Timeline for Why $e^x$ is always greater than $x^e$?

Current License: CC BY-SA 3.0

13 events

when toggle format	what		by	license	comment
Feb 16, 2016 at 12:53	audit	Low quality posts
Feb 16, 2016 at 12:53
Jan 22, 2016 at 9:58	vote	accept	Neeraj
Jan 20, 2016 at 20:29	comment	added	Steven Stadnicki		A very good answer all around - and it's probably worth pointing out that answers can't really get much more intuitive than this simply because there isn't really an 'intuitive' definition of $x^e$ - any such definition is going to have to pass through a chunk of calculus (limits, in particular) and so the other tools of calculus 'should' be available for an answer.
Jan 20, 2016 at 19:39	history	edited	Brian Tung	CC BY-SA 3.0	edited body
Jan 20, 2016 at 19:34	comment	added	Neeraj		@BrainTung Thanks for explaining in so simple words. I wanted something more of explanation rather than just mathematical equation and this is exactly what you have been provided.
Jan 20, 2016 at 19:27	history	edited	Brian Tung	CC BY-SA 3.0	added 40 characters in body
Jan 20, 2016 at 19:18	comment	added	Brian Tung		@user254665: Yup, just added it as you were typing, I think.
Jan 20, 2016 at 19:18	history	edited	Brian Tung	CC BY-SA 3.0	added 1788 characters in body
Jan 20, 2016 at 19:16	comment	added	DanielWainfleet		The derivative of $ g(x)=\log (x^{1/x})=(\log x)/x$ is $(1-\log x)/x^2$, which is positive for $x<e$ and negative for $x>e$. So $g(x)$ ha a unique maximum at $x=e$ . So $x^{1/x}=e^{g(x)}$ has a unique maximum at $x=e$.
Jan 20, 2016 at 19:12	history	edited	Brian Tung	CC BY-SA 3.0	added 1788 characters in body
Jan 20, 2016 at 18:56	comment	added	Neeraj		@BrainTung Yes, Please explain why this is so ?
Jan 20, 2016 at 18:53	history	edited	Brian Tung	CC BY-SA 3.0	added 89 characters in body
Jan 20, 2016 at 18:39	history	answered	Brian Tung	CC BY-SA 3.0