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#