1. **State the problem:** We have viral load data $V$ measured at different times $t$ (days) after treatment started. We want to model $V$ as a function of $t$, find the half-life (time to decrease by 50%), and predict $V$ at $t=20$ days. 2. **Modeling viral load with exponential decay:** The viral load decreases over time, suggesting an exponential decay model: $$V(t) = V_0 e^{-kt}$$ where $V_0$ is the initial viral load and $k$ is the decay constant. 3. **Find $V_0$ and $k$ using data:** Using the first data point at $t=1$, $V(1) = 76.0$, so approximately $V_0 e^{-k} = 76.0$. Take natural logarithm of viral load values to linearize: $$\ln V = \ln V_0 - kt$$ Calculate $\ln V$ for each $t$: - $t=1$, $\ln 76.0 \approx 4.3307$ - $t=4$, $\ln 41.8 \approx 3.7320$ - $t=8$, $\ln 22.99 \approx 3.1355$ - $t=11$, $\ln 12.6445 \approx 2.5373$ - $t=15$, $\ln 6.95448 \approx 1.9403$ 4. **Fit a line $y = mx + b$ to points $(t, \ln V)$:** Using two points for slope $m = -k$: Between $t=1$ and $t=15$: $$m = \frac{1.9403 - 4.3307}{15 - 1} = \frac{-2.3904}{14} \approx -0.1707$$ Intercept $b = \ln V_0$ can be found using $t=1$: $$4.3307 = b - 0.1707 \times 1 \Rightarrow b = 4.3307 + 0.1707 = 4.5014$$ 5. **Write the model:** $$\ln V = 4.5014 - 0.1707 t \Rightarrow V(t) = e^{4.5014} e^{-0.1707 t}$$ Calculate $e^{4.5014} \approx 90.2$, so $$V(t) = 90.2 e^{-0.1707 t}$$ 6. **Find half-life $t_{1/2}$:** Time when $V(t)$ is half of $V_0$: $$\frac{V_0}{2} = V_0 e^{-k t_{1/2}} \Rightarrow \frac{1}{2} = e^{-k t_{1/2}}$$ Take natural log: $$\ln \frac{1}{2} = -k t_{1/2} \Rightarrow t_{1/2} = \frac{\ln 2}{k} = \frac{0.6931}{0.1707} \approx 4.06 \text{ days}$$ 7. **Predict viral load at $t=20$ days:** $$V(20) = 90.2 e^{-0.1707 \times 20} = 90.2 e^{-3.414} \approx 90.2 \times 0.0329 = 2.97$$ **Final answers:** - Model: $$V(t) = 90.2 e^{-0.1707 t}$$ - Half-life: approximately 4.06 days - Viral load at 20 days: approximately 2.97 RNA copies/mL