Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Section 1.4 - Exponential Functions - 1.4 Exercises - Page 54: 31

Answer

a) $25 mg$ b) $y=200(\frac{1}{2})^{\frac{1}{5}t}$ c) $\approx10.88\thinspace mg$ d $\approx 38.22 days$. See graph.

Work Step by Step

a) We start with 200 mg after 0 days. After 5 days, we halve it to 100 mg. After 10 days, we have 50 mg. After 15 days, we have 25 mg. b) We start with 200 mg and halve it, so we have the equation $y=200(\frac{1}{2})^{kt}$. We know $y(5)=100$ and $y(10)=50$ from part (a). $100=200(\frac{1}{2})^{5k}$ $k=\frac{1}{5}$ Our equation is $y=200(\frac{1}{2})^{\frac{1}{5}t}$. c) 2 weeks = 21 days. Plug this into our equation from part (b). $y=200(\frac{1}{2})^{(\frac{1}{5})(21)}\approx 10.88$ d. Graph $y=200(\frac{1}{2})^{\frac{1}{5}t}$ and $y=1$ and see where the graph intersects.
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