1. **Problem Statement:** We are given the function describing the amount of medication in Carlos's bloodstream over time: $$M(t) = 20 \cdot e^{-0.8t}$$. We want to find the time $$t$$ in hours when $$M(t) = 1$$ mg. 2. **Set up the equation:** Set $$M(t) = 1$$ and solve for $$t$$: $$1 = 20 \cdot e^{-0.8t}$$ 3. **Isolate the exponential term:** Divide both sides by 20: $$\frac{1}{20} = e^{-0.8t}$$ 4. **Take the natural logarithm:** Apply the natural logarithm $$\ln$$ to both sides: $$\ln\left(\frac{1}{20}\right) = \ln\left(e^{-0.8t}\right)$$ 5. **Simplify using logarithm properties:** $$\ln\left(\frac{1}{20}\right) = -0.8t$$ 6. **Solve for $$t$$:** $$t = -\frac{1}{0.8} \cdot \ln\left(\frac{1}{20}\right)$$ Since $$\ln\left(\frac{1}{20}\right) = -\ln(20)$$, then $$t = -\frac{1}{0.8} \cdot (-\ln(20)) = \frac{\ln(20)}{0.8}$$ 7. **Calculate the numerical value:** Using $$\ln(20) \approx 2.995732$$, $$t \approx \frac{2.995732}{0.8} \approx 3.744665$$ 8. **Round the answer:** Rounded to the nearest hundredth, $$t \approx 3.74$$ hours. **Final answer:** Carlos will have approximately 3.74 hours before the medication amount decays to 1 mg.