數學__Indices and Iogarithms

Let Ia is the intensity of b1 and Ib is the intensity of b0.
so 105 = 10 log (Ia/I0) or 10.5 = log (Ia/I0) ...........(1) and
80 = 10 log (Ib/I0) or 8 = log (Ib/I0)...........(2)
(1) - (2) we get
2.5 = log (Ia/I0) - log (Ib/I0)
2.5 = log [(Ia/I0)/(Ib/I0)]
2.5 = log (Ia/Ib)
Ia/Ib = 10 ^ 2.5 = 316.2



2009-04-04 12:27:41 補充：
Correction: 1st line should be Ib is the intensity of b2, not b0.

b1 = 105 = 10log(I1 / I0)...(1)
b2 = 80 = 10log(I2 / I0)...(2)
(1)-(2):
105 - 80 = 10[log(I1 / I0) - log(I2 / I0)]
25/10 = log (I1 / I2)
I1 / I2 = 10 ^ 2.5 = 100√10 = 316.227766...
So b1 = 105dB is 316.227766...times greater than b2 = 85dB