Problem 49 - College Algebra

Question

Round all answers to the nearest thousandth. Recall the formula for calculating the magnitude of an earthquake, \(M=\frac{2}{3} \log \left(\frac{S}{S_{0}}\right)\) . One earthquake has magnitude 3.9 on the MMS scale. If a second earthquake has 750 times as much energy as the first, find the magnitude of the second quake. Round to the nearest hundredth.

Step-by-Step Solution

Answer

The magnitude of the second earthquake is 5.82.

Step 1: Understand the formula

The provided formula is for the magnitude of an earthquake: \[M = \frac{2}{3} \log\left(\frac{S}{S_0}\right)\]where \(S\) is the energy released by the earthquake, and \(S_0\) is a reference energy.

Step 2: Set up the equation for the first earthquake

For the first earthquake with magnitude \(M_1 = 3.9\), we use the formula\[3.9 = \frac{2}{3} \log\left(\frac{S_1}{S_0}\right)\]This equation represents the energy \(S_1\) of the first earthquake.

Step 3: Set up the equation for the second earthquake

The second earthquake has energy \(S_2 = 750S_1\). Therefore, its magnitude \(M_2\) is:\[M_2 = \frac{2}{3} \log\left(\frac{750S_1}{S_0}\right)\]We need to find \(M_2\).

Step 4: Relate the two equations

Using logarithm properties, express the equation for the second earthquake in terms of \(M_1\):\[M_2 = \frac{2}{3} \log\left(\frac{750}{1}\right) + \frac{2}{3} \log\left(\frac{S_1}{S_0}\right)\]Since \(M_1 = \frac{2}{3} \log\left(\frac{S_1}{S_0}\right)\), substitute \(M_1\) into the equation:\[M_2 = \frac{2}{3} \log(750) + 3.9\]

Step 5: Simplify and compute \(M_2\)

Calculate the logarithm term and substitute it back:\[\log(750) \approx 2.875\]Thus,\[M_2 = \frac{2}{3} \times 2.875 + 3.9 \approx 5.817\]

Step 6: Round to the nearest hundredth

The calculated value \(5.817\) is rounded to the nearest hundredth, resulting in \(5.82\).

Key Concepts

Logarithms

Logarithms are mathematical tools used to handle exponential relationships efficiently. In simpler terms, a logarithm tells us how many times we must multiply a base number to reach another number.
For example, if we say \(\log_{10}(100)\), we ask how many times we need to multiply 10 to get 100. The answer is 2, because 10 times 10 equals 100.
When it comes to earthquake magnitudes, we often use base-10 logarithms because of the vast energy range they cover. The logarithm in the earthquake formula is used to compare the energy released to some reference energy. Using logarithms simplifies this comparison to a linear scale, which is easier to understand.

Energy Calculation

Energy calculation in the context of earthquakes refers to the amount of energy released during an earthquake. In the exercise, we have two earthquakes to consider, with one releasing 750 times more energy than the other.
To find this, we start by defining the energy of the first earthquake (\(S_1\)) and expressing the second earthquake’s energy as \(S_2 = 750S_1\).The energy released gives us a direct clue about the magnitude. Since it's on a logarithmic scale, even a small increase in magnitude can mean a huge increase in energy. This highlights the incredible power behind each step on the magnitude scale.

Magnitude Scale

The magnitude scale used in this exercise is likely the Moment Magnitude Scale (MMS), which is the go-to scale for measuring large earthquakes. Magnitude is a number that gives us an idea of the earthquake's size by quantifying its energy release. This scale is logarithmic, meaning each whole number increase on the magnitude scale represents a tenfold increase in measured amplitude and roughly 31.6 times more energy release.
In practice, this means if we know the magnitude of one earthquake and how much more energy another earthquake releases, we can find the second earthquake's magnitude using the magnitude formula. The calculated magnitude measures just how intense that second earthquake would feel compared to the first.