College Algebra (10th Edition)

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

Chapter 6 - Section 6.8 - Exponential Growth and Decay Models; Newton's Law: Logistic Growth and Decay Models - 6.8 Assess Your Understanding - Page 486: 9



Work Step by Step

We know that $x=2xe^{k\cdot1690}\\0.5=e^{1690k}\\1690k=\ln0.5\\k=\frac{\ln0.5}{1690}\approx-0.000410146$ Thus $N(50)=10e^{-0.000410146\cdot50}\approx9.797$
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.