Calculus problems please?

[Alvaro]

I know that one(e^pi or pi^e).
e^pi can be written as e^pi*e/e
pi^e can be written as pi^e*pi/pi
Now thats in the form x^k/x.
Differentiate x^1/x and find maxima, which turns out to be at e.
So e^pi is greater.
That's a popular one.

Yeah, you seem to have seen it before, but you got the details wrong.

The function that has a maximum at e is log(x)/x. So
log(e)/e > log(pi)/pi
log(e)*pi > log(pi)*e
log(e^pi) > log(pi^e)
e^pi > pi^e

[Jetru]

No...i said x^(1/x) has a max at e. I just graphed it. I don't know your technique, but this one works too.

[Alvaro]

No...i said x^(1/x) has a max at e. I just graphed it. I don't know your technique, but this one works too.

Oh, ok. It was the lack of parenthesis that confused me. My function is just log of your function. It's the same solution in essence.