SOLUTION: Hi, I'd love some help on this if possible. M=0.67 log(0.36E)+1.46 the function above is used to define the magnitude of an earthquake. M is the magnitude and E is the energy in k

Question 921925: Hi, I'd love some help on this if possible. M=0.67 log(0.36E)+1.46
the function above is used to define the magnitude of an earthquake. M is the magnitude and E is the energy in kilowatt-hours. I need to explain this 'rule of thumb' by proper generalization. Using mathematical process. I hope this is understandable. Thank you.
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
We are given,
M = 0.67*log(0.36E) + 1.46
suppose we have a magnitude 6.0 earthquake, then we ask how many kilowatt hours(E) are released?
6.0 = 0.67*log(0.36E) + 1.46
0.67*log(0.36E) = 6.0 - 1.46 = 4.54
log(0.36E) = 4.54 / 0.67 = 6.776119403
now use product of logarithms rule
log(0.36) + log(E) = 6.776119403
-4.4369749923e-1 + log(E) = 6.776119403
log(E) = 6.776119403 + 4.4369749923e-1 = 7.219816902
now use definition of logarithm
E = 10^7.219816902 = 16588873.760056074
E = 1.6588874e7 kilowatt hours were released

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