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Thread: 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; Mar 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:
    $\displaystyle \ln e^{-2x} = e^{-2x}$

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


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

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


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

    How do we get $\displaystyle \ln 3$ as an answer?


    Thanks.


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

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

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


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


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




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

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

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

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

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

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

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


    I hope it's helpful.
    Last edited by wingless; Mar 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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