On Jun 23, 2013, at 11:04 AM, Joe Niederberger <niederberger@comcast.net> wrote:

Wrong. The reasoning doesn't pass muster.

The error itself is kind of interesting, it could be used

to create all sorts of mischievously incorrect proofs.

Maybe this will help...

The original problem is actually based on this equality...

e^x = x^e

We know this is true at x = e, but is that the only point at which it is true? If it is the only point then for all x > e, e^x > x^e and pi > e thus e^pi > pi^e.

We can take the log() of both sides to get...

x = elog(x)

We know this works for x = e, let us now assume that it holds for another point, k*e where k != 1.

ke = elog(ke)

k = log(ke)

k = log(k) + 1

This was what I actually saw at point P.

We know this works for k = 1, but what about for some other value of k, like 1 + d?

1 + d = log(1 + d) + 1

d = log(1 + d)

Remember, at d = 0, they are equal. For d > 0, which increases faster? d or log(1 + d)? If two functions are equal and one function increases faster than the other, can they ever be equal again?

I attached a picture to help.

Bob Hansen