# a^b and b^a

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

Yes.

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How about a = 1

b = -1

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

Sorry.

Edited by studiot

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a=2

b=4

2^4=4^2

a=-2

b=-4

also works.

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

Edited by Carrock

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Infinitely many solutions with a surprising graph.

Edited by wtf

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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.

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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.

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

Examples:
$x=3\textup{, }y\approx 2.47805 \dots$
$x=4\textup{, }y=2$
$x=5\textup{, }y\approx 1.76492 \dots$

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

Edited by renerpho

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Thanks for all your help.

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