which of the functions is bigger?

e^pi

pi^e= 22.459157718361045473427152204544
e^pi= 23.140692632779269005729086367949

the idea is to prove it without using any calculator.

You could graph e^x and x^e and show that when x = pi (or x> wherever they cross that is less than pi) one graph is higher than the other.

no, i mean prove it by algebra.

What class is this for?

discrete maybe??
how about this:

question,
pi^e ? e^pi meaning that they have some relation but we dont know yet

take the log of both sides (since doing so shouldnt change anything),

ln( pi^e ) ? ln( e^pi )

doing some rearranging we get,

e*ln( pi ) ? pi*ln( e )

and since the base of the log is e we get,

e*ln( pi ) ? pi

and taking the log of both sides one more time should yield,

ln( e*ln( pi ) ) ? ln( pi )

and some more algebra should tell us which side is larger,

ln( ln( pi ) ) + 1 ? ln( pi )

therfore the right side was larger, meaning that e^pi is > pi^e

How do you conclude that ln(ln(pi))+1 < ln(pi)?

the reason is that the ln( ln ( pi ) ) term is a factor of e less than that of just one ln. its like this. you have a number x and you take the ln of it and then you do it again. that is like taking the e-root (square root but in terms of e, like cube root etc) twice vs just taking it once on the other side.

ex: ln( ln( 27 ) ) = 1.1926601162848....
and
ln( 27 ) = 3.2958368660043...

so e^e^1.1926601162848 = e^3.2958368660043 = 27

the ln(ln( term is 1 factor of e away from the ln term.
and in terms of the question at hand, to get back to pi you must take e^e^ to the ln( ln( pi ) ) + 1 to get it back to pi^e state.

another way to look at it is by taking the whole equation to the e. so we get,
ln( pi )*e ? pi. divide be e to get
ln( pi ) ? pi/e. and since in one instance you are only dividing by e and not taking the e-root or taking it to the 1/e power, it should be clear the right hand is larger.

i hope that clears stuff up.

Then why did God create calculators?

No, it doesn't.

I agree that ln(ln(pi)) < ln(pi)

Other than saying it's so, you haven't demonstrated that ln(ln(pi)) + 1 < ln(pi)

I know that it is. I just don't see the reason in your proof.

markr is right, your proof is not right. you have to get a term that shows obviously which of the functions is bigger.

well the only thing i can think of to do would be this.

e^( ln( ln( pi ) ) + 1 ) ? e^( ln( pi ) ).

ln( pi )*e ? pi.

ln( pi ) ? pi/e.

e^( ln( pi ) ) ? e^( pi/e ).

pi ? e^( pi/e ).

that would be the only thing i can think of at this moment. sorry its not coming to me but i was just trying to help. the difference between the 2 is soo miniscule that i dont think the answer will pop right out at you like say: 2x ? 2x^30000000 ∀x>1 , x<-1            would be

Don't be sorry. You were the only one who offered up anything, and it was so close.