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 $t=\frac{ln(0.6)}{-0.0578}\approx8.8\ hrs$
