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 the formula: $ M = \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, we have $ \log_{10} \left( \frac{I}{S} \right) = 7.9 $. Therefore, the magnitude of the 1906 earthquake is $ 7.9 + \log_{10}(12) \approx 7.9 + 1.08 = 8.98 $, which rounds to 9.

The Geologist C log⁡10(IS)\log _ { 10 } \left( \frac { I } { S } \right)log10​(SI​)

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