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anonymous

  • 5 years ago

e^log3 - e^log 2 =

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  1. myininaya
    • 5 years ago
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    e^log3-e^log2=3-2=1

  2. myininaya
    • 5 years ago
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    thats assuming thats log base e

  3. anonymous
    • 5 years ago
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    oh i see is it the same as doing this e^log3 - e^log 2 = e^log1 =e^0 =1

  4. myininaya
    • 5 years ago
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    e^x and lnx are inverse functions so e^ln3=3 and e^ln2=2

  5. anonymous
    • 5 years ago
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    oh so they cancel out

  6. myininaya
    • 5 years ago
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    yea! are we to assume that the logs are base e?

  7. anonymous
    • 5 years ago
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    yes it says it in the text book also one other question \[\int\limits_{\log6}^{\log4} e ^{-x}\]

  8. myininaya
    • 5 years ago
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    the antiderivative of e^(-x) is -e^(-x)+C

  9. anonymous
    • 5 years ago
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    what ive done is this [\[[-e^{-x}\]

  10. anonymous
    • 5 years ago
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    so you get -e^-log6+e^-log4

  11. anonymous
    • 5 years ago
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    then using what you said before the answer should be 10

  12. myininaya
    • 5 years ago
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    no upper limit is log4 right? lower limit is log6 right? it looks like you plugged in the lower limit first

  13. myininaya
    • 5 years ago
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    and also if we simplify what you have we have -6^(-1)+4^(-1)=......

  14. myininaya
    • 5 years ago
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    =-1/6+1/4

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

  16. myininaya
    • 5 years ago
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    ok then what you have is right but you still simplified wrong

  17. myininaya
    • 5 years ago
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    do you remember rlogx=logx^r

  18. anonymous
    • 5 years ago
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    yeah my answer is wrong according to the back of the book the answer should be 1/12

  19. anonymous
    • 5 years ago
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    yes i rember that

  20. myininaya
    • 5 years ago
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    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
    • 5 years ago
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    and, you still there?

  22. myininaya
    • 5 years ago
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    do you got it from here? i'm fixing to go to bed

  23. anonymous
    • 5 years ago
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    ok yeah thanks i get it lol thanks :) good night

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