## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 9 - Section 9.2 - The Hyperbola - Exercise Set - Page 984: 94

#### Answer

$8.8\ hrs$

#### Work Step by Step

Step 1. Use the given model equation $A=A_0e^{kt}$. At $t=12\ hrs$, $A=\frac{A_0}{2}$ and thus $e^{12k}=\frac{1}{2}$ Step 2. Solving for $k$, we have $k=\frac{ln(1/2)}{12}\approx-0.0578$ Step 3. For a decay to $60\%$, we have $A=0.6A_0$ and $e^{-0.0578t}=0.6$. Thus $t=\frac{ln(0.6)}{-0.0578}\approx8.8\ hrs$

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