If an earthquake has magnitude 3.9 with mms formula $M=\frac{2}{3}\log(\frac{S}{S_o})$ then what is the magnitude of an earthquake 750 larger? $M=\frac{2}{3}\log(\frac{S}{S_o})$
From my online textbook:

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 ?
 A: $\frac{2}{3}log(\frac{S}{S_0})=3.9$
$\frac{2}{3}log(\frac{750 S}{S_0})=\frac{2}{3}log(\frac{S}{S_0}) + \frac{2}{3}log(750)$
$\frac{2}{3}log(\frac{750 S}{S_0})= 3.9 + 1.92 = 5.82$
edit: where $S$ is the power and $S_0$ is the reference power.
A: First you have to find the value of energy in the first earthquake.
3.9 = 2/3 * log ( S1 / S0 )
Note, 3.9 is the M value, magnitude, in the original formula provided by the textbook.  We need to find the energy value, then later we will manipulate it to provide us with the answer of the question.
S1 is what I have replaced S with to make it clearer, this is the amount of energy to produce the magnitude, M, or in this case, 3.9.
Isolate the log by itself on the right, by dividing everything by 2/3
3.9 / (2/3) =  ( (2/3) * log ( S1 / S0 ) ) / (2/3)
The right cancels out 2/3 on the top and bottom, equals log ( S1 / S0 ) ).  The left is calculated to be 5.85.
5.85 = log (S1 / S0)
Now rewrite this log into power form:
10^5.85 = (S1 / S0)
Then isolate S1 by multiplying both sides by S0. (Remember, S0 is just a constant, it is not the value of the energy).
(10^5.85) * S0  = S1
Now we finally have a value for S1, or as it is called, S, in the original equation.
This value of energy, times 750, is the magnitude we want to find for the solution.
M  = 2/3 * log (  750 * ( (10^5.85) * S0 ) / S0 )
Cancel the S0 on top and bottom:
M  = 2/3 * log (  (10^5.85) * 750 )
Now evaluate using a calculator: 5.81670750893.  Rounding to nearest hundredth is 5.82.
Sorry, I don't know how to format it properly but I hope this helps anyone else who stumbled upon this question looking for the solution!
