## College Algebra (6th Edition)

a. $D=10\displaystyle \log\frac{I}{I_{o}}$ b. 20 decibels
a. A difference of logarithms appears in the Quotient Rule: $\displaystyle \log_{\mathrm{b}}(\frac{\mathrm{M}}{\mathrm{N}})=\log_{\mathrm{b}}\mathrm{M}-\log_{\mathrm{b}}\mathrm{N}$ so, $10(\displaystyle \log I-\log I_{o})=10\log\frac{I}{I_{o}}$ $D=10\displaystyle \log\frac{I}{I_{o}}$ ----------- $b.\displaystyle \quad D_{1}=10\log(\frac{100I}{I_{0}}) \qquad$... apply quotient rule, $=10[\log(100I)-\log I_{0}) \qquad$... apply product rule, $=10[\log 100+\log I-\log I_{0}] \qquad$. .. apply $\log 100=\log_{10}10^{2}=2$ $=10(2)+10(\log I-\log I_{0}) \qquad$ ... the second term equals D $=20+D$ This is the loudness level of the softer sound, increased by 20 (decibels). This means that the 100 times louder sound will be larger (louder ) by 20 decibels.