APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS

	APPLICATIONS OF EXPONENTIAL
	AND
	LOGARITHMIC FUNCTIONS

EARTHQUAKE WORD PROBLEMS:

As with any word problem, the trick is convert a narrative statement or question to a mathematical statement.

Before we start, let's talk about earthquakes and how we measure their intensity.

In 1935 Charles Richter defined the magnitude of an earthquake to be

$\begin{eqnarray*}M &=&\log \displaystyle \frac{I}{S} \end{eqnarray*}$

where I is the intensity of the earthquake (measured by the amplitude of a seismograph reading taken 100 km from the epicenter of the earthquake) and S is the intensity of a ''standard earthquake'' (whose amplitude is 1 micron =10^-4 cm).

The magnitude of a standard earthquake is

$\begin{eqnarray*}M &=&\log \displaystyle \frac{S}{S}=\log 1=0 \end{eqnarray*}$

Richter studied many earthquakes that occurred between 1900 and 1950. The largest had magnitude of 8.9 on the Richter scale, and the smallest had magnitude 0. This corresponds to a ratio of intensities of 800,000,000, so the Richter scale provides more manageable numbers to work with.

Each number increase on the Richter scale indicates an intensity ten times stronger. For example, an earthquake of magnitude 6 is ten times stronger than an earthquake of magnitude 5. An earthquake of magnitude 7 is $10\times 10=100$ times strong than an earthquake of magnitude 5. An earthquake of magnitude 8 is $10\times 10\times 10=1000$ times stronger than an earthquake of magnitude 5.

Problem 3: If one earthquake is 31 times as intense as another, how much larger is its magnitude on the Richter scale?

Answer: 1.49

Solution: I₁ = 31 I₂

We are looking for the quantity M₁ -- M₂.

$\begin{eqnarray*}M_{1} &=&\log \displaystyle \frac{I_{1}}{S} \\ && \\ M_{2} &=&\log \displaystyle \frac{I_{2}}{S} \end{eqnarray*}$

I₁ = 31 I₂

$\begin{eqnarray*}M_{1}-M_{2} &=&\log \displaystyle \frac{I_{1}}{S}-\log \display... ...}}{S}\right) -\left( \log \displaystyle \frac{I_{2}}{S }\right) \end{eqnarray*}$

$\begin{eqnarray*}M_{1}-M_{2} &=&\left( \log I_{1}-\log S\right) -\left( \log I_{... ...eft( \log 31I_{2}-\log S\right) -\left( \log I_{2}-\log S\right) \end{eqnarray*}$

$\begin{eqnarray*}M_{1}-M_{2} &=&\left( \log 31+\log I_{2}-\log S\right) -\left( ... ...& \\ M_{1}-M_{2} &=&\log 31+\log I_{2}-\log S-\log I_{2}+\log S \end{eqnarray*}$

$\begin{eqnarray*}M_{1}-M_{2} &=&\log 31 \\ && \\ M_{1}-M_{2} &\approx &1.49 \end{eqnarray*}$

The larger earthquake was 1.49 larger on the Richter scale.

Let's check our answer.

$\begin{eqnarray*}M_{1}-M_{2} &=&1.49=\log \displaystyle \frac{I_{1}}{S}-\log \di... ...1}}{S}\right) -\left( \log \displaystyle \frac{I_{2}}{S} \right) \end{eqnarray*}$

$\begin{eqnarray*}1.49 &=&\left( \log I_{1}-\log S\right) -\left( \log I_{2}-\log S\right) \\ && \\ 1.49 &=&\log I_{1}-\log I_{2} \end{eqnarray*}$

$\begin{eqnarray*}1.49 &=&\log \displaystyle \frac{I_{1}}{I_{2}} \\ && \\ 10^{1.49} &=&\displaystyle \frac{I_{1}}{I_{2}} \end{eqnarray*}$

$\begin{eqnarray*}\displaystyle \frac{I_{1}}{I_{2}} &\approx &30.9029543251 \\ &... ...ac{I_{1}}{I_{2}} &\approx &31 \\ && \\ I_{1} &\approx &31I_{2} \end{eqnarray*}$

The check will not be exact because we rounded the answer. However, it is close enough for checking.

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[Exponential Rules] [Logarithms]

[Algebra] [Trigonometry ] [Complex Variables]

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Author: Nancy Marcus