Chapter 4, Problem 105RE

To determine

To Find: Themagnitude of another quake that is $35$ times as intense.

Answer to Problem 105RE

  $M=8.044$

Explanation of Solution

Given: If one earthquake has magnitude $6.5$ on Richter scale, what is the magnitude of another quake that is $35$ times as intense?

FORMULA USED:

  $\begin{array}{l}M=\log \frac{I}{S}\\ {\log }_a\frac{A}{B}={\log }_{a}A-{\log }_{a}B\\ {\log }_a{A}^c=C{\log }_{a}A\end{array}$

The Richter scale measures the intensity of the earthquake is given by

  $M=\log \frac{I}{S}$

The magnitude of the one earthquake is 6.5.

Thus

  $\begin{array}{l}{M}_1=\frac{I_1}{S}\\ 6.5=\log \frac{I_1}{{10}^{-4}}\\ 6.5=\log I_1-\log {10}^{-4}\\ 6.5=\log I_1+4\log 10\\ 6.5=\log I_1+4\end{array}$

Add $-4$ both sides

  $\begin{array}{l}6.5-4=\log I_1+4-4\\ 2.5=\log I_1\\ I_1={10}^{2.5}\end{array}$

Now, another quake is $35$ times intense

Thus

  $\begin{array}{l}{I}_2=35\times {10}^{2.5}\\ M=\log \frac{35\times {10}^{2.5}}{{10}^{-4}}=8.044\\ M=8.044\end{array}$

Hence, the magnitude of another quake is $M=8.044$