Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.6 Applications and Models of Exponential Growth and Decay - 4.6 Exercises - Page 479: 14

$ k=\frac{1}{4}ln(\frac{1}{9})$

Step 1. State the model equation: $y(t)=y_0e^{kt}$ Step 2. For $y_0=8.1\ kg$ and $y(4)=0.9$, we have $0.9=8.1e^{4k}$ Step 3. Solve the above: $e^{4k}=\frac{1}{9}\longrightarrow 4k=ln(\frac{1}{9}) \longrightarrow k=\frac{1}{4}ln(\frac{1}{9})$