Question

(1/e log(1/e) )-1

Answer 1

\[\frac{1}{e\ln(\frac{1}{e})}-1\]?

Answer 2

http://www.wolframalpha.com/input/?i=(1%2Fe+log(1%2Fe)+)-1

Answer 3

\[\ln(\frac{1}{e})=-1\] so substitute

Answer 4

i don't get their solution for their exact value

Answer 5

ok

Answer 6

exact result sorry

Answer 7

\[\frac{1}{e}\times \ln(\frac{1}{e})-1=\frac{1}{e}\times -1-1\]
\[=-\frac{1}{e}-1\]\]

Answer 8

because
\[\ln(\frac{1}{e})=\ln(e^{-1})=-1\]

Answer 9

not sure that helped, but that is how it got from "input" to "exact result"

Answer 10

i don't get how ln(e^-1) =-1

Answer 11

oh i can explain that

Answer 12

as functions
\[\ln(x)\] and \[e^x\] are inverses which means
\[\ln(e^x)=x\] and
\[e^{\ln(x)}=x\]

Answer 13

so for sure you must have
\[\ln(e^{-1})=-1\]

Answer 14

oh wait i get it, THANKS A MILLION SATELLITE73

Answer 15

btw how do you determine the nature of a stationary point?