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Math Help - logs

  1. #1
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    Cool logs

    Let y = loga(10) + loga(12) - loga(6)


    Find a if y=1
    Find a if y=-1
    Find a if y=2


    If you can show me any of the obove please let me know.


    THANKS
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  2. #2
    Senior Member
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    Quote Originally Posted by Mr_Green View Post
    Let y = loga(10) + loga(12) - loga(6)


    Find a if y=1
    Find a if y=-1
    Find a if y=2


    If you can show me any of the obove please let me know.


    THANKS
    log(10)/log(a) + log(12)/log(a) - log(6)/log(a) = log(20)/log(a)

    1 = log(20)/log(a)

    log(a) = log(20)

    ln of both sides; a = 20

    You do them now; answers, for:

    y = -1, you should get: a = 1/20

    y = 2, you should get: a = 2*sqrt(5)
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  3. #3
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    Quote Originally Posted by Mr_Green View Post
    Let y = loga(10) + loga(12) - loga(6)


    Find a if y=1
    Find a if y=-1
    Find a if y=2
    Combine the logarithms,
    y=\log_a (10 \cdot 12/6)=\log_a 20
    Thus, by definition of logarithms,
    a^y=20 (since a>0)
    Thus, (if y\not = 0)
    a=20^{1/y}
    So just substitute the value of "y" to take the value of "a"
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