Question

How many times more intense is an earthquake that measures 7.5 on the Richter scale than one that measures 6.7? (Recall that the Richter scale defines magnitude of an earthquake with the equation M=log i/s , where i is the intensity of the earthquake being measured, and S is the intensity of a standard earthquake

Answer 1

richter scale is log base ten scale

Answer 2

so you are comparing something of order of magnitude \(10^{7.5}\) with something of magnitude \(10^{6.7}\)

Answer 3

compare by division
\[\frac{10^{7.5}}{10^{6.7}}=10^{.8}\] or so

Answer 4

since \(10^8=6.3\) rounded it is about six times as strong, depending on how accurate you wish to be

Answer 5

i got it this way, does this work??


log(I2/S)=7.5
log(I2/S)=6.7
log(I1/S)-log(I2/S)=7.5-6.7
[log(I1)-log(S)]-[log(I2)-log(S)]=0.8
log(I1)-log(S)-log(I2)+log(S)=0.8
log(I1)-log(I2)=0.8
log(I1/I2)=0.8
10^[log(I1/I2)]=10^(0.8)
I1/I2=6.31

@satellite73

Answer 6

man that is a lot of work to say "subtract 6.7 from 7.5 and then raise 10 the the power of the result"

Answer 7

yeah, i don't know thats the way it shows in my book, i just wanted to learn a better way and i think yours is faster

Answer 8

so 6.3 would be the accurate answer then?

Answer 9

i'll say
there is no end to the obfuscation of the painfully obvious by textbooks

Answer 10

agreed

Answer 11

thanks so much, could you possibly help me with another problem?

Answer 12

6.3 times as large would be my answer

Answer 13

i like your method better then the ones i'm taught

Answer 14

great, i'll put that