The geologist C. F. Richter defined the magnitude of an earthquake to be $\log _ { 10 } \left( \frac { I } { S } \right)$ where $I$ is the intensity of the quake (measured by the amplitude of a seismograph $100 \mathrm {~km}$ from the epicenter) and $S$ is the intensity of a "standard" earthquake (where the amplitude is only 1 micron $= 10 ^ { - 4 }$ $\mathrm { cm }$ ). The 1989 Loma Prieta earthquake that shook San Francisco had a magnitude of $7.9$ on the Richter scale. The 1906 San Francisco earthquake was 12 times as intense. What was its magnitude on the Richter scale?
A) 12
B) 11
C) 9
D) 14
E) 13

Question

Quizplus · Accepted Answer

The magnitude of the 1906 earthquake can be calculated using Richter's formula: $\log _ { 10 } \left( \frac { 12I } { S } \right) = \log _ { 10 } \left( \frac { I } { S } \right) + \log _ { 10 } (12)$. Given that the 1989 earthquake had a magnitude of 7.9, and using the property of logarithms that $\log(a) + \log(b) = \log(ab)$, the magnitude of the 1906 earthquake is $7.9 + \log _ { 10 } (12)$. Since $\log _ { 10 } (12) \approx 1.08$, the magnitude of the 1906 earthquake is approximately $7.9 + 1.08 = 8.98$, which rounds to 9 on the Richter scale.

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The Geologist C $\log _ { 10 } \left( \frac { I } { S } \right)$