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Math Help - Property of Logs

  1. #1
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    Property of Logs

    This question has been bugging me lately

    If a>0, use the functional equation for the logarithm to prove that log (a^r) = r log a for every rational number r.
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  2. #2
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    Quote Originally Posted by HallsFB View Post
    This question has been bugging me lately

    If a>0, use the functional equation for the logarithm to prove that log (a^r) = r log a for every rational number r.
    Hello,

    I've attached a screenshot of my calculations:
    Attached Thumbnails Attached Thumbnails Property of Logs-log_funktionalglg.gif  
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  3. #3
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    Earboth only proved it for positive integers.
    I will do it for rational numbers,

    Let,
    log (a^r) = n (1)
    log a = m (2)

    We want to show that,
    n = r*m (3)

    By definition equations (1) and (2) become,
    10^n = a^r (4)
    10^m = a (5)

    Raise (5) to the power of r to get,
    10^{r*m} = a^r
    Now this is equal to (4) thus,
    10^n = 10^{r*m}
    Thus,
    n = r*m
    And we have shown (3)
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by HallsFB View Post
    This question has been bugging me lately

    If a>0, use the functional equation for the logarithm to prove that log (a^r) = r log a for every rational number r.
    How about this one. It doesn't explicitly say r is rational as TPH's does, but I see no problem with it, since the proof should work for any real r.

    Note: I will use the notation log[a]b to mean "log to the base a of b" where a and b represent any arbitrary variables or numbers.


    We know by the definition of the logarithm that:

    log[a]b = c
    => a^c = b


    It seems that your problem is for log to the base 10, but it will work for any base, so i shall use dase x to represent any arbitrary base.


    Proof:
    We show log[x](a^r) = rlog[x]a

    if r = 0, the proof is immediate and we have the desired result (since we would have log[x]1 = 0*log[x]a = 0). Now assume r is not zero

    Let log[x](a^r) = y

    => x^y = a^r
    => (x^y)^(1/r) = (a^r)^(1/r) ........we can divide by r since it is not zero
    => x^(y/r) = a

    rewriting this using the definition of logarithms again, we get:

    log[x]a = y/r
    => rlog[x]a = y

    But we said y = log[x](a^r)

    Therefore, we have:
    log[x](a^r) = rlog[x]a
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    since the proof should work for any real r.
    The definition for real exponents is:
    Given a,b>0 we define:
    a^b

    As, exp(ln(a)*b)

    The difficulty here is that we need to introduce the natural logarithm function ln x, and the its inverse function, the exponention e^x.

    This is the difficutly with real exponents which is not covered in school.
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