Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 7 Exponential and Logarithmic Functions - 7.5 Apply Properties of Logarithms - 7.5 Exercises - Problem Solving - Page 512: 70



Work Step by Step

$L(I)=10\log\frac{I}{I_0}$ 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 $L(I)=10\log\frac{1.6\cdot10^{-3}}{10^{−12}}=10\log{1.6\cdot10^9}\approx92$
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