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A sound's noise level is measured in decibels (dB) by comparing the sound's intensity, $I$ , to a benchmark sound that has an intensity of $10^{-16}$ watts/cm $^2$ . In particular, $\text{dB}=10\log(I/10^{-16})=10\log(10^{16}I).$

a) The intensity of Joel's whisper is approximately $10^{-13}$ watts/cm $^{2}$ . Compute the noise level of the whisper in decibels.

b) A certain sound with a noise level of 96 dB has been known to damage hearing. Compute the sound's intensity at this noise level.

watts/cm $^2$

c) Kansas City Chiefs' fans recently broke the world record for "loudest roar" at a sports event. The fans reached a noise level of 142.2 dB. The previous world record was held by Seattle Seahawks' fans, with a noise level of 137.6 dB. About many times more intense was the roar of the Chiefs' fans than the roar of the Seahawks' fans?

Hint

a) We are given an equation that uses intensity as an input, and we are given an intensity. So, we can evaluate the equation at the given intensity of Joel's whisper to find out how many decibels the noise level of his whisper was.

b) Here, we want to solve for $I$ in the decibel equation, given the decibels of a certain noise level. That is, we want to solve for $I$ in $96=10\log(10^{16}I)$

c) We calculate this by first finding the intensity of the noise levels of both fan groups. This can be done by following the steps in part (b) but using the decibels given in this part, solving for $I_1$ in $137.6=10\log(10^{16}I_1)$ and for $I_2$ in $142.2=10\log(10^{16}I_2)$ Once we have these, we can find how many times more intense the Chiefs' fans' noise was than the Seahawks' fans' noise by dividing the Chiefs' fans' intensity by the Seahawks' fans' intensity. This will tell us what we would need to multiply the Seahawks' fans' intensity by to get the Chiefs' fans' intensity, which is another way of saying how many times more intense the Chiefs' fans' noise was than the Seahawks' fans' noise.