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Math Help - Prove a natural logarithmic property using ... ?

  1. #1
    Member Miss's Avatar
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    Prove a natural logarithmic property using ... ?

    The Question:
    Use the definition of the natural logarithm function to prove that :
    ln (ab) = ln(a) + ln(b)
    for any positive number a and b.


    my problem is that I do not know how to prove it using the definition

    BTW, the definition is \int_{1}^{x} \dfrac{dt}{t} = ln|x|
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Miss View Post
    The Question:
    Use the definition of the natural logarithm function to prove that :
    ln (ab) = ln(a) + ln(b)
    for any positive number a and b.


    my problem is that I do not know how to prove it using the definition

    BTW, the definition is \int_{1}^{x} \dfrac{dt}{t} = ln|x|
    By use of the chain rule we get that

    \displaystyle\frac{d}{db}\ln(a\cdot b)=\frac{d}{db}\int_1^{a\cdot b}\frac{dt}{t}=\frac{1}{a\cdot b}\cdot a=\frac{1}{b}

    Hence, by integration, we conclude that

    \text{(*)}\qquad \ln(a\cdot b)=\ln(b)+c

    For some value of c. To determine the value of c we set b=1 in (*) to find that

    \ln(a\cdot 1)=\ln(1)+c=0+c=c, i.e. c=\ln(a)

    Plugging this information into (*) we can finally say, with confidence, that

    \ln(a\cdot b)=\ln(b)+\ln(a)

    must hold.
    Last edited by Failure; July 11th 2010 at 07:28 AM. Reason: fixed lower limit of integral
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  3. #3
    Member Miss's Avatar
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    Thanks
    but you have a small mistake
    the lower limit for integral should be 1 not 0

    thanks again
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by Miss View Post
    Thanks
    but you have a small mistake
    the lower limit for integral should be 1 not 0

    thanks again
    Terrible mistake, as if the lower limit of every integral were 0! - I'm going to correct that mistake right away.
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  5. #5
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    A slight variation, a bit more direct, I think, since it uses the integral only and not the derivative:

    If a is positive, then \frac{1}{a} is also positive and, by that definition, ln(1/x)= \int_1^{1/a} \frac{1}{t}dt.

    Let u= at so that t= u/a, dt= (1/a)du. When t= 1, u= a, when t= 1/a, u= 1. Now we have ln(1/a= \int_{a}^1 (a/u)(1/a)du. The "a"s cancel giving
    ln(1/a)= \int_a^1 \frac{1}{u} du= -\int_1^a \frac{1}{u}du= -ln(a).

    If a and b are positive then ab is also positive and, by that definition,
    ln(ab)= \int_1^{ab}\frac{1}{t}dt.

    Now, let u= t/a so that t= au, dt= adu, (1/t)dt= (1/au)(adu)= (1/u)du. When t= 1, u= 1/a and when t= ab, u= b. The integral becomes \int_{1/a}^b \frac{1}{u}du=  \int_{1/a}^1 \frac{1}{u}du+ \int_1^b \frac{1}{u}du = -\int_1^{1/a}\frac{1}{u}du+ \int_1^b \frac{1}{u}du= -ln(1/a)+ ln(b)= ln(a)+ ln(b).
    Last edited by HallsofIvy; July 11th 2010 at 08:49 AM.
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  6. #6
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    Another way.
    Here is the way Leonard Gillman does this problem.
    Because x>0 we use \int_1^x {\frac{{du}}{u}}  = \ln (x).

    Using u-substitution \int_a^{ab} {\frac{{dx}}{x}}  = \int_1^b {\frac{{du}}{u}} ;\;u = \frac{x}{a}.

    Now consider that  \int_1^{ab} {\frac{{dx}}{x}}  = \int_1^a {\frac{{dx}}{x}}  + \int_a^{ab} {\frac{{dx}}{x}}  = \int_1^a {\frac{{dx}}{x}}  + \int_1^b {\frac{{dx}}{x}}
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  7. #7
    Member Miss's Avatar
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    HallsOfIvy :
    Sir, you should preview your reply before post it
    there are 45545 replies for you with latex errors and non-complete latex codes
    Thanks

    Plato:
    Thanks
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