1. **Stating the problem:** We have a drug dosage problem where a 20 mg tablet is taken once daily, and the drug amount decreases by 30% each day. The amount left after $n$ days is given by: $$20(0.7)^n$$ We need to find: (i) The amount left after 5 days. (ii) The number of days until less than 1 mg remains. (iii) Explain why the total amount after the 5th tablet is a geometric series. 2. **Formula and rules:** The drug amount after $n$ days is modeled by exponential decay: $$A_n = 20(0.7)^n$$ where $0.7$ is the decay factor (since 30% is lost daily). 3. **(i) Amount after 5 days:** Calculate: $$20(0.7)^5 = 20 \times 0.16807 = 3.3614$$ Rounded to 2 decimal places: $$3.36\text{ mg}$$ 4. **(ii) Days until less than 1 mg remains:** Set up inequality: $$20(0.7)^n < 1$$ Divide both sides by 20: $$\cancel{20}(0.7)^n < \cancel{20} \times \frac{1}{20}$$ $$ (0.7)^n < 0.05$$ Take natural logarithm on both sides: $$\ln((0.7)^n) < \ln(0.05)$$ Use log power rule: $$n \ln(0.7) < \ln(0.05)$$ Divide both sides by $\ln(0.7)$ (negative number, so inequality reverses): $$n > \frac{\ln(0.05)}{\ln(0.7)}$$ Calculate values: $$n > \frac{-2.9957}{-0.3567} = 8.4$$ So, after about 9 days, less than 1 mg remains. 5. **(iii) Explanation of geometric series:** Each day a 20 mg tablet is taken, but the previous doses decay by 30% daily. After 5 tablets, the total amount is the sum of the remaining amounts from each tablet: $$20 + 20(0.7)^1 + 20(0.7)^2 + 20(0.7)^3 + 20(0.7)^4$$ This is a geometric series with first term $a=20$ and common ratio $r=0.7$. **Final answers:** (i) $3.36$ mg (ii) Approximately $9$ days (iii) The total amount is a geometric series because each dose decays by a constant factor $0.7$ each day, so the sum of remaining doses forms a geometric progression.