APPLICATIONS OF EXPONENTIAL
AND
LOGARITHMIC FUNCTIONS



EARTHQUAKE WORD PROBLEMS:



As with any word problem, the trick is to 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 1: Early in the century an earthquake measured 8.0 on the Richter scale. In the same year, another earthquake was recorded that was six time stronger. What was the magnitude of the earthquake of the stronger earthquake?



Answer: 8.78 on the Richter Scale.


Solution: Convert the first sentence to an equivalent mathematical sentence or equation where

\begin{eqnarray*}M_{1} &=&\log \displaystyle \frac{I_{1}}{S}=8.0
\end{eqnarray*}


Convert the second sentence to an equivalent mathematical sentence or equation:

I2=6 I1


Convert the third sentence into an equivalent mathematical sentence or equation:

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

Replace I2 with I1

\begin{eqnarray*}&=&\log \left( 6I_{1}\right) -\log S \\
&& \\
&=&\log 6+\log I_{1}-\log S
\end{eqnarray*}

\begin{eqnarray*}&=&\log 6+(\log I_{1}-\log S) \\
&& \\
&=&\log 6+\log \displaystyle \frac{I_{1}}{S}
\end{eqnarray*}

\begin{eqnarray*}&=&\log 6+8.0 \\
&& \\
&=&0.778151250384+8.0
\end{eqnarray*}

\begin{eqnarray*}&=&8.778151250384 \\
&& \\
M_{2} &=&8.78
\end{eqnarray*}


The intensity of the second earthquake was 8.78 on the Richter scale.



Let's check our answer.

\begin{eqnarray*}8.78 &=&\log \displaystyle \frac{I_{2}}{S}=\log I_{2}-\log S \\...
... \\
8.0 &=&\log \displaystyle \frac{I_{1}}{S}=\log I_{1}-\log S
\end{eqnarray*}

\begin{eqnarray*}8.78-8.0 &=&\left( \log I_{2}-\log S\right) -\left( \log I_{1}-...
... S\right)
\\
&& \\
0.78 &=&\log I_{2}-\log S-\log I_{1}+\log S
\end{eqnarray*}

\begin{eqnarray*}0.78 &=&\log I_{2}-\log I_{1} \\
&& \\
0.78 &=&\log \displaystyle \frac{I_{2}}{I_{1}}
\end{eqnarray*}

\begin{eqnarray*}10^{0.78} &=&\displaystyle \frac{I_{2}}{I_{1}} \\
&& \\
6.02559586074 &\approx &\displaystyle \frac{I_{2}}{I_{1}}
\end{eqnarray*}



The check will not be exact because we rounded the answer. However, it is close enough to see if we are in the ball park with our answer.



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.

[Exponential Rules] [Logarithms]

[Algebra] [Trigonometry ] [Complex Variables]

S.O.S MATHematics home page

Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.

Author: Nancy Marcus

Copyright 1999-2017 MathMedics, LLC. All rights reserved.
Contact us
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA
users online during the last hour