Solve the problem.
-The local magnitude $M _ { L }$ (on the Richter scale) of an earthquake of intensity $I$ is given by $M _ { L } = \log \left( \frac { I } { I _ { 0 } } \right)$ where $I _ { 0 }$ is a minimum reference intensity of a "zero-level" earthquake against which the intensities of other earthquakes may be compared. How many times more intense is an earthquake of magnitude $5.8$ than an earthquake of magnitude $2.9$ ? Round to the nearest whole number.
A) Approximately 2 times more intense.
B) Approximately 794 times more intense.
C) Approximately 481 times more intense.
D) Approximately 18 times more intense.

Question

Quizplus · Accepted Answer

To find how many times more intense one earthquake is compared to another, we use the given formula for local magnitude and solve for the intensity ratio. For earthquakes of magnitudes $M_{L1} = 5.8$ and $M_{L2} = 2.9$, we have:$$M_{L1} = \log\left(\frac{I_1}{I_0}\right) = 5.8$$$$M_{L2} = \log\left(\frac{I_2}{I_0}\right) = 2.9$$Subtracting the second equation from the first gives:$$5.8 - 2.9 = \log\left(\frac{I_1}{I_0}\right) - \log\left(\frac{I_2}{I_0}\right) = \log\left(\frac{I_1}{I_2}\right)$$$$2.9 = \log\left(\frac{I_1}{I_2}\right)$$To solve for $\frac{I_1}{I_2}$, we raise both sides as powers of 10:$$10^{2.9} = \frac{I_1}{I_2}$$$$794 \approx \frac{I_1}{I_2}$$Therefore, an earthquake of magnitude 5.8 is approximately 794 times more intense than an earthquake of magnitude 2.9.

Solve the Problem MLM _ { L }ML​ (On the Richter Scale) of an Earthquake of Intensity

Solve the Problem $M _ { L }$ (On the Richter Scale) of an Earthquake of Intensity