Answer
a. $D=10\displaystyle \log\frac{I}{I_{o}}$
b. 20 decibels
Work Step by Step
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, $10(\displaystyle \log I-\log I_{o})=10\log\frac{I}{I_{o}}$
$D=10\displaystyle \log\frac{I}{I_{o}}$
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$b.\displaystyle \quad D_{1}=10\log(\frac{100I}{I_{0}}) \qquad$... 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.
