Please help with Physics: What is the sound intensity level 37 m from the speaker?

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Nichole

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A rock band playing an outdoor concert produces sound at 130 decibels 6.0 m away from their single working loudspeaker.

Can you explain how you do it please? Thanks in advance!
 
decibel> dB
1 dB = 10 ln[I/Io] = 10 ln I - 10 ln Io
I = intensity of source (W/m^2)
Io = reference intensity for human ear (10^-12 W/m^2)
==================
I = k/r^2 [intensity decreases with increasing distance)
dB = 10 ln[k/r^2] - 10 ln Io
dB = 10 ln[k] - 20 ln (r) - 10 ln Io -------------------- (A)
---------------------------- we have 2 cases
130 = 10 ln[k] - 20 ln (6) - 10 ln Io
x = 10 ln[k] - 20 ln (37) - 10 ln Io
---------------------------------------- subtract
130 - x = 20 [ln37] - 20 ln[6] = 20 ln[37/6] = 36.38
x = 130 - 36.38
x = 93.6 decibels
 
Sound intensity difference in db is 10*log(I1/I2). Since intensity varies inversely as the square of the distance, I1/I2 = (6/37)²

10*log(6/37)² = 20*log(6/37) = -15.8 db

The intensity at 37 m is then 130 - 15.8 = 114.2 db
 
Sound intensity difference in db is 10*log(I1/I2). Since intensity varies inversely as the square of the distance, I1/I2 = (6/37)²

10*log(6/37)² = 20*log(6/37) = -15.8 db

The intensity at 37 m is then 130 - 15.8 = 114.2 db
 
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