Without finding the numerical values, show which is greater, e^π or π^e.

Let f(x) = (1/x)*ln(x)

Then, f'(x) = [(1/x)^2]*[1 - ln(x)]; and:

f"(x) = - [(1/x)^3]*[3 - 2*ln(x)]

Thus, setting f'(x) = 0, we obtain x = e.

Since, 3 - 2*ln(e) = 1, it follows that:

f"(e) < 0, whenever x> 0

Thus, it follows that f(e)> f(x) whenever |x-e|> 0 with x>0

Accordingly:

(1/e)> (1/pi)*ln(pi)

Or, e^(1/e) > pi^(1/pi)

Or, e^(pi) > (pi)^e

Consequently e^(pi) is greater.

*Edited on ***May 25, 2007, 11:37 am**

*Edited on ***July 6, 2008, 12:37 pm**