Chapter 5 - 5.3 - Properties of Logarithms - 5.3 Exercises - Page 386: 79

$\beta=10(\log I+12)$ $\beta(10^{-6})=60\text{ decibels}$

$\beta=10\log \frac{I}{10^{-12}}$ $~~~=10(\log I-\log 10^{-12})$ $~~~=10\log I-(-12)\log 10$ $~~~=10(\log I+12(1))$ $\beta=10(\log I+12)$ Substitute $I=10^{-6}$ into the equation: $\beta=10(\log 10^{-6}+12)$ $~~~=10(-6\log 10+12)$ $~~~=10(-6+12)$ $~~~=10(6)$ $\beta=60\text{ decibels}$