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# About $$x^y=y^x$$

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?

Note by Janardhanan Sivaramakrishnan
1 year, 11 months ago

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## Comments

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There are solutions that do not fit. There will be a negative real solution as long as $$y=\frac{a}{b}>1$$, and there might be 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 $$y=\frac{a}{b}$$, $$k_c<y<1$$, $$0.752688172043<k_c<0.757894736842$$

- 1 year, 11 months ago

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I do not even restricting $$x,y$$ to be real numbers.

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

- 1 year, 11 months ago

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Related: soumava's algorithm

- 1 year, 8 months ago

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Comment deleted Dec 15, 2015

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That's the same as the second condition, just substitute $$\alpha=\frac{1}{\beta -1}$$

- 1 year, 11 months ago

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

- 1 year, 11 months ago

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$$\Large 2^4=4^2$$

- 1 year, 11 months ago

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This also fits the pattern. Take $$\alpha = 2$$, $$x=2^{\frac{1}{2-1}}=2,y=2*2=4$$.

- 1 year, 11 months ago

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