Without finding the numerical values, show which is greater, e^π or π^e.
Consider the value e^pi - pi^e. If positive, e^pi is greater, and if negative, pi^e is greater.
Consider the function f(x)=e^x-x^e
The function is defined on the interval [0,infinity) and is smooth along the interval (0,infinity). Since there are no singularities, the lowest value must occur at one of x=0, lim x to infinity, or at one of the local extrema.
f'(x) = e^x - e*x^(e-1)
f'(x) = 0 only if x=1 or x=e.
f(0) = 1, f(1) = e-1, f(e) = 0, and lim x to infinity of f(x) = infinity. The lowest of these values is f(e)=0, therefore for all x>0, the function is nonnegative. It follows e^pi-pi^e>0.
Therefore e^pi > pi^e.
Edited on September 26, 2003, 2:32 pm