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March 4, 2015

March 4, 2015

Posted by **Anonymous** on Tuesday, December 11, 2007 at 10:30pm.

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 11:00pmIn 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 11:04pmSo, you don't use the equation:

M = log(I_i / I_o) ?

- Math - Working with Logs!! -
**Reiny**, Tuesday, December 11, 2007 at 11:21pmmost 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 6:07pmi dont 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 12:47pmYou 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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