College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.5 - Page 505: 31

Answer

about 5.5 hours

Work Step by Step

The decay model: $A=A_{0}e^{kt}$ ($ k < 0 )$ We find k from the half life information $0.5A_{0}=A_{0}e^{k\cdot 36}\qquad/\div A_{0}$ $0.5=e^{36k}\qquad $ ... apply ln( ) to both sides $\ln 0.5=36k$ $ k=\displaystyle \frac{\ln 0.5}{36}\approx$-0.0192540883489$\approx-0.0193$ The decay model is : $A=A_{0}e^{-0.0193t}$ We want t (in hours) when $A=0.90A_{0}$ $0.9A_{0}=A_{0}e^{-0.0193t}\qquad/\div A_{0}$ $0.9=e^{-0.0193t}\qquad $ ... apply ln( ) to both sides $\ln 0.9=-0.0193t\qquad \div(-0.0193)$ $ t=\displaystyle \frac{\ln 0.9}{-0.0193}\approx$5.459094075548$\approx 5.5$ hours
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.