## anonymous 5 years ago e^log3 - e^log 2 =

1. myininaya

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

2. myininaya

thats assuming thats log base e

3. anonymous

oh i see is it the same as doing this e^log3 - e^log 2 = e^log1 =e^0 =1

4. myininaya

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

5. anonymous

oh so they cancel out

6. myininaya

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

7. anonymous

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

8. myininaya

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

9. anonymous

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

10. anonymous

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

11. anonymous

then using what you said before the answer should be 10

12. myininaya

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

13. myininaya

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

14. myininaya

=-1/6+1/4

15. anonymous

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

16. myininaya

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

17. myininaya

do you remember rlogx=logx^r

18. anonymous

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

19. anonymous

yes i rember that

20. myininaya

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

21. myininaya

and, you still there?

22. myininaya

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

23. anonymous

ok yeah thanks i get it lol thanks :) good night

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