Friday, February 27, 2015

College Algebra, Chapter 5, 5.5, Section 5.5, Problem 36

In 1994, the Northridge, California earthquake had a magnitude of 6.8 on the Richter scale. A year later, a 7.2-magnitude earthquake struck Kobe, Japan. Determine how many times more intense was the Kobe earthquake than the Northridge earthquake.



Recall that the Richter Scale is defined as

$\displaystyle M = \log \frac{I}{S}$

where

$M$ = magnitude of the earthquake

$I$ = intensity of the earthquake

$S$ = intensity of a standard earthquake

For Kobe, Japan


$
\begin{equation}
\begin{aligned}

M_1 =& \log \frac{I_1}{S}
\\
\\
10^{M_1} =& \frac{I_1}{S}
\\
\\
S =& \frac{I_1}{10^{M_1}}
\qquad \text{Equation 1}

\end{aligned}
\end{equation}
$


For Northridge,


$
\begin{equation}
\begin{aligned}

M_2 =& \log \frac{I_2}{S}
\\
\\
10^{M_2} =& \frac{I_2}{S}
\\
\\
S =& \frac{I_2}{10^{M_2}}
\qquad \text{Equation 2}

\end{aligned}
\end{equation}
$


By using equations 1 and 2


$
\begin{equation}
\begin{aligned}

\frac{I_1}{10^{M_1}} =& \frac{I_2}{10^{M_2}}
&& \text{Multiply each side by } 10^{M_1}
\\
\\
I_1 =& \frac{10^{M_1}}{10^{M_2}} I_2
&& \text{Substitute given}
\\
\\
I_1 =& \frac{10^{7.2}}{10^{6.8}} I_2
&& \text{Evaluate}
\\
\\
I_1 =& 2.51 I_2 \text{ or } 2.50 I_2

\end{aligned}
\end{equation}
$



It shows that the earthquake had struck in Kobe, Japan is $2.50$ more intense than the earthquake that hit Northridge.

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