Math  Working with Logs!!
posted by Anonymous on .
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

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 
So, you don't use the equation:
M = log(I_i / I_o) ? 
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 
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 
You have it right in the second line in the laws of logarithms a*b ends up equaling log 3 + log 10^5