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Math Help - Logs

  1. #1
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    Logs

    I have two questions:


    1. If log_{14} 7 = a and log_{14} 5 = b calculate log_{35} 28
    2. Let log 36 = a and log 125 = b find log (1/12) in terms of a and b

    Thanks in advanced
    Last edited by mr fantastic; January 10th 2010 at 07:14 PM. Reason: Fixed log bases.
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  2. #2
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    Quote Originally Posted by ntrantrinh View Post
    I have two questions:


    1. If log_{14} 7 = a and log_{14} 5 = b calculate log_{35} 28
    [snip]
    Thanks in advanced
    It follows that \log_{14} 35 = a + b \Rightarrow  14^{a + b} = 35. From here it follows that (a + b) \log_{35} 14 = 1 \Rightarrow \log_{35} 14 = \frac{1}{a + b}. From this you should be able to get an expression for \log_{35} 28 using a simple log rule.
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  3. #3
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    Hello, ntrantrinh!

    If the second one is base-10, I have a solution.


    Let \log 36 = a and \log 125 = b.

    Find \log\left(\tfrac{1}{12}\right) in terms of a and b.

    \log(36) \,=\,a\quad\Rightarrow\quad \log(6^2) \,=\,a \quad\Rightarrow\quad 2\log(6) \,=\,a \quad\Rightarrow\quad \log(6) \,=\,\frac{a}{2}

    \log(125) \,=\,b \quad\Rightarrow\quad \log(5^3)\,=\,b \quad\Rightarrow\quad 3\log(5) \,=\,b \quad\Rightarrow\quad \log(5) \,=\,\frac{b}{3}


    \log\left(\frac{1}{12}\right) \;\;=\;\;\log\left(\frac{5}{60}\right) \;\;=\;\;\log(5) - \log(60) \;\;=\;\;\log(5) - \log(6\cdot10)

    . . . . =\;\; \log(5) - \bigg[\log(6) + \log(10)\bigg] \;\;=\;\;\log(5) - \bigg[\log(6) + 1\bigg]

    . . . . =\;\; \log(5) - \log(6) - 1 \;\;=\;\;\frac{b}{3} - \frac{a}{2} - 1

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