Posted by Anonymous on Tuesday, December 11, 2007 at 10:30pm.
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?
I would post my answer, but in text, it's just confusing and I can't post picture links to show my work...
Btw, the answer is 5.477

Math  Working with Logs!!  Reiny, Tuesday, December 11, 2007 at 11:00pm
In 1935 C.F. Richter set up the scale of earthquakes by
R = log(I) where R is the Richter scale number and I is the intensity of the earthquake
so from your data
5 = log(I)
I = 10^5
so an earthquake 3 times as intense would be 3I
so multiplying the above equation by 3
3I = 3(10^5)
log (3I) = log(3*10^5)
= log 3 + log 10^5
= .47712 + 5
= 5.47712

Math  Working with Logs!!  Anonymous, Tuesday, December 11, 2007 at 11:04pm
So, you don't use the equation:
M = log(I_i / I_o) ?

Math  Working with Logs!!  Reiny, Tuesday, December 11, 2007 at 11:21pm
most texts, and the one that I used last, use the formula
R = log(I/I_{0}) where _{0} is a constant, the minimum intensity corresponding to R = 0
so if you want
5 = log (I/I_{0})
5^10 = I/I_{0}
3*5^10 = 3I/I_{0}
log(3*5^10 = log(I/I_{0})
notice it has no effect on the answer

Math  Working with Logs!!  Anonymous, Friday, March 19, 2010 at 6:07pm
i don't get where the 3 goes in the second line of this part :
log (3I) = log(3*10^5)
= log 3 + log 10^5

Math  Working with Logs!!  Anonymous, Sunday, July 27, 2014 at 12:47pm
You have it right in the second line in the laws of logarithms a*b ends up equaling log 3 + log 10^5
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