If an earthquake measured 6.1 on the Richter scale, what was the intensity of the earthquake? $\mathrm { R } = \log \left( \frac { \mathrm { I } } { \mathrm { I } _ { 0 } } \right)$ Round to the nearest whole number.
A) 1,258,925
B) 125,893
C) 1,000,000
D) 446

Question

Quizplus · Accepted Answer

The Richter scale formula is $ R = \log \left( \frac { I } { I_0 } \right) $. Given $ R = 6.1 $, we solve for $ I $ as follows: $ 6.1 = \log \left( \frac { I } { I_0 } \right) $. Therefore, $ I = I_0 \times 10^{6.1} $. Assuming $ I_0 = 1 $, $ I = 10^{6.1} \approx 1,258,925 $.

If an Earthquake Measured 6 R=log⁡(II0)\mathrm { R } = \log \left( \frac { \mathrm { I } } { \mathrm { I } _ { 0 } } \right)R=log(I0​I​)

If an Earthquake Measured 6 $\mathrm { R } = \log \left( \frac { \mathrm { I } } { \mathrm { I } _ { 0 } } \right)$