ln(e^pi)≈3.142, ln(pi^e)=eln(pi)≈2.718×1.13=3.0752, 3.142>3.075, so also e^pi≈20+pi, pi^e≈21.46 so easy natural logs are so powerful
melanie.campbell1 week, 4 days ago
Key approach to solve this problem -: 1. e^x > 1+x 2. Replace x with π/e - 1
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wandamyst11 week, 5 days ago
Amazing how math is mathing
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brenda.padilla1 week, 6 days ago
Wow, I was so wrong! I was rounding pi to 3 and e to 2 and assuming that would be close enough, but it turns out that will not give you the correct answer!!
micheal_santiago2 weeks ago
What made you think to let x= pi/e - 1? Sure it works, but how did you know to choose that particular substitution?
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erick_pimenta2 weeks ago
Wow
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carrie.chambers2 weeks ago
y = e^x /x^e 0 < y(e) = 1 **** ln y = x - e ln x y'/y = 1 - e/x y' = (1 - e/x)(e^x / x^e) > 0 for x > e y' = 0 for x =e. Therefore y is monotonic increasing for x>e In particular as pi > e y(pi) > y(e) = 1 implies e^pi / pi^e > 1 implies e^ pi > pi^e
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damien_davies2 weeks ago
I use a calculator 😀
pedrolucas.abreu2 weeks ago
e^π=απ^e Lne^π=αln(π^e) π=αeln(π) π/eln(π)>1
helena_novais2 weeks ago
Cool, I personally came to the solution using the function f(x) = 1/x and the integral from e to pi of f(x) which is ln(pi) - 1 and from the graph we can see that (pi-e)*f(e) must be greater then the integral, and from there we can also find that e^pi > pi^e
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nicholas_bell2 weeks ago
Tbh pi^e is better cuz its yummy
matthewmist722 weeks ago
Bjt wouldnt the answer change depending on the value of e like if e=2 vs e=3
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hugo.ozuna2 weeks ago
Решил за 3,14с по теореме степень пизже основания
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daniel.cantu2 weeks ago
The posteev powur
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maríadelcarmenuribe8012 weeks ago
Very clever solution!
utkarsh.kalita2 weeks, 1 day ago
Elegant and understandable..... and sneaky way to insert in later in progress turns out exactly π ^e <e^π.... Thank you 👍
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steven.leon2 weeks, 1 day ago
Slick!
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sallysimmons3452 weeks, 1 day ago
This was an elegant solution for sure definitely better than using ln
ross_shepherd2 weeks, 1 day ago
I was expecting to see you use the derivative of x^(1/x). This was a nice surprise.
ln(e^pi)≈3.142, ln(pi^e)=eln(pi)≈2.718×1.13=3.0752, 3.142>3.075, so also e^pi≈20+pi, pi^e≈21.46 so easy natural logs are so powerful
Key approach to solve this problem -: 1. e^x > 1+x 2. Replace x with π/e - 1
Amazing how math is mathing
Wow, I was so wrong! I was rounding pi to 3 and e to 2 and assuming that would be close enough, but it turns out that will not give you the correct answer!!
What made you think to let x= pi/e - 1? Sure it works, but how did you know to choose that particular substitution?
Wow
y = e^x /x^e 0 < y(e) = 1 **** ln y = x - e ln x y'/y = 1 - e/x y' = (1 - e/x)(e^x / x^e) > 0 for x > e y' = 0 for x =e. Therefore y is monotonic increasing for x>e In particular as pi > e y(pi) > y(e) = 1 implies e^pi / pi^e > 1 implies e^ pi > pi^e
I use a calculator 😀
e^π=απ^e Lne^π=αln(π^e) π=αeln(π) π/eln(π)>1
Cool, I personally came to the solution using the function f(x) = 1/x and the integral from e to pi of f(x) which is ln(pi) - 1 and from the graph we can see that (pi-e)*f(e) must be greater then the integral, and from there we can also find that e^pi > pi^e
Tbh pi^e is better cuz its yummy
Bjt wouldnt the answer change depending on the value of e like if e=2 vs e=3
Решил за 3,14с по теореме степень пизже основания
The posteev powur
Very clever solution!
Elegant and understandable..... and sneaky way to insert in later in progress turns out exactly π ^e <e^π.... Thank you 👍
Slick!
This was an elegant solution for sure definitely better than using ln
I was expecting to see you use the derivative of x^(1/x). This was a nice surprise.
The one with the biggest exponent. Always.