## Precalculus (6th Edition) Blitzer

a. $10log\frac{I}{I_0}$ b. $20\ db$
a. We can rewrite the formula as $D=10(logI-logI_0)=10log\frac{I}{I_0}$ b. Given $\frac{I_1}{I_2}=100$ where $1$ is for the louder sound and $2$ for the softer sound, we have $D_1-D_2=10log\frac{I_1}{I_0}-10log\frac{I_2}{I_0}=10log\frac{I_1}{I_2}=10log100=20\ db$