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Sriman Dutta

a^b and b^a

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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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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 [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 by renerpho

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