## Algebra 2 (1st Edition)

$\approx92$
$L(I)=10\log\frac{I}{I_0}$ where $I_0=10^{−12}$. The three conversations have a combined intensity of $I=5\cdot3.2\cdot10^{-4}=16\cdot10^{-4}=1.6\cdot10^{-3}$ Hence here $L(I)=10\log\frac{1.6\cdot10^{-3}}{10^{−12}}=10\log{1.6\cdot10^9}\approx92$