The decibel rating D is related to the sound intensity I by the formula 0=10 log

Question

The decibel rating D is related to the sound intensity I by the formula 0=10 log_10 (I-10^-16) for the noise level in decibels. Let D and d represent the decibel ratings of sounds of intensity I and i. respectively. Using properties of logarithms, find a simplified formula for the difference between the two ratings. D - d, in terms of the two intensities I and i. D - d = help (logarithms) (Enter log 10 or log ten for the base 10 logarithms.) It a sound's intensity triples, how many decibels does the sound become? decibels help (numbers)

Explanation / Answer

D=10log10(I/10-16)

d=10log10(i/10-16)

a)D-d=10log10(I/10-16)-10log10(i/10-16)

D-d=10(log10(I/10-16)-log10(i/10-16))

log(m)-log(n)=log(m/n)

D-d=10log10((I/10-16)/(i/10-16))

D-d=10log10(I/i)

b)sound intensity triples=>I=3i

D-d=10log10(3i/i)

D-d=10log103

D-d=4.77 decibles

sound becomes louder by 4.77 decibles

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