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Math Help - Log simplification problem.

  1. #1
    Junior Member
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    Log simplification problem.

    I am struggling to see the step below in a log problem.
    I cannot see how you get to \frac{3}{2}

    \log_3 x=1-\frac{1}{2} \log_3 4
    \log_3 x=1-\log_3 2
    x = 3^{(1-\log_3 2)}

    x = [3][3^{(-log_3 2)}]

    x = \frac{3}{3^{(log_3 2)}}

    x = \frac{3}{2}

    I don't understand how the last step is done?
    How does that equal one and a half?

    \log_3 2 => 3^?=2 confused.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by craigmain View Post
    I am struggling to see the step below in a log problem.
    I cannot see how you get to \frac{3}{2}

    \log_3 x=1-\frac{1}{2} \log_3 4
    \log_3 x=1-\log_3 2
    x = 3^{(1-\log_3 2)}

    x = [3][3^{(-log_3 2)}]

    x = \frac{3}{3^{(log_3 2)}}

    x = \frac{3}{2}

    I don't understand how the last step is done?
    How does that equal one and a half?

    \log_3 2 => 3^?=2 confused.
    Conceptually, if y=\log_3 2, then y is the exponent to which you need to raise 3 in order to get 2. So whatever y is, you have 3^y, so by the definition of logs, you get out 3^y=2.
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  3. #3
    Junior Member
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    Just to clarify.

    Are you saying:

    we have:
    x=\frac{3}{3^{(\log_3 2)}}

    let y=\log_3 2 (1)

    Therefore
    x=\frac{3}{3^y}

    but from (1)
    3^y=2

    Therefore
    x = \frac{3}{2}.

    Sorry, but it seems like there's some circular logic here, so I want to be sure I am understanding you.
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  4. #4
    Super Member redsoxfan325's Avatar
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    Right.
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