```Date: Jun 21, 2013 5:44 PM
Author: Dave L. Renfro
Subject: Re: new tutor here

Joe Niederberger wrote (in part):http://mathforum.org/kb/message.jspa?messageID=9139298>> Here's the pic:>> Simply the log (ln) curve, with points where it intersects>> (e,1) and (pi,log(pi)) noted.Robert Hansen wrote:http://mathforum.org/kb/message.jspa?messageID=9139925> Ok, so let me ask you: How does your picture show that e^pi > pi^e?I don't know what Joe's explanation is (when more hand-holdingdetails are included), although I haven't tried very hard tofigure it out. However, one way is to note that the numericalorder of e^pi and pi^e is unchanged if e^pi is replaced with(ln e)/e and pi^e is replaced with (ln pi)/pi. This followsfrom the fact that the natural logarithm function is strictlyincreasing:e^pi # pi^e  [# is < or = or >, whichever is correct]is equivalent toln(e^pi) # ln(pi^e)which is equivalent topi*(ln e) # e*(ln pi)which is equivalent to(ln e)/e # (ln pi)/pi  [Note: This suggests considering f(x) = (ln x)/x.]which is equivalent to1/e # (ln pi)/piOff hand, I don't see how we can tell by precalculus methodswhich of 1/e and (ln pi)/pi is greater (or equivalently, whichof pi/e and (ln pi) is greater), but I do know that taking thederivative of f(x) = (ln x)/x and setting the result equal tozero (and paying attention to where the derivative is positiveand where the derivative is negative), we find that f(x) has amaximum (local and global) at x = e. Thus,(ln e)/e > (ln pi)/piand so we gete^pi > pi^eSee the following web page for several proofs of this inequality,including a usefully labeled graph of f(x) = (ln x)/x.http://math.stackexchange.com/questions/7892/comparing-pie-and-e-piDave L. Renfro
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