Question 1190491: On September 25, 2003 an earthquake that
measured 7.4 on the Richter scale shook Hokkaido, Japan.
How much energy (joules) did the earthquake emit?
I get 10^15.5 joules using book formula of M=(2/3)log(E/10^4.4). The answer in my book says 3.16 x 10^15 joules.
Found 2 solutions by Alan3354, math_tutor2020:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! On September 25, 2003 an earthquake that
measured 7.4 on the Richter scale shook Hokkaido, Japan.
How much energy (joules) did the earthquake emit?
I get 10^15.5 joules using book formula of M=(2/3)log(E/10^4.4). The answer in my book says 3.16 x 10^15 joules.
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What is the E?
What is the base of the log?
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
M = (2/3)*log(E/10^4.4)
7.4 = (2/3)*log(E/10^4.4)
7.4*(3/2) = log(E/10^4.4)
11.1 = log(E/10^4.4)
E/10^4.4 = 10^11.1
E/10^4.4 = 125,892,541,179.417
E = (25,118.8643150959)*(125,892,541,179.417)
E = (2.5118 * 10^4)*(1.2589 * 10^11)
E = (2.5118*1.2589) * (10^4*10^11)
E = 3.16210502 * 10^(4+11)
E = 3.16 * 10^15
Answer: 3.16 * 10^15 joules approximately
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