Sriman Dutta Posted October 2, 2016 Posted October 2, 2016 Does there exists any two distinct numbers a and b, such that a^b=b^a ?
studiot Posted October 2, 2016 Posted October 2, 2016 (edited) How about a = 1 b = -1 Edit I think this is wrong (1)-1 is not equal to (-1)1 Sorry. Edited October 2, 2016 by studiot
Carrock Posted October 2, 2016 Posted October 2, 2016 (edited) a=2 b=4 2^4=4^2 a=-2 b=-4 also works. (-2)^-4 = (-4)^-2 Edited October 2, 2016 by Carrock
wtf Posted October 2, 2016 Posted October 2, 2016 (edited) Infinitely many solutions with a surprising graph. https://www.wolframalpha.com/input/?i=x^y+%3D+y^x Edited October 2, 2016 by wtf 2
John Cuthber Posted October 2, 2016 Posted October 2, 2016 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.
mathematic Posted October 2, 2016 Posted October 2, 2016 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.
renerpho Posted October 3, 2016 Posted October 3, 2016 (edited) 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_functionExamples:[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, 2016 by renerpho 3
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