Y^x = x^y

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  • #1
jcsd
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For the equation:

[tex]y^x = x^y[/tex]

There are several trivial solutions:

such as:

[tex]y = x[/tex]

and a further set of trivial solutions for:

[tex]y^{y^y} = y^{y^2}[/tex]

Now I know for a fact there are also sevral non-trivial solutions and there is a method thta could possibly obtain a
general solution, so does anyone here know how to obtian a general solution or if not suggest other sets of solutions?
 

Answers and Replies

  • #2
MathematicalPhysicist
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can you show the method that possibly give a general solution?

i know for sure that some users at scinceforums.net will be glad to know the solution.
 
  • #3
jcsd
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I have seen the method before and a few non-trivial solutions obtained from it, but unfortunately I cannot remember it or even it's name.

As I said before their most definetely does exist non-trivial solutions to this equation which some might be interetsed to know.
 
  • #4
NateTG
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Originally posted by jcsd
For the equation:

[tex]y^x = x^y[/tex]

does anyone here know how to obtian a general solution or if not suggest other sets of solutions?
[tex]y^x=x^y[/tex]
[tex]xlny=ylnx[/tex] (assuming x and y are > 0 )
[tex]\frac{ln x}{x}=\frac{ln y}{y}[/tex] (x and y are non-zero by assumption above)
[tex]x^{\frac{1}{x}}=y^{\frac{1}{y}}[/tex]

Now , in the reals, for positive [tex]x[/tex]
[tex]f(x)=x^{\frac{1}{x}}[/tex] has two inverses for any [tex]f(x) \in (1,e^{\frac{1}{e}})[/tex] which leads to a family of non-trivial solutions. For example:
2 and 4 or 3 and 2.47779..

Fractional roots in the complex domain are multi-valued, so things get a bit more complicated if this approach is applied for x,y < 0.
[tex]x^y=y^x[/tex]
[tex]e^{ylnx}=e^{xlny}[/tex]
[tex]ylnx=xlny + n2\pi i[/tex]
Since the natural long in complex numbers has multiple branches, things get a bit sticky here, but there are probably additional complex solutions.
 

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