$M=\frac{2}{3}\log(\frac{S}{S_o})$
Recall the formula for calculating the magnitude of an earthquake,
$M=\frac{2}{3}\log(\frac{S}{S_o})$. One earthquake has magnitude 3.9 on the MMS scale. If a
second earthquake has 750 times as much energy as the first, find the
magnitude of the second quake. Round to the nearest hundredth.
I'm unsure how to approach this problem. I could not find anything about mms scale in the book and am unsure what $S$ and $S_o$ are supposed to be.
I'm guessing that my first step would be:
$3.9=\frac{2}{3}\log(\frac{S}{S_o})$
From here, I really don't know where to go. I could try to represent $S$ and $S_o$ but am unsure if that's even needed, e.g.
$$3.9=\frac{2}{3}\log(\frac{S}{S_o})$$
$$3.9=\frac{2}{3}\log S-\frac{2}{3}\log(S_o)$$
$$\frac{3.9}{\frac{2}{3}}=\log S-\log S_o$$
$$5.85=\log S-\log S_o$$
(From this point I'm pretty lost already and not sure if below or above are even correct or the right path)
$$\log S=5.85+\log S_o$$
$$10^x=S=5.85+\log S_o$$
How can I calculate the magnitude of the second earthquake if I know that it was 750 times more powerful than the one at 3.9 magnitude ?
Best Answer
$\frac{2}{3}log(\frac{S}{S_0})=3.9$
$\frac{2}{3}log(\frac{750 S}{S_0})=\frac{2}{3}log(\frac{S}{S_0}) + \frac{2}{3}log(750)$
$\frac{2}{3}log(\frac{750 S}{S_0})= 3.9 + 1.92 = 5.82$
edit: where $S$ is the power and $S_0$ is the reference power.