Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.4* General Logarithmic and Exponential Functions - 6.4* Exercise - Page 464: 58



Work Step by Step

Loudness of a sound with intensity $I, L=10log_{10}(\frac{I}{I_{0}})$ Consider intensity of the rock music =$ I$ And intensity of the mower = $ I'$ As per the given problem, we have $10log_{10}(\frac{I}{I_{0}})=120$ dB $log_{10}(\frac{I}{I_{0}})=12$ dB or $I=(10)^{12}{I_{0}}$ ...(1) Amplified rock music is measured at 120 dB, whereas the noise from a motor-driven lawn mower is measured at 106 dB. $10log_{10}(\frac{I'}{I_{0}})=106$ dB $log_{10}(\frac{I'}{I_{0}})=10.6$ dB or $I=(10)^{10.6}{I_{0}}$ ...(2) From equations 1 and 2, we get $\frac{I}{I'}=\frac{(10)^{12}}{(10)^{10.6}}$ This implies, $I:I'=10^{1.4}:1$ Hence, the ratio of the intensity of the rock music to that of the mower is $I:I'=10^{1.4}:1$.
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