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Math Help - Differential: Richter Scale

  1. #1
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    Differential: Richter Scale

    On Richter scale, magnitude M of an earthquake of intesity I is:
    M = (ln(I)) - ln(I)0)/(ln(10))
    Where I0 is the min intensity used for comparison, and assume that it equals 1.

    a. Find intensity of 106 San Fran earthquake (M = 8.3)
    This is what I did:
    I plugged in the givens.
    8.3 = (ln I - ln(1))/ln(10)
    I know that ln(1) = 0 so that gets taken out
    I get left with:
    8.3 = (ln I)/(ln 10)

    Not sure if I did this correctly... But I multiplied both sides by ln 10.
    But then I got lost when there was another ln on the other side.
    I think I forgot something that was crucial in order to find the variable of something with a natural log.

    Can someone remind me or help me out?
    Thanks
    Answer: 108.3=199,526,231.5
    I think this is what they did:
    I converted the last thing I did into exponential (the part where I multiplied both sides by 10)
    I=eln10*8.3... Oh wait I just figured it out, it equals the answer, but it doesn't look like the 108.3 thing xD. But yeah, if anyone wants to explain to me how
    eln10*8.3 = 108.3 that would be nice


    Also
    b. Find factor by which intensity is increased if Richter scale measurement is doubled
    I'm not really sure what to first do.
    The answer is: 10R
    So I'm assuming it was the base to the power of R?
    Is it really just like that, or was there more steps before it?

    Thanks
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  2. #2
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    Re: Differential: Richter Scale

    So you want to solve ln(I)/ln(10)= 8.3 so that ln(I)= 8.3 ln(10).
    Now, use the "laws of logarithms": [itex]ln(I)= ln(10^{83})[/itex]

    If the Richter scale is doubled, that is 16.6 rather than 8.3, then you need to solve ln(I)/ln(10)= 16.6 so [itex]ln(I)= 16.6 ln(10)= ln(10^{16.6}[/itex]. Once you have found this I, divide by the previous one to find the factor.
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  3. #3
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    Re: Differential: Richter Scale

    Quote Originally Posted by HallsofIvy View Post
    So you want to solve ln(I)/ln(10)= 8.3 so that ln(I)= 8.3 ln(10).
    Now, use the "laws of logarithms": [itex]ln(I)= ln(10^{83})[/itex]

    If the Richter scale is doubled, that is 16.6 rather than 8.3, then you need to solve ln(I)/ln(10)= 16.6 so [itex]ln(I)= 16.6 ln(10)= ln(10^{16.6}[/itex]. Once you have found this I, divide by the previous one to find the factor.
    Oh right!
    I forgot!
    When there's a number in front of a log, which is the coefficient, then it becomes the exponent!
    But then, you would just divide ln from both sides right? If that's possible.
    So yeah, thanks!

    And I assume you do the same thing for letter b.?
    Have it equal to 16.6 instead of 8.3
    Then go through the same process?

    By the way, it's actually 10M, my bad, accidently put R.
    But wait, the answer has the M in it, so would I plug anything in?
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