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.
Friday, February 27, 2015
College Algebra, Chapter 5, 5.5, Section 5.5, Problem 36
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