# Y^x = x^y

1. Feb 22, 2004

### jcsd

For the equation:

$$y^x = x^y$$

There are several trivial solutions:

such as:

$$y = x$$

and a further set of trivial solutions for:

$$y^{y^y} = y^{y^2}$$

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?

2. Feb 23, 2004

### MathematicalPhysicist

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. Feb 23, 2004

### jcsd

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. Feb 23, 2004

### NateTG

$$y^x=x^y$$
$$xlny=ylnx$$ (assuming x and y are > 0 )
$$\frac{ln x}{x}=\frac{ln y}{y}$$ (x and y are non-zero by assumption above)
$$x^{\frac{1}{x}}=y^{\frac{1}{y}}$$

Now , in the reals, for positive $$x$$
$$f(x)=x^{\frac{1}{x}}$$ has two inverses for any $$f(x) \in (1,e^{\frac{1}{e}})$$ 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.
$$x^y=y^x$$
$$e^{ylnx}=e^{xlny}$$
$$ylnx=xlny + n2\pi i$$
Since the natural long in complex numbers has multiple branches, things get a bit sticky here, but there are probably additional complex solutions.