The Intensity Level of a sound of intensity I, in dB, is defined as follows:
IL \equiv (10 dB) \log_{10}(I/I_0) ...... (1)
where I_0 is the intensity at the threshold of hearing.
Comparing the intensity levels of two different sounds of intensities I_1 and I_2,
IL_2 - IL_1 = (10 dB)[\log_{10}(I_2/I_0) - \log_{10}(I_1/I_0)]
But, \qquad \log_{10}(a) - \log_{10}(b) = \log_{10}(a/b)
IL_2 - IL_1 = (10 dB)\log_{10}(I_2/I_1)
IL_1 = IL_2 - (10 dB)\log_{10}(I_2/I_1)
The intensity of sound wave produced by two identical fire crackers is twice the intensity of one : I_2 = 2I_1; \qquad I_2/I_1 = 2
IL_2 - IL_1 = (10 dB)\log_{10}(2) = 3.01 dB
IL_1 = IL_2 - 3.01 dB = 85 dB - 3.01 dB = 81.99 dB
