Solution to the e^(Pi) problem

Consider the function f(x) = e^{x} - x^{e}. Its derivative is f'(x) = e^{x} - ex^{e-1}: since this is an exponential and a power function, they will intersect at most twice on the positive real line. We observe that f'(1) = 0 and f'(e) = 0, so 1 and e are the two points of intersection. The exponential function grows to infinity faster than the power function, so f'(x) >= 0 for x > e. This means that f is an increasing function, and f(e) = 0. Thus, f(Pi) > 0, so e^{Pi} - Pi^{e} > 0. This implies that e^{Pi} is greater than Pi^{e}.


Consider the function f(x) = x - e(ln(x)). Its derivative is f'(x) = 1 - e/x. Clearly, this is positive for x>e, so f(x) > f(e) = 0 for all x >e. Since Pi > e, we have that Pi - e(ln(Pi)) > 0, so Pi > e(ln(Pi)). Since e^x is a monotone increasing function, this implies that e^Pi > Pi^e. Note that when this is done on a calculator, e^Pi = 23.14 and Pi^e = 22.46.