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Math Help - Log + Ritcher Scale

  1. #1
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    Log + Ritcher Scale

    Word Problem:
    On September 26 2001, an earthquake in North Bay measured 5.0 on the Ritcher scale. What is the magnitude of an earthquake 3 times as intense as North Bay's earthquake?



    My Work So Far:
    Let I_B represent the earthquake in North Bay
    Let I_F represent the earthquake in ________?

    North Bay:
     5 = log(\frac {I_B}{I_o})

    (\frac {I_B}{I_o}) = 10^{5}

    I_B = 10^{5} I_o


    Somewhere Else:
    3r = log(\frac {I_F}{I_o})

    (\frac {I_F}{I_o}) = 10^{3r}

    I_F = 10^{3r} I_o

    Comparison:
    \frac {I_B}{I_F} = \frac {10^{5} I_o}{10^{3r} I_o}

    \frac {I_B}{I_F} = \frac {10^{5}}{10^{3r}}

    {I_B}= 10^{5-3r}I_o


    Am I doing this right? If I am, what should I do next? If I'm not, where did I make my mistake and how should I correct it?

    --
    Answer: 5.477
    Last edited by Macleef; December 11th 2007 at 07:35 PM.
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  2. #2
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    Quote Originally Posted by Macleef View Post
    Word Problem:
    On September 26 2001, an earthquake in North Bay measured 5.0 on the Ritcher scale. What is the magnitude of an earthquake 3 times as intense as North Bay's earthquake?



    My Work So Far:
    Let I_B represent the earthquake in North Bay
    Let I_F represent the earthquake in ________?

    ...
    Hello,

    you know that I_F = 3 \cdot I_B . Thus you have:

    r_{I_F} = \log\left( \frac{I_F}{I_0}\right) = \log\left( \frac{3 \cdot I_B}{I_0} \right) =  \log\left(3 \cdot  \frac{ I_B}{I_0} \right) = \log(3) + \log\left( \frac{ I_B}{I_0} \right) = \log(3) + 5 \approx 5.477121254...
    Last edited by earboth; December 11th 2007 at 11:02 PM.
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