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    This is wrong (what did I do wrong?):



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    (Original post by Bealzibub)
    This is wrong (what did I do wrong?):


    You're essentially saying that -\log 2 - \log y = \log \frac{2}{y} but this is not true, since the negative signs distribute.

    i.e: -\log \frac{2}{y} = -(\log 2 - \log y) = -\log 2 + \log y \neq - \log 2 - \log y
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    (Original post by Bealzibub)
    This is wrong (what did I do wrong?):



    This is correct:

    \displaystyle -\log 2 - \log y = -\log\left(\frac{2}{y}\right)

    This step is wrong.

    If you work backwards it may become clearer why:

    \displaystyle -\log\left(\frac{2}{y}\right) = -\left(\log 2 - \log y\right) = -\log 2 + \log y
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    A minus log you are dividing by. When you divide a number say x by y you are effectively saying x * 1/y. When you have 2 numbers dividing another you basically have the following x * 1/y * 1/z.

    What you have done is instead of 1/y * 1/z is y * 1/z and so basically made logy positive rather than negative.

    The positive log number should be the numerator and the negative log numbers in the denominator.

    Another way to think about it is that -logy is equal to +logy^-1 which is the same as what I put above.
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    (Original post by Zacken)
    You're essentially saying that -\log 2 - \log y = \log \frac{2}{y} but this is not true, since the negative signs distribute.

    i.e: -\log \frac{2}{y} = -(\log 2 - \log y) = -\log 2 + \log y \neq - \log 2 - \log y
    I think I've found my identical maths twin

    You're just 10 seconds quicker!
 
 
 
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