Consider the equation \[x^y=y^x\] and it is required to find ALL solutions \((x,y)\) for this equation,

It is obvious that all ordered pairs \((x,x)\), i.e., \(y=x \ne 0\) is a trivial set of solutions. (Here, I am not sure if \(x=y=0\) can be considered as a solution!!)

In addition, the set \(x=\alpha^{\frac{1}{\alpha-1}},y=\alpha x\), for some real (not sure if complex works) value of \(\alpha \ne 1,0\), would also fit in as solutions.

Are there any solutions which do not fit into the above two categories? Is it possible to prove otherwise?

## Comments

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TopNewestThere are solutions that do not fit. There will be a negative real solution as long as \(y=\frac{a}{b}>1\), and there

mightbe 2 negative real solutions when \(y=\frac{a}{b}<1\), where \(a\) is an even number and \(b\) is an odd number. I haven't actually studied this in detail yet.From what I got so far, numerically, there would be 2 negative real solutions when \(y=\frac{a}{b}\), \(k_c<y<1\), \(0.752688172043<k_c<0.757894736842\) – Julian Poon · 1 year, 3 months ago

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Take \(\alpha = -1\), we get \(x=i=e^{\frac{\pi}{2}i},y=-i=e^{\frac{-\pi}{2}i}\) and

thus \(x^y=e^{\frac{\pi}{2}i \times (-i)}=e^{\frac{\pi}{2}}\)

and also \(y^x=e^{\frac{-\pi}{2}i \times (i)}=e^{\frac{\pi}{2}}\) – Janardhanan Sivaramakrishnan · 1 year, 3 months ago

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Related: soumava's algorithm – Agnishom Chattopadhyay · 1 year ago

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– Julian Poon · 1 year, 3 months ago

That's the same as the second condition, just substitute \(\alpha=\frac{1}{\beta -1}\)Log in to reply

– Sharky Kesa · 1 year, 3 months ago

LOL, was trying to figure out why it seemed similar. Thanks.Log in to reply

\(\Large 2^4=4^2\) – Nihar Mahajan · 1 year, 3 months ago

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– Janardhanan Sivaramakrishnan · 1 year, 3 months ago

This also fits the pattern. Take \(\alpha = 2\), \(x=2^{\frac{1}{2-1}}=2,y=2*2=4\).Log in to reply