Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.6 Applications and Models of Exponential Growth and Decay - 4.6 Exercises - Page 479: 13

$ k=\frac{1}{2}ln(\frac{1}{4})$

Step 1. State the model equation: $y(t)=y_0e^{kt}$ Step 2. For $y_0=2.4\ lb$ and $y(2)=0.6$, we have $0.6=2.4e^{2k}$ Step 3. Solve the above: $e^{2k}=\frac{1}{4}\longrightarrow 2k=ln(\frac{1}{4}) \longrightarrow k=\frac{1}{2}ln(\frac{1}{4})$