College Algebra (10th Edition)

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

Chapter 6 - Section 6.4 - Logarithmic Functions - 6.4 Assess Your Understanding - Page 451: 121


See below.

Work Step by Step

a) Our equation is: $320=760e^{-0.145h}\\\frac{320}{760}=\frac{8}{19}=e^{-0.145h}\\\ln{\frac{8}{19}}=-0.145h\\h=\ln{\frac{8}{19}}\frac{1}{-0.145}\approx5.97$ b) Our equation is: $667=760e^{-0.145h}\\\frac{667}{760}=e^{-0.145h}\\\ln{\frac{667}{760}}=-0.145h\\h=\ln{\frac{667}{760}}\frac{1}{-0.145}\approx0.9$
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