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Thread: Why is -ln8 equal ln1/8?

  1. #1
    Junior Member BugzLooney's Avatar
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    Why is -ln8 equal ln1/8?

    I solved this problem and couldn't get the last step but apparently it went like this. First e^-x=8
    Then ln(8)=-x
    So x=-ln(8)

    Apparently the answer is then x=ln(1/8)

    I just don't know why the negative ln8 is equal to 1/8.
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  2. #2
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    pickslides's Avatar
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    $\displaystyle x=-\ln(8) \implies x=\ln(8)^{-1} \implies x=\ln\left(\frac{1}{8}\right)$

    All good?
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  3. #3
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    Quote Originally Posted by BugzLooney View Post
    I solved this problem and couldn't get the last step but apparently it went like this. First e^-x=8
    Then ln(8)=-x
    So x=-ln(8)

    Apparently the answer is then x=ln(1/8)

    I just don't know why the negative ln8 is equal to 1/8.
    That last line is not true. You need to use the following facts:

    $\displaystyle \ln{1} = 0$

    $\displaystyle \ln{a} - \ln{b} = \ln{\left(\frac{a}{b}\right)}$

    Then

    $\displaystyle - \ln{8} = 0 - \ln{8} = \ln{1} - \ln{8} = \ln{\left(\frac{1}{8}\right)}$


    edit: oops, beaten to it! My Latex skills are too slow.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by BugzLooney View Post
    I solved this problem and couldn't get the last step but apparently it went like this. First e^-x=8
    Then ln(8)=-x
    So x=-ln(8)

    Apparently the answer is then x=ln(1/8)

    I just don't know why the negative ln8 is equal to 1/8.
    The technical reason...is well technical. But think about it like this, $\displaystyle \ln(x)$ is just the solution to $\displaystyle e^y=x$. So let $\displaystyle \alpha=\ln\left(\frac{1}{8}\right)$ then $\displaystyle e^{\alpha}=\frac{1}{8}$ or equivalently $\displaystyle 8=\frac{1}{e^{\alpha}}=e^{-\alpha}$ so that $\displaystyle \ln(8)=-\alpha\implies -\ln(8)=\alpha=\ln\left(\tfrac{1}{8}\right)$
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  5. #5
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    e^(i*pi)'s Avatar
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    Of course neither $\displaystyle -ln(8)$ nor $\displaystyle ln\left(\frac{1}{8}\right)$ are fully simplified

    $\displaystyle -3ln(2) = 3ln\left(\frac{1}{2}\right)$
    Last edited by e^(i*pi); Jan 11th 2010 at 01:18 PM. Reason: that's what I get for being a smart-arse >.<
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  6. #6
    Super Member bigwave's Avatar
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    why would you say that is simplified??
    it doesn't look any like any reduced form?

    just being cranky.
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  7. #7
    Junior Member BugzLooney's Avatar
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    Woah how does ln(1/8) simplify to 3ln(2) and so forth?
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by BugzLooney View Post
    Woah how does ln(1/8) simplify to 3ln(2) and so forth?
    $\displaystyle \ln\left(ab\right)=\ln(a)+\ln(b)$ and $\displaystyle \ln\left(a^n\right)=n\ln(a)$
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  9. #9
    Junior Member BugzLooney's Avatar
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    I understand everything except for the last post as it relates to mine.

    ln(1/8) simplifies to 3ln(2) by that rule? I just don't see it. Though the explanations have been great thus far.
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by BugzLooney View Post
    I understand everything except for the last post as it relates to mine.

    ln(1/8) simplifies to 3ln(2) by that rule? I just don't see it. Though the explanations have been great thus far.
    $\displaystyle \ln(8)=\ln\left(2^3\right)=3\ln(2)$
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  11. #11
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    Quote Originally Posted by BugzLooney View Post
    I solved this problem and couldn't get the last step but apparently it went like this. First e^-x=8
    Then ln(8)=-x
    So x=-ln(8)

    Apparently the answer is then x=ln(1/8)

    I just don't know why the negative ln8 is equal to 1/8.
    Since $\displaystyle e^{-x}= 8$, $\displaystyle e^x= \frac{1}{8}$.
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