X^Y = Y^XDate: 11/13/2002 at 19:38:47 From: Jeanella Clark Subject: x^y=y^x I'm in an anaysis class and as part of our assignments for the semester, we're supposed to find (with proof) all pairs of integers {x,y} that satisfy x^y=y^x. Obviously if x=y then the answer would be all integers, but when x does not equal y then the only pairs are {2,4} or {4,2} (not counting negative integers). I know this from inspection. How can I prove this? I've gotten as far as ln x/x=ln y/y. Date: 11/14/2002 at 10:22:21 From: Doctor Floor Subject: Re: x^y=y^x Hi, Jeanella, Thanks for your question. If we consider the graph of f(x) = ln(x)/x, then we can observe the following: * f(x)>0 if and only if x>1. * the derivative of f(x) is f'(x) = (1 - ln(x)) / x^2, so that f'(x) = 0 gives x = e. From that we conclude that f(x) is increasing for 0<x<e and decreasing for x>e. * Each horizontal line intersects the graph of f(x) twice. * We know that f(2)=f(4). * For any integer n>4 the horizontal line y=f(n) is "lower" than the line y=f(2) and thus intersects the graph of f(x) as second time for some 1<x<2, so that there is no integer m with f(m)=f(n). * In the same way the line y=f(3) intersects the graph of f(x) a second time for some 2<x<e, so that there is no integer m with f(m)= f(3). This yields that there are no unequal positive integer solutions other than 2^4=4^2. When we started with x/ln(x) = y/ln(y), we restricted ourselves to positive x and y (by definition of ln). But the original question does not exclude negative solutions. It is easy to show that there are no solutions with one positive and one negative integer. Also, when x^y = y^x, then (-x)^(-y) = +/- (-y)^(-x).... Can you finish? See also: Solutions to X^Y = Y^X - Dr. Math archives http://mathforum.org/library/drmath/view/53229.html If you have more questions, just write back. Best regards, - Doctor Floor and Sarah, The Math Forum http://mathforum.org/dr.math/ |
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