Let's solve the problem step-by-step to determine the magnitude [tex]\( M \)[/tex] of an earthquake releasing [tex]\( 2.5 \cdot 10^{15} \)[/tex] joules of energy using the given formula:
[tex]\[ M = \frac{2}{3} \log \frac{E}{E_0} \][/tex]
where [tex]\( E \)[/tex] is the energy released by the earthquake in joules and [tex]\( E_0 = 10^{4.4} \)[/tex] is the assigned minimal measure released by an earthquake.
1. Identify [tex]\( E \)[/tex] and [tex]\( E_0 \)[/tex]:
[tex]\[
E = 2.5 \cdot 10^{15} \, \text{joules}
\][/tex]
[tex]\[
E_0 = 10^{4.4}
\][/tex]
2. Calculate the argument of the logarithm ([tex]\( \frac{E}{E_0} \)[/tex]):
[tex]\[
\frac{E}{E_0} = \frac{2.5 \cdot 10^{15}}{10^{4.4}}
\][/tex]
By computing this, we get:
[tex]\[
\frac{2.5 \cdot 10^{15}}{10^{4.4}} \approx 99526792638.37422
\][/tex]
3. Calculate the logarithm base 10 of the argument:
[tex]\[
\log \left( 99526792638.37422 \right) \approx 10.997940008672037
\][/tex]
4. Calculate the magnitude [tex]\( M \)[/tex]:
[tex]\[
M = \frac{2}{3} \log \left( 99526792638.37422 \right)
\][/tex]
Substituting the value of the logarithm:
[tex]\[
M \approx \frac{2}{3} \times 10.997940008672037
\][/tex]
Simplifying this, we get:
[tex]\[
M \approx 7.331960005781358
\][/tex]
Finally, observing the options provided in the given question:
[tex]\[
\begin{array}{l}
2.5 \cdot 10^{15}=\frac{2}{3} \log \frac{E}{10^{4.4}} \\
10^{4.4}=\frac{2}{3} \log \frac{E}{2.5 \cdot 10^{15}} \\
M=\frac{2}{3} \log \left( 9.95 \cdot 10^9 \right)
\end{array}
\][/tex]
We see that the valid step in the process that matches our detailed calculation is:
[tex]\[
M=\frac{2}{3} \log \left( 9.95 \cdot 10^9 \right)
\][/tex]
Hence, the correct answer is:
[tex]\[
M = \frac{2}{3} \log \left( 9.95 \cdot 10^{9} \right)
\][/tex]
