Question 693604
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log\left(\frac{I_G}{I_0}\right)\ =\ 6.7]


Since the definition of logarithms says:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = \log_b(x) \ \ \Rightarrow\ \ b^y = x]


and we are assuming the base of the given log to be 10 because it is not specified, we can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \left(\frac{I_G}{I_0}\right)\ =\ 10^{6.7}]


And then


*[tex \LARGE \ \ \ \ \ \ \ \ \ I_G\ =\ I_0\,\cdot\,10^{6.7}]


Similarly


*[tex \LARGE \ \ \ \ \ \ \ \ \ I_C\ =\ I_0\,\cdot\,10^{8}]


and finally:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \frac{I_C}{I_G}\ =\ \frac{I_0\,\cdot\,10^{8}}{I_0\,\cdot\,10^{6.7}}\ =\ 10^{8\,-\,6.7}\ =\ 10^{1.3}\ \approx\ 20]


You can usually do relative size of earthquakes in your head.  A whole number is 10 times the intensity of the next lower whole number.  So two whole numbers is 100 times, etc.  0.1 is a 1.25 multiplier, 0.2 is approximately a 1.6 multiplier, and 0.3 is very close to a 2 multiplier.  So your 1.3 difference is worth 10 (for the whole number) times 2 (for the 0.3) for a total of 20.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \