Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Review - Exercises - Page 390: 108

Answer

$10^{10.4}$

Work Step by Step

Here, we have $B_1=132 =10 \log \dfrac{I_1}{I_0}$ This gives: $I_1=I_0 10^{13.2}$ ...(1) Now, $B_2=28 =10 \log \dfrac{I_2}{I_0}$ This gives: $I_2=I_0 10^{2.8}$ ...(2) From equations (1) and (2), we have $R=\dfrac{I_2}{I_1}=10^{13.2-2.8}=10^{10.4}$
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