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### Comparing x^y and y^x

Date: 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/
Associated Topics:
College Calculus

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