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Math Help - Logs: Solving for b

  1. #1
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    Logs: Solving for b

    -\frac{2}{3} = Log_b\frac{1}{a}

    I need help solving for b.T-T

    Thanks a bunch!
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  2. #2
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    Quote Originally Posted by hydride View Post
    -\frac{2}{3} = Log_b\frac{1}{a}

    I need help solving for b.T-T

    Thanks a bunch!
    -\frac{2}{3} = Log_b\frac{1}{a}

    = Log_b{a^-1}

    = -Log_b{a}

    \frac{2}{3} = Log_b{a}

    a = b^\frac{2}{3}

    Now find b = ......
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  3. #3
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    You'll need this: \log_{b}(a)\;=\;c\;\iff\;b^{c}\;=\;a
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  4. #4
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    what if c was a negative fraction? i.e b^{-\frac{2}{3}} = 1/9
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  5. #5
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    Quote Originally Posted by hydride View Post
    what if c was a negative fraction? i.e b^{-\frac{2}{3}} = 1/9
    It should be

    b^{-\frac{2}{3}} = 1/a
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  6. #6
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    oh sorry, i meant 1/9 to begin with.
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  7. #7
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    then

    b^{-\frac{2}{3}} = \frac{1}{9}

    \frac{1}{b^{\frac{2}{3}}} = \frac{1}{9}

    b^{\frac{2}{3}} = {9}
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