中四數學!!!

(a)
E=10^(11.4+1.5M)
E=10^[11.4+1.5(7.3)]
E=2*10^22J (correct to 1 significant figure)
(b)
E'=10^[11.4+1.5(M+1)]
E'=10^[(11.4+1.5M)+1.5]
E'=[10^(11.4+1.5M)]*10^(1.5)
Therefore,
E'/E
={[10^(11.4+1.5M)]*10^(1.5)}/E=10^(11.4+1.5M)
=10^(1.5)
=32 times (correct to 2 significant figure)

1st of all, we have to clarify what log E means
1. when we write log E, we usually mean log to the base 10
2. if we wish to express log to the base e, we'll write ln E
3. however, most of the scientific equations are using log to the base e. Seldom using base 10

Therefore, your question may have 2 answers

Case 1 (log to the base 10)
log E = 1 1.4+1.5 x 7.3
log E = 22.35
E = 10 to the power 22.35
E = 10 ^ 22.35
E = 2 x 10^22 J

Case 2 (log to the base e)
log E = 1 1.4+1.5 x 7.3
log E = 22.35
E = e to the power 22.35
E = e ^ 22.35
E = 5,087 x 10^6 J
E = 5 GJ


2007-12-30 01:19:55 補充：
let energy be E(0) when magnitude M; andenergy be E(1) when magnitude M+1Thenlog E(0) = 11.4 + 1.5ME(0) = e^11.4 x e^1.5Mlog E(1) = 11.4 + 1.5 (M+1) = 12.9 + 1.5ME(1) = e^12.9 x e^1.5MSo, E(1) / E(0) = e^(12.9 - 11.4) = e^1.5= 4.5