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 \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 2: A recent earthquake measured 6.8 on the Richter scale. How many times more intense was this earthquake than an earthquake that measured 4.3 on the Richter scale?



Answer: 316


Solution:Let I1 refer to the earthquake that measured 6.8 on the Richter scale, and I2 refer to the earthquake that measured 4.3 on the Richter scale.

\begin{eqnarray*}6.8 &=&\log \displaystyle \frac{I_{1}}{S}\\
&&\\
4.3 &=&\log \displaystyle \frac{I_{2}}{S}
\end{eqnarray*}


Solve for $\displaystyle \frac{I_{2}}{I_{1}}$.

\begin{eqnarray*}6.8-4.3 &=&\log \displaystyle \frac{I_{1}}{S}-\log \displaystyl...
...1}}{S}\right) -\left( \log \displaystyle \frac{I_{2}}{S}
\right)
\end{eqnarray*}

\begin{eqnarray*}2.5 &=&\left( \log I_{1}-\log S\right) -\left( \log I_{2}-\log S\right) \\
&& \\
2.5 &=&\log I_{1}-\log S-\log I_{2}+\log S
\end{eqnarray*}

\begin{eqnarray*}2.5 &=&\log I_{1}-\log I_{2} \\
&& \\
2.5 &=&\log \displaystyle \frac{I_{1}}{I_{2}}
\end{eqnarray*}

\begin{eqnarray*}10^{2.5} &=&\displaystyle \frac{I_{1}}{I_{2}} \\
&& \\
316.227766017 &=&\displaystyle \frac{I_{1}}{I_{2}}
\end{eqnarray*}

\begin{eqnarray*}\displaystyle \frac{I_{1}}{I_{2}} &\approx &316 \\
&& \\
I_{1} &\approx &316I_{2}
\end{eqnarray*}



The intensity of the larger earthquake was earthquake as 316 times as intense as the smaller earthquake.



Let's check our answer.

\begin{eqnarray*}4.3 &=&\log \displaystyle \frac{I_{2}}{S}=\log I_{2}-\log S \\ ...
...\
M_{1} &=&\log \displaystyle \frac{I_{1}}{S}=\log I_{1}-\log S
\end{eqnarray*}

\begin{eqnarray*}M_{1} &=&\log 316I_{2}-\log S \\
&& \\
M_{1} &=&\log 316+\log I_{2}-\log S
\end{eqnarray*}

\begin{eqnarray*}M_{1} &=&\log 316+\log \displaystyle \frac{I_{2}}{S} \\
&& \\
M_{1} &=&\log 316+M_{2}
\end{eqnarray*}

\begin{eqnarray*}M_{1} &=&\log 316+4.3 \\
&& \\
M_{1} &=&2.49968708262+4.3=6.79968708262\\
&&\\
M_{1} &\approx &6.8
\end{eqnarray*}



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



If you would like to review problem 2, click on problem.


If you would like to go back to the table of contents, click on contents.

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Author: Nancy Marcus

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