac{-\ln(x)}{x}\right)}$$
where $k$ is the branch. From graphing, you can see that only two branches, $k=0,-1$, (I think) produce real values for $x,y$. All other branches will produce complex values.
Simply Beautiful Art
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