Sriman Dutta Posted October 2, 2016 Share Posted October 2, 2016 Does there exists any two distinct numbers a and b, such that a^b=b^a ? Link to comment Share on other sites More sharing options...
John Cuthber Posted October 2, 2016 Share Posted October 2, 2016 Yes. Link to comment Share on other sites More sharing options...
studiot Posted October 2, 2016 Share 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 Link to comment Share on other sites More sharing options...
Carrock Posted October 2, 2016 Share 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 Link to comment Share on other sites More sharing options...
wtf Posted October 2, 2016 Share 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 Link to comment Share on other sites More sharing options...
John Cuthber Posted October 2, 2016 Share 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. Link to comment Share on other sites More sharing options...
mathematic Posted October 2, 2016 Share 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. Link to comment Share on other sites More sharing options...
renerpho Posted October 3, 2016 Share 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 Link to comment Share on other sites More sharing options...
Sriman Dutta Posted October 4, 2016 Author Share Posted October 4, 2016 Thanks for all your help. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now