# x^y=y^x

Discussion in 'Physics & Math' started by BloodSuckingGerbile, Jul 5, 2002.

1. ### James RJust this guy, you know?Staff Member

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31,445
What's ProductLog?

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41

5. ### DinosaurRational SkepticValued Senior Member

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4,829
I wish I had Mathematica, but cannot afford it.

I have MathCad 7 and PSI Plot 5, both of which will do surface plots, as will Mathematica. I never use these capabilities and do not want to spend the time now on this problem.

Perhaps a Mathematica owner might want to use the surface plot capabilities to check for all real solutions.

Superimpose the following three surface plots.

z = x^y

z = y^x

The plane y = x

Each surface intersects the plane at possible solutions. If both surfaces intersect at the same height off the XY-Plane, you have a solution.

It should not be difficult to pick out possible solutions, verify them numerically, and be pretty sure you found them all.

BTW: Has anybody considered complex solutions? I have not studied all the posts carefully. Perhaps this has already been done. If not, it is an interesting extension of the problem.

7. ### BloodSuckingGerbileMaster of PuppetsRegistered Senior Member

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440
Well, x,y,z, are N, but it really is interesting...

OKAY EVERYBODY!!

UPDATE!!!!

X,Y and Z ARE COMPLEX!!!

Thanx, Dinosaur

8. ### ChristCrusherRegistered Senior Member

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you can't use complex numbers, as given in the problem statement

anyways, the formal solution is anything that satisfies the following system

k=1+ ln(k)/ln(x) ;
where
y=k*x; and y and x are non-zero integers (hence is k) , as that is a trivial solution space anyways

given that this system is underspecified, only iterative solution seeking can work, as evidenced earlier

9. ### Han BaumerMemberRegistered Senior Member

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Complex powers

Is x^y defined for every x and y? According to my math-book (Thomas M. Apostol) x^y is defined to be exp(y*Log(x)) for x,y>0. The Exp and Log funtion are defined for complex numbers, so maybe x^y can be defined for complex x and y. Does anyone know this?

Greetings,

Han.

10. ### ChristCrusherRegistered Senior Member

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Re: Complex powers

duh.

11. ### Han BaumerMemberRegistered Senior Member

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Re: Re: Complex powers

I take it you are a real Baevis&Butthead fan!

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13. ### Han BaumerMemberRegistered Senior Member

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Correct! I think you mean that x^y is defined to be Exp(y*Log(x)) for Complex x and y where x,y <>0. I suspected so, but I couldn't get any confirmation.

Well, if that is the case the solution to the problem x^y = y^x is as follows:

x^y = y^x
= { definition of x^y, x, y<>0}
Exp(y*Log(x)) = Exp(x*Log(y))
= { Exp(x) = Exp(y) <=> x=y }
y*Log(x) = x*Log(y)
= { x,y <>0, division by x and by y}
1/x*Log(x) = 1/y*Log(y)
= { Log(x) = -Log(1/x)}
1/x*Log(1/x) = 1/y*Log(1/y)
= { x = Exp(Log(x)) }
Exp(Log(1/x)))*Log(1/x) = 1/y*Log(1/y)
= { Definition of ProductLog: x*Exp(x) = y <=> x=ProductLog(y) }
Log(1/x) = ProductLog(1/y*Log(1/y))
= { Exp(x) = Exp(y) <=> x=y, x = Exp(Log(x)) }
1/x = Exp(ProductLog(1/y*Log(1/y)))
= { 1/x = Exp(y) <=> x=Exp(-y) }
x=Exp(-ProductLog(1/y*Log(1/y)))

There has to be a step in the proof where a '<=' should be instead of '='. I cannot find it. This has to be the case since the last formula doesn't describe all the solutions, since clearly x=y is also a solution.

Greetings,

Han.