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Math Help - Product rule for logarithms

  1. #1
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    Product rule for logarithms

    Have a silly question.

    In the proof that ln(ax)=ln(a) + ln(x)

    we show that ln(ax) and ln(x) have the same derivative, therefore they differ by a constant; ln(a)=ln(x) + C

    The book then sets x=1, and says C=ln(a), and the rule is proved. But I get C=ln(a)-ln(x), and that x=1 is a special case.

    What's my mistake?
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  2. #2
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    Look at it this way.
    \ln(x)=\ln(a)-C so \ln(1)=0=\ln(a)-C.
    Therefore C=\ln(a).
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  3. #3
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    What I'm not understanding is how we can set x= any number, without loosing the generality of the rule. I see that ln(a)=ln(x) + C is true for any value of x, but if we choose x=2 instead of x=1 then we get C=ln(a)-ln(2) instead of C=ln(a)
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  4. #4
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    Is it because that C + ln (2) is just another constant?

    C+ln(2)=ln(a), which equals C=ln(a) ????????
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  5. #5
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    No that is not the point. (BTW: I don't like that proof).
    But the fact is \ln(1)=0. Therefore, it must be the case that C=\ln(a).
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  6. #6
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    Maybe the diagram will help show why x=1 is chosen (for the case of a=2).

    Remember, in order to discover the constant C, you must eliminate x.
    Setting x=1 causes ln(x) to be zero, eliminating x.

    Also, you have a typo in your post

    ln(a)=ln(x)+C should read ln(ax)=ln(x)+C.

    Hence C=ln(ax)-ln(x) but only "removing" x shows the value of C.

    Setting x to 1,

    ln(a)-ln(1)=C

    ln(a)-0=C
    Attached Thumbnails Attached Thumbnails Product rule for logarithms-ln-2x-.jpg  
    Last edited by Archie Meade; August 15th 2010 at 06:36 AM.
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  7. #7
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    Resolved. I woke up this morning and can't even remember what my confusion was over.
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