# Y^x = x^y

1. Jun 14, 2005

### cronxeh

An interesting function. If plotting you would probably omit the 0,0 point too.

How do I get the general solution for this? like y = [function of x]? Is it just y=x ?

What about the boundary conditions near origin? How does it look like there??

2. Jun 14, 2005

### HallsofIvy

If you mean y being defined implicitely as a function of x, then it certainly is not "y= x". I doubt that it could be solved in terms of elementary function but you might be able to use a version the the Lambert's W function.

3. Jun 14, 2005

### saltydog

Hello guys. I don't think it's just y=x. I've attached a rough plot of a curve I think all real solutions (in first quadrant) are on but can't prove it.

Oh yea, I think it would be an interesting exercise to figure out the complex solutions.

#### Attached Files:

• ###### sol1.JPG
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Last edited: Jun 14, 2005
4. Jun 15, 2005

### cronxeh

Some particularly interesting similar function was $$y=x^x$$ When I plotted it from 0 to 1 there was what seemed like an inflection point - although I dont think its really called that but from a graph it does look pretty neat.

Could this be the plot for y^x = x^y from 0 to 1 ?

Last edited: Oct 8, 2005
5. Jun 15, 2005

### Icebreaker

It's not just y=x: here's a counterexample! 2^4 = 4^2

6. Jun 15, 2005

### jdavel

there are an infinite number of counterexamples. slatydog graphed them in #3

7. Jun 15, 2005

### BicycleTree

8. Jun 15, 2005

### Icebreaker

Yes, I know. I never said it was the only counterexample, now did I?

9. Jun 15, 2005

### Hurkyl

Staff Emeritus
I think there should be a straightforward proof that there are exactly two (real) solutions for y, for each x, except for when x is... (reaches into my magic hat)... e. I just can't remember what the proof is.

If you happen to recall the proof that 2^y = y^2 has exactly two solutions, that should generalize to this problem.

10. Jun 15, 2005

### saltydog

Hello Bicycle. I checked out your reference. Thanks. I found it an interesting exercise to convert:

$$x^y=y^x$$

using the substitution y=kx yielding the parametric equations:

$$x=k^{\frac{1}{k-1}}$$

$$y=k^{\frac{k}{k-1}}$$

However, when I plotted this, I did not obtain the line y=x. Attached is the parametric plot.

#### Attached Files:

• ###### x^y=y^x.JPG
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11. Jun 16, 2005

### Hurkyl

Staff Emeritus
The reason you don't get the y=x solution is that you assumed k was not 1 when you took the (k-1)-th root of the equation.

12. Jun 16, 2005

### mepcotterell

my simple attempt at proving nothing...

y^x = x^y

xlogy=ylogx

(logy) = (ylogx)/x

10^(ylogx) = y

ylogxlog10 = y

logxlog10 = 1

log(x+10) = 1

10^1 = x+10

10 = x + 10

x = 0

y^0 = 0^y

y = 0

0^0 = 0^0

1 = 1

so, the only solutions I can find for y and x are = 0

(don't flame, i'm a noob and i'm probably wrong)

however...

y^0 = 0^y

0logy = ylog0 <-- that doesn't work...

0 = undefined ?

13. Jun 16, 2005

### Icebreaker

y=x=0 is the only point on y=x that's not a solution.

14. Jun 16, 2005

### cronxeh

z = f(x,y) = x^y - y^x = 0

Can anyone plot this. I got some weird looking graph, maybe someone else will have a better one

Oh man this is exciting. Check out this plot!

Last edited: Oct 8, 2005
15. Jun 16, 2005

### Zurtex

Given some value fixed value of x then a solution of y will be:

$$y = - \frac{x \text{ProductLog} \left( -\frac{\log x}{x} \right)}{\log x}$$

According to mathematica. This gives you a graph like so:

#### Attached Files:

• ###### Clipboard03.jpg
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16. Jun 16, 2005

### cronxeh

Forget the positive side its too boring! Plot all stuff in QIII !

Same interesting effect but without the positive side can be observed in $$x^y=-y^x$$

Last edited: Jun 16, 2005
17. Jun 26, 2005

### tamalkuila

I think if x^y=y^x and x,y both integer then y=x

If u need the proof please tell

18. Jun 26, 2005

### Hurkyl

Staff Emeritus
A proof would be interesting to see, since your claim is false...

19. Jun 26, 2005

### cronxeh

20. Jun 26, 2005

### rachmaninoff

$$2^4=4^2=16$$

edit: Just noticed Icebreaker said it before me in this same thread, giving him the credit.