The magnitude of the earthquake is related to the released energy E via the equation
[tex]\log _{10}E=4.4+1.5M[/tex]where M is the magnitude of the earthquake. Given that M is equal to 4.6, the equation can be simplified as:
[tex]\begin{gathered} \log _{10}E=4.4+1.5(4.6) \\ \log _{10}E=4.4+6.9 \\ \log _{10}E=11.3 \end{gathered}[/tex]We both raise the two sides of the equation by 10 to eliminate the log on the left-hand side and solve, as follows:
[tex]\begin{gathered} E=10^{11.3} \\ E=1.99526231\times10^{11} \end{gathered}[/tex]Since the question wants us to write the answer in at least 3 decimal points, the final answer would be
[tex]E=1.995\times10^{11}J[/tex]Answer: The energy Thor created is about 1.995 x 10^11 Joules.