Timeline for Without using a calculator and logarithm, which of $100^{101} , 101^{100}$ is greater?

8 events

when toggle format	what		by	license	comment
Dec 5, 2017 at 5:46	audit	First posts
Dec 5, 2017 at 5:47
Dec 3, 2017 at 20:26	audit	First posts
Dec 3, 2017 at 20:26
Nov 28, 2017 at 10:06	comment	added	Dehbop		I don't see what good is multiplying $\left(1+\frac1{100}\right)^{100}$ with $\left(1-\frac1{100}\right)^{100}$, arrive at $\left(1-\frac1{10000}\right)^{100}$???
Nov 28, 2017 at 9:04	comment	added	ilkkachu		About taming the approximation, doesn't $(1+1/n)^n = e'$ approach $e$ from below, so that $e' < e$ always? And then $e'/100 < 1$
Nov 27, 2017 at 21:21	history	edited	Hagen von Eitzen	CC BY-SA 3.0	added 493 characters in body
Nov 27, 2017 at 20:16	comment	added	Zereges		Didn't you meant $\frac{1}{100}$ as first term after first equality?
Nov 27, 2017 at 19:54	review	Suggested edits
Nov 27, 2017 at 20:04
Nov 27, 2017 at 19:42	history	answered	Hagen von Eitzen	CC BY-SA 3.0