1 decibel = A*log(P/P0) = A*log(1.122)
Where P0 is the initial pressure, P is the pressure
1 decibel = 0.05*A
A = 20
L = 20*log(P/P0)
8 = 20*log(P/P0)
Solve for P/P0
Taking a decibel to be an increase in pressure of 12.2%, by what factor has the pressure changed when the sound level has fallen by 8.00 decibels?
2 answers
Jennifer's answer is correct, but you have to use -8