Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.6 Applications and Models of Exponential Growth and Decay - 4.6 Exercises - Page 479: 11

$ k=\frac{1}{100}ln(\frac{1}{2})$

Step 1. State the model equation: $y(t)=y_0e^{kt}$ Step 2. For $y_0=10\ mg$ and $y(100)=10/2=5$, we have $5=10e^{100k}$ Step 3. Solve the above: $e^{100k}=\frac{1}{2}\longrightarrow 100k=ln(\frac{1}{2}) \longrightarrow k=\frac{1}{100}ln(\frac{1}{2})$