Solve x^y=y^x

[AvgStudent]

Inspired by the other thread, [imath]x^x=x[/imath], I'm trying to solve the following equation.
For [imath]x,y \in \R+[/imath], solve [math]\boxed{x^y=y^x}[/math]I tried to apply a similar technique by taking the log on both sides, but it doesn't work. The nature of the problem is probably different from the other question. Maybe the problem is way over my head, and I'm poking around the wrong bush, without knowing what's going to come out of it. ?

What I have so far...
[math]x^y=y^x\\ y\log x =x\log y\\ \frac{y}{\log y}=\frac{x}{\log x}[/math]How to continue or it's not a viable approach in the first place?

 

[JeffM]

Have not worked it through, but I might think about the change of base formula.

 

[Dr.Peterson]

Inspired by the other thread, [imath]x^x=x[/imath], I'm trying to solve the following equation.
For [imath]x,y \in \R+[/imath], solve [math]\boxed{x^y=y^x}[/math]I tried to apply a similar technique by taking the log on both sides, but it doesn't work. The nature of the problem is probably different from the other question. Maybe the problem is way over my head, and I'm poking around the wrong bush, without knowing what's going to come out of it. ?

What I have so far...
[math]x^y=y^x\\ y\log x =x\log y\\ \frac{y}{\log y}=\frac{x}{\log x}[/math]How to continue or it's not a viable approach in the first place?

It's an entirely different kind of problem, since there are two variables! Consider the difference between solving x^2 = 16 for x (two solutions) and "solving" x^2 = y (infinitely many points on a curve you can graph). You'll have to decide what your goal is. (Presumably you want to exclude the trivial solutions y=x.)

Spoiler

https://en.wikipedia.org/wiki/Equation_xy_=_yx

 

[AvgStudent]

It's an entirely different kind of problem, since there are two variables! Consider the difference between solving x^2 = 16 for x (two solutions) and "solving" x^2 = y (infinitely many points on a curve you can graph). You'll have to decide what your goal is. (Presumably you want to exclude the trivial solutions y=x.)

Spoiler

https://en.wikipedia.org/wiki/Equation_xy_=_yx

Thank you for the link. I’ll see if I can understand. Often Wikipedia is Greek to me.

 

[AvgStudent]

This is my interpretation of the Wikipedia page minus the Lambert's function which I know nothing about...
[math]x^y=y^x\\ \underbrace{x\cdot x\cdot \dots x}_{y\medspace times}=\underbrace{y\cdot y \cdot \dots y}_{x \medspace times}\\ u\cdot x=v\cdot y\\ {\text{ (where u is multiple of x and v is multiple of y})} \\ (1)-\boxed{y=\frac{u}{v}x=mx}\\ (\text{where } m =\frac{u}{v})\\ \text{Substitute (1) back into the original equation}\\ x^{mx}=(mx)^x\\ (x^{mx})^{1/x}=[(mx)^x]^{1/x}\\ x^m=mx\\ m=x^{m-1}\\ m^{\frac{1}{m-1}}=x^{\frac{m-1}{m-1}}\\ (2)- \boxed{x=m^{\frac{1}{m-1}}}\\ \text{Substitute into equation (2) into (1):}\\ (3)-\boxed{y=mx=mm^{\frac{1}{m-1}}=m^{\frac{m}{m-1}}}\\ \therefore \fcolorbox{red}{aqua}{$ \text{Solutions} \begin{cases} x=m^{\frac{1}{m-1}} \\ y=m^{\frac{m}{m-1}} \end{cases}$}[/math]