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

Answer

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: $667=760e^{-0.145h}\\\frac{667}{760}=e^{-0.145h}\\\ln{\frac{667}{760}}=-0.145h\\h=\ln{\frac{667}{760}}\frac{1}{-0.145}\approx0.9$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.