A machine shop has 120 equally noisy machines that together produceand intensity

Question

A machine shop has 120 equally noisy machines that together produceand intensity level of 92 dB. If the intensity level must bereduced to 82 dB, how many machines must be turned off?
first I tried to find the intensity level:
92dB=10 log (120I/10EE-12)    I got : I=1.32EE-4
then I tried to calculate how many machines it would take toproduce 82dB:
82 dB = 10 log (n1.32EE-4/10EE-12) I got : n = 12 machines Then taking 12 from 120 I got that the total number ofmachines to shut off would be 108.
This answer feels awfully big to me but I cannot figure outwhere I went wrong. I would appreciate any help. Thanks!
first I tried to find the intensity level:
92dB=10 log (120I/10EE-12)    I got : I=1.32EE-4
then I tried to calculate how many machines it would take toproduce 82dB:
82 dB = 10 log (n1.32EE-4/10EE-12) I got : n = 12 machines Then taking 12 from 120 I got that the total number ofmachines to shut off would be 108.
This answer feels awfully big to me but I cannot figure outwhere I went wrong. I would appreciate any help. Thanks!

Explanation / Answer

Let the intensity of the one mechine be I Given intensity level of 120 mechines L = 92 dB                             10log 10 ( 120I /Io) = 92 dB                               log 10 ( 120I /Io) = 9.2                                           120 I / Io = 10^9.2                                                         120I = Io * 10 ^ 9.2                                                         = 10^ -12 * 10^ 9.2                                                          =10^-2.8                                                       I= (10^-2.8 )/ 120 Let the intensity level of N mechines be 82 dB then   10 log 10 ( NI /Io) = 82 dB                               log 10 ( 120I /Io) = 8.2                                            NI / Io = 10^8.2                                                         NI = Io * 10 ^ 8.2                                                         = 10^ -12 * 10^ 8.2                                                          =10^-3.8                                                      N= ( 1/ I ) * 10 ^ -3.8                                                         =( 120 / 10^-2.8 ) * 10 ^ -3.8                                                         = 120 * 10 ^ -1                                                          =12 Therefore No.of mechines turned off = 120 -12 = 108                               log 10 ( 120I /Io) = 8.2                                            NI / Io = 10^8.2                                                         NI = Io * 10 ^ 8.2                                                         = 10^ -12 * 10^ 8.2                                                          =10^-3.8                                                      N= ( 1/ I ) * 10 ^ -3.8                                                         =( 120 / 10^-2.8 ) * 10 ^ -3.8                                                         = 120 * 10 ^ -1                                                          =12 Therefore No.of mechines turned off = 120 -12 = 108

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