precalc question

Question

The MMS magnitude M of an earthquake with energy S is given by the formula M=23log(SS0).

Earthquake A had an MMS magnitude of 4.5. The energy released in Earthquake B was 1280 times the energy in Earthquake A. What was the MMS magnitude of Earthquake B? Round your answer to the nearest tenth.

Answer 1

Hello Ashley,

You are probably missing fractions in the MMS equation which should be M = (2/3)* log(S/S0)

Ma = 4.5 = (2/3)*log(Sa/S0)

To find Mb: Mb = (2/3)*log(Sb/S0)

We can replace Sb by 1280*Sa based on the info given in the problem, which allows us to rewrite Mb as:

Mb = (2/3)*log(1280*Sa/S0)

We can use the product rule for logs to expand log(1280*Sa/S0) as follows:

log(1280*Sa/S0) = log(1280) + log(Sa/S0)

Therefore: Mb = (2/3)*[log(1280) + log(Sa/S0)] = (2/3) *log 1280 + (2/3)*log(Sa/S0)

Recall (2/3)*log(Sa/S0) is Ma which the problem tells you is 4.5.

Therefore: Mb = (2/3)*log(1280) + 4.5 = (2/3)*(3.10720996...) + 4.5 = 6.57147331...

Mb = 6.57147331... Now all you have to do is round is to the nearest tenth.