Comparing x^y and y^xDate: 10/17/2002 at 20:55:14 From: Katy Subject: Diffeq application My question is: Given x>y>e, is y^x > x^y or is x^y > y^x (or neither)? I know the answer is not neither, but I am honestly not even sure where to start. The class is differential equations, so the way to get the answer obviously has something to do with diffeq, but I don't see it. My classmates said they figured it out but only with a lot of number crunching, and I am sure there is a better way. If you could just give me a place to start, that would be more than enough. Thank you! Date: 10/18/2002 at 05:20:57 From: Doctor Schwa Subject: Re: Diffeq application Hi Katy, This is a very interesting problem! The best idea I have come up with so far is to start by taking the 1/(xy) power of both quantities. That is, instead of comparing x^y and y^x, you can compare x^(1/x) and y^(1/y). Whichever is greater of these last two will determine which of the original two quantities is greater. This helps a lot, because now you can figure out if x^(1/x) is increasing or decreasing by taking a derivative. If it's increasing, then the bigger value (x) makes it greater, so x^y is the winner; if it's decreasing, then the smaller value (y) makes it greater. I hope that makes some sense. - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
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