The sound level at a point is increased by 30 dB. What is factor is the pressure amplitude increased?

Explanation

The sound level in dB is
`L = (10 log_(10))(I)/(I_(0))`
if `L_(1)` and `L_(2)` are the sound levels and `I_(1)` and `I_(2)` are the corresponding intensities in the two cases,
`L_(2) - L_(1) = 10[(log_(10)) (I_(2))/(I_(0)) - (log_10) (I_(1))/(I_(0))]`
`30 = (10log)(I_(2))/(I_(0))`
`(I_(2))/(I_(1)) = 10^(3)`
As the intensity is proportional to the square of the pressure amplitude, thus we have
`(DeltaP_(2))/(DeltaP_(1)) = sqrt((I_(2))/(I_(1))) = sqrt(1000) approx 32`