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Math Help - a few questions about lne

  1. #1
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    a few questions about lne

    Attached are a few questions I have about operations concerning ln and ln e.
    Any help would be greatly appreciated.
    1.
    The book has:
    lne^(-2x)=e^(-2x)
    I thought lne^(-2x)=-2x
    2.
    3e^(ln8/3)=3e^(ln2) How does Ln8/3=ln2?
    3.1/2(ln27-ln3)=Ln3
    How do we get ln3 as an answer.
    Attached Files Attached Files
    Last edited by kid funky fried; March 29th 2008 at 10:08 AM. Reason: avoid dl a file
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  2. #2
    Super Member wingless's Avatar
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    kid funky fried posted:

    1.
    The book has:
    \ln e^{-2x} = e^{-2x}

    I thought:
    \ln e^{-2x} = 2x ?


    2.
    3e^{\frac{\ln 8 }{3}} = 3e^{\ln 2}

    How does (\ln 8) /3)=\ln 2 ?


    3.
    \frac{1}{2}(\ln 27 - \ln 3) = \ln 3

    How do we get \ln 3 as an answer?


    Thanks.


    ----------------

    Firstly, \log_c a^b=b\log_c a

    1. \ln e^{-2x}
    The exponent will become the coefficient
    {-2x}\ln e
    As ln e = 1, it's -2x


    2. \frac{\ln 8 }{3}
    \ln 8 = \ln 2^3
    This is the same situation, the exponent will become the coefficient.
    \ln 8 = \ln 2^3 = 3\ln 2
    And \frac{3\ln 2}{3} = \ln 2


    3. \frac{1}{2}(\ln 27 - \ln 3)
    \frac{1}{2}(\ln 3^3 - \ln 3)
    \frac{1}{2}(3\ln 3 - \ln 3)
    \frac{1}{2}(2\ln 3)
    \ln 3




    There are also other ways to find these. For example,

    3. \frac{1}{2}(\ln 27 - \ln 3)

    Rule: \log_c a - \log_c b = \log_c \frac{a}{b}

    So, \frac{1}{2}(\ln \frac{27}{3})

    \frac{1}{2}(\ln 9)

    \frac{1}{2}(\ln 3^2)

    \frac{1}{2}(2\ln 3) = \ln 3


    I hope it's helpful.
    Last edited by wingless; March 29th 2008 at 10:14 AM.
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  3. #3
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    thx

    thx wingless, I appreciate you help!
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