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: 30

Answer

$t \approx 6.2$ $hours$

Work Step by Step

$$A = A_{0}e^{kt}$$ Let $A$ represent the percentage of the dosage that remains after $t$ hours; $A_{0}$ is therefore $1$. Since we know that Aspirin's half-life is $12$ hours, we can find the value for $k$ as follows: $$0.5 = \frac{1}{2} = e^{12k}$$ $$\ln \frac{1}{2} = \ln 2^{-1} = -\ln 2 = 12k$$ $$-\frac{\ln 2}{12} = k$$ To find the value of $t$ when $A = 0.7$, then: $$0.7 = e^{-\frac{\ln 2}{12} t} = (e^{\ln 2})^{\frac{-t}{12}} = 2^{\frac{-t}{12}}$$ $$\frac{t}{-12} = \log_{2}0.7$$ $$t = -12\log_{2}0.7 \approx 6.2$$
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.