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Thread: ln

  1. #1
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    ln

    Hi

    I got the following question:

    make y the subject of the following:
    ln(x) + ln(y) = 0

    This is what I have done so far, could someone help please?:

    ln(x) + ln(y) = 0
    ln(y) = -ln(x)
    e^(ln(y) = e^(-ln(x))
    y = e^(-ln(x))

    Does that seem right?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by taurus View Post
    Hi

    I got the following question:

    make y the subject of the following:
    ln(x) + ln(y) = 0

    This is what I have done so far, could someone help please?:

    ln(x) + ln(y) = 0
    ln(y) = -ln(x)
    So:

    $\displaystyle
    \exp(\ln(y))=\exp(-\ln(x)) = [\exp(\ln(x))]^{-1}
    $

    so:

    $\displaystyle
    y=1/x
    $


    RonL
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  3. #3
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    Thumbs up

    so basically:

    e^(-ln(x)) = 1/x

    ?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    why was writing both sides in base e necessary? could we have just droped the logs when appropriate?

    $\displaystyle \ln y + \ln x = 0$

    $\displaystyle \Rightarrow \ln y = - \ln x$

    $\displaystyle \Rightarrow \ln y = \ln \left( x^{-1} \right)$

    $\displaystyle \Rightarrow y = x^{-1}$

    OR

    $\displaystyle \ln y + \ln x = 0$

    $\displaystyle \Rightarrow \ln xy = 0$

    $\displaystyle \Rightarrow xy = e^0 = 1$

    $\displaystyle \Rightarrow y = \frac 1x$


    Oh well, I guess it was just a matter of taste
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by taurus View Post
    so basically:

    e^(-ln(x)) = 1/x

    ?
    yes

    $\displaystyle e^{- \ln x} = e^{\ln \left( x^{-1} \right)} = \frac 1x$

    since $\displaystyle a^{\log_a b} = b$
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