Does there exists any two distinct numbers a and b, such that a^b=b^a ?

Yes.

How about a = 1

b = -1

 

Edit I think this is wrong (1)^-1 is not equal to (-1)¹

 

Sorry.

Edited October 2, 20169 yr by studiot

a=2

b=4

 

2^4=4^2

 

a=-2

b=-4

also works.

 

(-2)^-4 = (-4)^-2

Edited October 2, 20169 yr by Carrock

Infinitely many solutions with a surprising graph.

 

https://www.wolframalpha.com/input/?i=x^y+%3D+y^x

Edited October 2, 20169 yr by wtf

Interesting plot: I don't understand it.

I can see how there's an infinite set where a=b but that's not what was asked for (And I don't know what happens in the complex plane)

I spotted 2,4 by inspection and I should have spotted -2,-4 by thinking about it.

If you want only positive integer solutions then (2,4) is the only one. The graph indicates that there are infinite number of real solutions for other y.

For real numbers [math]x,y>1[/math], the solutions to the equation [math]x^y=y^x[/math] are given by the trivial [math]x=y[/math] and the more interesting [math]y=\frac{-x}{\ln(x)}W\left ( \frac{-x}{\ln(x)} \right )[/math], where [math]W[/math] is the product log function, see https://en.wikipedia.org/wiki/Lambert_W_function

Examples:
[math]x=3\textup{, }y\approx 2.47805 \dots[/math]
[math]x=4\textup{, }y=2[/math]
[math]x=5\textup{, }y\approx 1.76492 \dots[/math]

 

For [math]0<x \leqslant 1[/math], there is only the trivial solution. For negative [math]x[/math] the term [math]x^y[/math] does not define a real number unless [math]x[/math] and [math]y[/math] are both integers. The only nontrivial solutions for [math]x<0[/math] are [math](x,y)=(-2,-4)[/math] and [math](x,y)=(-4,-2)[/math] (assuming you are only interested in real solutions). This is equivalent with saying that [math]2^4=4^2[/math] is the only nontrivial pair of solutions in [math]\mathbb{N}[/math].
A proof of the formula involving the product log function can be found here: http://mathforum.org/library/drmath/view/66166.html

Edited October 3, 20169 yr by renerpho

Thanks for all your help.