2$ both positive integers, which of $a^b$ and $b^a$ is larger?

14 events

when toggle format	what		by	license	comment
May 3, 2022 at 1:55	history	edited	Bill Dubuque		edited tags
Jan 2, 2020 at 9:54	comment	added	Henry		related: math.stackexchange.com/questions/517555/…
May 16, 2016 at 13:26	history	edited	Martin Sleziak		edited tags
Dec 15, 2015 at 16:55	comment	added	user249332		$a\gt b\gt 2$, does not state everything. If specifically the order with respect to $e$ is said, like $e\gt a\gt b\gt 2$ or $a\gt b\gt e\gt 2$, then it would be more clear.
Dec 15, 2015 at 16:45	answer	added	S. Wilson		timeline score: 0
Apr 7, 2014 at 12:20	vote	accept	John Marty
Apr 7, 2014 at 12:20	vote	accept	John Marty
Apr 7, 2014 at 12:20
Jun 4, 2013 at 17:25	answer	added	awllower		timeline score: 1
Jun 4, 2013 at 7:56	comment	added	user641		If you let $a=b+k$, then you want to solve $(b+k)^b < b^{(b+k)}$, which is equivalent to $(1+\frac{k}{b})^b < b^k$. If you know the definition that $e^k$ is the limit of the increasing sequence $(1+\frac{k}{n})^n$, then this implies $b>e$, or equivalently that $b\ge3$.
Jun 4, 2013 at 4:02	answer	added	Ivan Loh		timeline score: 11
Jun 4, 2013 at 3:24	answer	added	Shuhao Cao		timeline score: 13
Jun 4, 2013 at 3:23	answer	added	André Nicolas		timeline score: 3
Jun 4, 2013 at 3:19	comment	added	Maesumi		what if you took root $1/ab$ from both sides and studied $y=x^{1/x}$.
Jun 4, 2013 at 3:13	history	asked	John Marty	CC BY-SA 3.0