Question

e^log3 - e^log 2 =

Answer 1

e^log3-e^log2=3-2=1

Answer 2

thats assuming thats log base e

Answer 3

oh i see is it the same as doing this

e^log3 - e^log 2

= e^log1

=e^0

=1

Answer 4

e^x and lnx are inverse functions
so e^ln3=3 and e^ln2=2

Answer 5

oh so they cancel out

Answer 6

yea!
are we to assume that the logs are base e?

Answer 7

yes it says it in the text book also one other question
\[\int\limits_{\log6}^{\log4} e ^{-x}\]

Answer 8

the antiderivative of e^(-x) is -e^(-x)+C

Answer 9

what ive done is this [\[[-e^{-x}\]

Answer 10

so you get -e^-log6+e^-log4

Answer 11

then using what you said before the answer should be 10

Answer 12

no
upper limit is log4 right?
lower limit is log6 right?
it looks like you plugged in the lower limit first

Answer 13

and also if we simplify what you have we have -6^(-1)+4^(-1)=......

Answer 14

=-1/6+1/4

Answer 15

\[\int\limits_{\log4}^{\log6}e ^{-x}\] it should be that lol sorry i did it the wrong way round.

Answer 16

ok then what you have is right but you still simplified wrong

Answer 17

do you remember rlogx=logx^r

Answer 18

yeah my answer is wrong according to the back of the book the answer should be 1/12

Answer 19

yes i rember that

Answer 20

so we have -e^[-log6]=-e^log6^(-1)=-6^(-1)=-1/6
and we have e^[-log4]=e^log4^(-1)=4^(-1)=1/4

Answer 21

and, you still there?

Answer 22

do you got it from here?
i'm fixing to go to bed

Answer 23

ok yeah thanks i get it lol thanks :)

good night