Algebra 2 (1st Edition)

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

Chapter 7 Exponential and Logarithmic Functions - Standardized Test Practice - Extended Response - Page 547: 22b


See below

Work Step by Step

Given: $P=14.7e^{-0.0004h}$ Solving $h$ for $P=12$, we have: $12=14.7e^{-0.0004h}\\e^{-0.0004h=\frac{12}{14.7}}\\\ln e^{-0.00044h}=\ln\frac{12}{14.7}\\-0.0004h=\ln\frac{12}{14.7}\\h\approx507.35 ft$
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.