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Thread: Logarithms

  1. #1
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    Logarithms

    Hi,

    I don't see where I"m going wrong in these. Forgive me for the simple errors

    Express in terms of log a, log b, and log c.

    $\displaystyle \log \bigg(\sqrt[a]{\big(\frac{b}{c^3}\big)}\bigg)$

    $\displaystyle \log \bigg(\bigg(\frac{b}{c^3}\bigg)^\frac{1}{a}\bigg)$

    $\displaystyle \frac{1}{a} \log b - \frac{1}{a} 3 \log c$

    Answer given: $\displaystyle \log a + \frac{1}{2} \log b - \frac{3}{2} \log c$

    Express as a single logarithm

    $\displaystyle 3 \log_a 3 - \log_a 15 + 2 \log_a 5$

    $\displaystyle = \log_a 9 - \log_a 15 + \log_a 25$

    $\displaystyle = \log_a \big(\frac{9}{15}\big) + \log_a 25$

    $\displaystyle = \log_a \big(\frac{3}{5} \times 25\big)$

    $\displaystyle = \log_a 15$

    Answer given: $\displaystyle \log_a 45$
    Last edited by Hellbent; Nov 14th 2010 at 10:18 AM.
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  2. #2
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    Your answer to #1 is correct. The given answer is not.

    Note that: $\displaystyle 3\log_a(3)=\log_a(27).$
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  3. #3
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    Thank you

    $\displaystyle = \log_a 27 - \log_a 15 + 2 \log_a 25$

    $\displaystyle = \log_a \big(\frac{27}{15}\big) + \log_a 25$

    $\displaystyle = \log_a \big(\frac{9}{5} \times 25\big)$

    $\displaystyle = \log_a 45$
    Last edited by Hellbent; Nov 14th 2010 at 10:19 AM.
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  4. #4
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  5. #5
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    Quote Originally Posted by Hellbent View Post
    Hi,

    I don't see where I"m going wrong in these. Forgive me for the simple errors

    Express in terms of log a, log b, and log c.

    $\displaystyle \log \bigg(\sqrt[a]{\big(\frac{b}{c^3}\big)}\bigg)$

    $\displaystyle \log \bigg(\bigg(\frac{b}{c^3}\bigg)^\frac{1}{a}\bigg)$

    $\displaystyle \frac{1}{a} \log b - \frac{1}{a} 3 \log c$

    Answer given: $\displaystyle \log a + \frac{1}{2} \log b - \frac{3}{2} \log c$
    IF the problem were $\displaystyle log\left(a\sqrt{\frac{b}{c^3}}\right)$
    rather than $\displaystyle log\sqrt[a]{\frac{b}{c^3}}$ then the given answer would be correct.
    Express as a single logarithm

    [tex]3 \log_a 3 - \log_a 15 + \log_a 5[/MATH$\displaystyle = \log_a 9 - \log_a 15 + \log_a 25$]
    $\displaystyle 3^3= 27$, not 9. How did that last "5" become "25"?


    $\displaystyle = \log_a \big(\frac{9}{15}\big) + \log_a 25$

    $\displaystyle = \log_a \big(\frac{3}{5} \times 25\big)$

    $\displaystyle = \log_a 15$

    Answer given: $\displaystyle \log_a 45$
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    IF the problem were $\displaystyle log\left(a\sqrt{\frac{b}{c^3}}\right)$
    rather than $\displaystyle log\sqrt[a]{\frac{b}{c^3}}$ then the given answer would be correct.

    $\displaystyle 3^3= 27$, not 9. How did that last "5" become "25"?
    I fixed it. Thanks.
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