College Algebra, Chapter 5, 5.5, Section 5.5, Problem 34
Determine how much larger is the magnitude on the Richter scale if one earthquake is 20 times as intense as another.
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 San Francisco,
$
\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 Japan,
$
\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^{8.3}}{10^{4.9}} I_2
&& \text{Evaluate}
\\
\\
I_1 =& 2511.89 I_2
\end{aligned}
\end{equation}
$
It shows that the earthquake in San Francisco is $2511$ more intense than the Japanese earthquake.
Comments
Post a Comment