Chapter 6.3, Problem 54E

(a)

To determine

To calculate: The energy magnitude of the earthquake.

Answer to Problem 54E

The required energy magnitude of the earthquake is 9 ergs.

Explanation of Solution

Given information:

The expression for the magnitude of the earthquake is $M=\frac{2}{3}\log E-9.9$ .

Energy released $E=2.24\times {10}^{28}\text{ergs}$

Formula Used:

Logarithmic power rule is

If $a^b$ is an expression then logarithm form ${\log }_a{\left(a\right)}^b$ is equal to $b$ .

Logarithmic multiplication rule is

For any multiplication expression $a\times b$ , their logarithmic expression is $\log a+\log b$ .

Calculation:

Consider the expression for the magnitude of the earthquake is $M=\frac{2}{3}\log E-9.9$ .

Substitute the value $2.24\times {10}^{28}$ for $E$ .

  $M=\frac{2}{3}\times \log \left(2.24\times {10}^{28}\right)-9.9$

Find the energy magnitude of the earthquake. Use the logarithmic multiplication rule.

  $\begin{array}{c}M=\frac{2}{3}\times \log \left(2.24\times {10}^{28}\right)-9.9\\ =\frac{2}{3}\times \left(\log 2.24+\log {10}^{28}\right)-9.9\end{array}$

Since, in the given expression is in common logarithmic form thus, it contains base of 10.

Again use the logarithmic power rule in the expression $\log {10}^{28}$ .

  $\begin{array}{c}M=\frac{2}{3}\times \left(\log 2.24+28\right)-9.9\\ =\frac{2}{3}\times \log 2.24+\frac{2}{3}\times 28-9.9\\ =9.0001\dots \\ \approx 9\end{array}$

Hence, the required energy magnitude of the earthquake is 9 ergs.

(b)

To determine

To calculate: The inverse of the function.

Answer to Problem 54E

The required inverse of the function is $E={10}^{\frac{3\left(M+9.9\right)}{2}}$ .

Explanation of Solution

Given information:

The expression for the magnitude of the earthquake is $M=\frac{2}{3}\log E-9.9$ .

Calculation:

Consider the expression for the magnitude of the earthquake is $M=\frac{2}{3}\log E-9.9$ .

Rewrite the expression as

  $\begin{array}{c}M=\frac{2}{3}\log E-9.9\\ M+9.9=\frac{2}{3}\log E\\ \frac{3\left(M+9.9\right)}{2}=\log E\end{array}$

Convert the logarithmic form into exponential form.

Since the given logarithmic function is in common log form. Thus, the base of the log is 10.

Take exponential in each side with base 10.

  $\begin{array}{c}\log E=\frac{3\left(M+9.9\right)}{2}\\ {10}^{\log E}={10}^{\frac{3\left(M+9.9\right)}{2}}\\ E={10}^{\frac{3\left(M+9.9\right)}{2}}\end{array}$

Hence, the required inverse of the function is $E={10}^{\frac{3\left(M+9.9\right)}{2}}$ .