# 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}.**
### OR.......

**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.**