Sound intensity logarithm problem?

Question

The loudness of a sound L, is measured in dB (decibels) and defined as follows:

L=10log I/Io ,

where "Io" is the threshold of hearing and equal to 1 x 10^-12

and l is the sound intensity.

"The jackhammer noise is 1.5 thousand million times as intense as the softest sound."

Sound: and loudness in dB

Jackhammer : 90dB

Heavy Traffic: 75 dB

Conversational speech: 60 dB

Quiet living room: 20dB

1) the threshold of pain for hearing is 135dB. How many times as loud as a jackhammer is the pain threshold?

2) Compare the intensity of a the sound of conversation to that of heavy traffic?

3) How many times is the sound of a quiet living room as loud as that of the threshold for hearing?

Please help??, I'm having trouble. Please include working

Answer 1

1) the threshold of pain for hearing is 135dB. How many times as loud as a jackhammer is the pain threshold?

Sound: and loudness in dB

Jackhammer : 90dB

threshold of pain is 1.5 times as loud=135/90

2) Compare the intensity of a the sound of conversation to that of heavy traffic?

Heavy Traffic: 75 dB

75=10*log(I/10^-12)

7.5=log(I/10^-12)

10^7.5=I/10^-12

10^(7.5-12)=I

I=10^-4.5

Conversational speech: 60 dB

in a similar manner

I=10^-6

The ratio of the two intensities is 10^1.5

that means that the intensity of the sound from heavy traffic is a little more than 31 times the intensity of the sound of conversational speech.

3) How many times is the sound of a quiet living room as loud as that of the threshold for hearing?

Quiet living room: 20dB

Threshold of hearing intensity is 10^-12

loudness would thus be L=10*log(10^12/10^-12)=10

Thus the noise in a quiet living room is twice as loud as the noise that is just at the threshold of hearing.