College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 6 - Review Exercises - Page 500: 47

Answer

See below.

Work Step by Step

a) By plugging in $d=3.5$ into the formula: $L=9+5.1\log{3.5}\approx11.77$ b) By plugging in $L=14$ into the formula: $14=9+5.1\log{d}\\5=5.1\log d\\\frac{5}{5.1}=\log d\\d=10^{\frac{5}{5.1}}\approx9.77$
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