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?

Answer 1

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

Answer 2

Jennifer's answer is correct, but you have to use -8