e^pi vs pi^e

Well e is roughly 2 and pi is roughly 3 2^3 = 8 3^2 = 9 8 < 9 so e^pi < pi^e Nvm I’m stupid I forgot that e = 2.718 and rounds up to 3

What a smart solution!

e^π=3^3=27 π^e=3^3=27 They're equal

In fact, e is the boss! e^x > x^e for all x>0 where x is not equal to e

An 8 year old doesn't know taylor series

Short and elegant.

Trivial. e = 3 pi = 3 e^pi = 3^3 = 27 pi^e = 3^3 = 27 e^pi = pi^e

Elegant and understandable..... and sneaky way to insert in later in progress turns out exactly π ^e <e^π.... Thank you 👍

*I just believe in the power of exponents 😌🙏🏼*

Real ones remember Dr. Peyam’s song

Решил за 0.000000267 секунды по теореме степень пизже основания

Решил за 0,5 секунды по теореме степень пизже основания. Изи

If you have a^b vs b^a, the larger exponent will usually win if the exponents are greater than 2

Interestingly enough it works not only for pi but for all positive numbers greater than e Actually it also works for numbers between 0 and e, because for numbers between -1 and 0 not including 0 we also get that e^x > 1+x, just a bit harder to prove that the tail of the series is positive, but it has to do with negative absolute value of x being less than 1 so all negative terms (odd powers starting with 3) are way smaller than positive ones (odd powers starting with 1)... 0 is excluded because if x=0 then e^0=1+0 as all terms just cancel out... Negative numbers i am not covering because raising them to the power of e would give a complex result, so you cannot do a comparison

Why did you let x = (π/e) - 1?

e^π wins

Beautiful proof

This solution is so good❤.

Actually e^x is equal to x+1 for x=0

I win