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

Joe Niederberger wrote (in part):

>> Here's the pic:
>> Simply the log (ln) curve, with points where it intersects
>> (e,1) and (pi,log(pi)) noted.

Robert Hansen wrote:

> 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-holding
details are included), although I haven't tried very hard to
figure it out. However, one way is to note that the numerical
order 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 follows
from the fact that the natural logarithm function is strictly

e^pi # pi^e [# is < or = or >, whichever is correct]

is equivalent to

ln(e^pi) # ln(pi^e)

which is equivalent to

pi*(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 to

1/e # (ln pi)/pi

Off hand, I don't see how we can tell by precalculus methods
which of 1/e and (ln pi)/pi is greater (or equivalently, which
of pi/e and (ln pi) is greater), but I do know that taking the
derivative of f(x) = (ln x)/x and setting the result equal to
zero (and paying attention to where the derivative is positive
and where the derivative is negative), we find that f(x) has a
maximum (local and global) at x = e. Thus,

(ln e)/e > (ln pi)/pi

and so we get

e^pi > pi^e

See the following web page for several proofs of this inequality,
including a usefully labeled graph of f(x) = (ln x)/x.

Dave L. Renfro