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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
12 months ago

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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 $$y=\frac{a}{b}$$, $$k_c<y<1$$, $$0.752688172043<k_c<0.757894736842$$ · 12 months ago

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 $$y^x=e^{\frac{-\pi}{2}i \times (i)}=e^{\frac{\pi}{2}}$$ · 12 months ago

Related: soumava's algorithm · 8 months, 3 weeks ago

Comment deleted 11 months ago

That's the same as the second condition, just substitute $$\alpha=\frac{1}{\beta -1}$$ · 11 months, 4 weeks ago

LOL, was trying to figure out why it seemed similar. Thanks. · 11 months, 4 weeks ago

$$\Large 2^4=4^2$$ · 12 months ago

This also fits the pattern. Take $$\alpha = 2$$, $$x=2^{\frac{1}{2-1}}=2,y=2*2=4$$. · 12 months ago

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