Solving x^y = y^(x - y) for All Natural x, yDate: 11/18/2010 at 22:40:16 From: ilyas Subject: Find all natural numbers x, y so that x^y = y^(x-y) Find all natural numbers x, y such that x^y = y^(x - y) Date: 11/18/2010 at 23:22:35 From: Doctor Vogler Subject: Re: Find all natural numbers x, y so that x^y = y^(x-y) Hi Ilyas, Thanks for writing to Dr Math. This is a lot like the problem at ... http://mathforum.org/library/drmath/view/66166.html ... and a little like the problem at http://mathforum.org/library/drmath/view/66640.html First of all, any solution to your equation must have x >= y, because otherwise x - y < 0, and then -- since x^y is an integer -- that would mean that this must also be an integer: y^(x - y) = 1/y^(y - x) That, in turn, would mean that y has to equal 1. But if y = 1, then x >= 1 anyway, since x is a natural number. We conclude that x >= y for all solutions. In fact, if x = y, then x^y = y^(x - y) = y^0 = 1 So 1 = x^y = x^x >= x There, the only solution with x and y equal is x = y = 1. Otherwise, x > y. Now if we divide both sides of the equation by x^(x - y), then we get x^(2y - x) = (y/x)^(x - y) Since y/x is a non-integer (between 0 and 1), and it is raised to a positive power, the right side of the equation is a non-integer, which means that the exponent on the left side must be negative, which means that x > 2y. Now go back to your original equation and divide both sides by y^y: (x/y)^y = y^(x - 2y) Since x > 2y, the right side is an integer. Since the right side is an integer, the left side has to be an integer too. But if you raise a non-integer rational number to an integer power, you don't get an integer. So that means that this must be an integer (bigger than 2): k = x/y Now we re-write our equation as k^y = y^((k - 2)y) We take the positive real y'th root of both sides of the equation and get k = y^(k - 2) We already know that k > 2. If k = 3, then 3 = y^1, so y = 3 and x = ky = 9 So that is another solution. If k = 4, then 4 = y^2, so y = 2 and x = ky = 8 And that is another solution. But otherwise, k >= 5 -- which means that y = k^(1/(k - 2)) -- must be strictly between 1 and 2, which no integer can do. So all solutions are x = 1, y = 1 x = 8, y = 2 x = 9, y = 3 If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 11/19/2010 at 02:06:22 From: ilyas Subject: Thank you (Find all natural numbers x, y so that x^y = y^(x-y)) Thank you very much. I love math. I want to learn more and more, especially logic math.http://mathforum.org/library/drmath/view/66640.html |
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