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Screw you, I'm a cheater.
e^pi is greater.
work? just like all my other math tests
I ****ING GUESSED
What is e?
e is like 2.2 or something, im guessing e^pie
Doing it from my head and rounding off and stuff, I figured e^pi.
I forgot what I did exactly, but I checked google and it said I was right, so yeah.
lπ(ln(e^π)) =lπ(π*ln(e))= lπ(π)=1
lπ(ln(π^e)) = lπ(e*ln(π))=lπ(e)+lπ(lπ(π))=lπ(e)
e<π implies lπ(e)<lπ(π)=1 thus e^π>π^e
Think that is sound
lπ is a logarithm with a base of π if it wasn't clear
Hmm, I would say pi^e. e is around 2 and pi is around 3. 3^2 > 2^3
Edit, just tried it out and it turns out e^pi is greater :(
I guess it makes sense seeing as pi is a greater power.
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