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Thread: Power Law for Logarithms

  1. #1
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    Power Law for Logarithms

    Solve for x in each of the following:

    $\displaystyle log_2x = 3log_24$

    My answer: 64

    --

    $\displaystyle logx = \frac{1}{3}log8$

    My answer: 2


    Are my answers correct?



    Evaluate each of the following:

    1) $\displaystyle log_2 \sqrt [5] {16}$

    2) $\displaystyle log_525 \sqrt [3] {5}$

    Can you please show me how to solve these logs, I don't know where to start
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Macleef View Post
    Solve for x in each of the following:

    $\displaystyle log_2x = 3log_24$

    My answer: 64

    --

    $\displaystyle logx = \frac{1}{3}log8$

    My answer: 2


    Are my answers correct?
    Yes.

    Quote Originally Posted by Macleef View Post
    Evaluate each of the following:

    1) $\displaystyle log_2 \sqrt [5] {16}$

    2) $\displaystyle log_525 \sqrt [3] {5}$

    Can you please show me how to solve these logs, I don't know where to start
    We know that
    $\displaystyle log_b(a^x) = x \cdot log_b(a)$

    So for the first one:
    $\displaystyle log_2 ( \sqrt[5]{16} ) = \frac{1}{5} \cdot log_2(16) = \frac{1}{5} \cdot 4 = \frac{4}{5}$

    You do the second one.

    -Dan
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  3. #3
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    For the second equation, is the following what you do...

    $\displaystyle = \frac {1}{3} (log_525(5))$

    $\displaystyle = \frac {1}{3} (log_5 5^{2} (5^{1}))$

    $\displaystyle = \frac {1}{3} (2) (1)$

    $\displaystyle = \frac {2}{3}$
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Macleef View Post
    For the second equation, is the following what you do...

    $\displaystyle = \frac {1}{3} (log_525(5))$

    $\displaystyle = \frac {1}{3} (log_5 5^{2} (5^{1}))$

    $\displaystyle = \frac {1}{3} (2) (1)$

    $\displaystyle = \frac {2}{3}$
    No, here we have another rule coming into play as well:
    $\displaystyle log_b(xy) = log_b(x) + log_b(y)$

    So....
    $\displaystyle log_5(25\sqrt[3]{5})$

    $\displaystyle = log_5(25) + log_5(\sqrt[3]{5})$

    Can you finish from here?

    -Dan
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  5. #5
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    Quote Originally Posted by topsquark View Post
    No, here we have another rule coming into play as well:
    $\displaystyle log_b(xy) = log_b(x) + log_b(y)$

    So....
    $\displaystyle log_5(25\sqrt[3]{5})$

    $\displaystyle = log_5(25) + log_5(\sqrt[3]{5})$

    Can you finish from here?

    -Dan

    Is the answer $\displaystyle \frac {7}{3}$?


    My work:

    $\displaystyle = log_55^{2} + \frac {1}{3} log_55$

    $\displaystyle = 2(log_55) + \frac {1}{3} log_55$

    $\displaystyle = 2 + \frac {1}{3}$

    $\displaystyle = \frac {7}{3}$
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Macleef View Post
    Is the answer $\displaystyle \frac {7}{3}$?


    My work:

    $\displaystyle = log_55^{2} + \frac {1}{3} log_55$

    $\displaystyle = 2(log_55) + \frac {1}{3} log_55$

    $\displaystyle = 2 + \frac {1}{3}$

    $\displaystyle = \frac {7}{3}$
    Yup!

    -Dan
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