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Math Help - Logarithms..(solve)

  1. #1
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    Angry Logarithms..(solve)

    Hi I cannot wrap my head around these two problems.
    Some help would be greatly appreciated.


    and

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  2. #2
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    log laws

    these questions are a matter of applying log laws

    \log_{a} b = c
    a^c=b
    so for your case a=2 and c=-3, so sub them into the second equation

    the second question is a combination of log laws

    e \log_{a} b = \log_{a} b^e

    and \log_{a} b + \log_{a} d = \log_{a} (b*d)
    and \log_{a} b - \log_{a} d = \log_{a} (\frac{b}{d})
    so for the second question rewrite the equation in the form of \log_{a} b^e and then apply the addition and subtraction rules to get your answer
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  3. #3
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    thank you, i think i got it..

    \log_{2} x = -3

    2^{-3}=x

    x=1/8


    and

    {3/2} \log_{b} 4 - {2/3} \log_{b}8 + \log_{b}2= \log_{b} x

    \log_{b}4^{3/2}-\log_{b}8^{2/3} + \log_{b}2= \log_{b}x

    (4^{3/2}) / (8^{2/3})*2=x

    8/2*2

    x=4

    i hope i am correct..
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  4. #4
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    Quote Originally Posted by Macabro View Post
    thank you, i think i got it..

    ...

    {3/2} \log_{b} 4 - {2/3} \log_{b}8 + \log_{b}2= \log_{b} x

    \log_{b}4^{3/2}-\log_{b}8^{2/3} + \log_{b}2= \log_{b}x

    (4^{3/2}) / (8^{2/3})*2=x

    (8/4)*2

    x=4

    i hope i am correct..
    Your result is correct

    There is only a small typo ...
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  5. #5
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    yay, thanks, i caught that but didnt bother to fix it...
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