1. **Problem statement:** Radon-222 decays exponentially with a half-life of 4 days. 2. **Formula for exponential decay:** $$ N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T}} $$ where $N(t)$ is the amount remaining at time $t$, $N_0$ is the initial amount, and $T$ is the half-life. ### Part (a): Find initial amount $N_0$ given $N(11) = 1.34$ grams 3. Substitute known values: $$ 1.34 = N_0 \left(\frac{1}{2}\right)^{\frac{11}{4}} $$ 4. Solve for $N_0$: $$ N_0 = \frac{1.34}{\left(\frac{1}{2}\right)^{\frac{11}{4}}} $$ 5. Simplify the denominator: $$ \left(\frac{1}{2}\right)^{\frac{11}{4}} = 2^{-\frac{11}{4}} $$ 6. Rewrite $N_0$: $$ N_0 = 1.34 \times 2^{\frac{11}{4}} $$ 7. Calculate $2^{\frac{11}{4}} = 2^{2.75} = 2^{2+0.75} = 2^2 \times 2^{0.75} = 4 \times 2^{0.75} \approx 4 \times 1.68179 = 6.727 $$ 8. Multiply: $$ N_0 \approx 1.34 \times 6.727 = 9.012 $$ 9. **Answer (a):** The initial sample was approximately **9.012 grams**. ### Part (b): Find time $t$ when $N(t) = 0.01$ grams, given $N_0 = 50$ grams 10. Use decay formula: $$ 0.01 = 50 \left(\frac{1}{2}\right)^{\frac{t}{4}} $$ 11. Divide both sides by 50: $$ \frac{0.01}{50} = \left(\frac{1}{2}\right)^{\frac{t}{4}} $$ 12. Simplify fraction: $$ 0.0002 = \left(\frac{1}{2}\right)^{\frac{t}{4}} $$ 13. Take logarithm base $\frac{1}{2}$ of both sides: $$ \log_{\frac{1}{2}}(0.0002) = \frac{t}{4} $$ 14. Solve for $t$: $$ t = 4 \log_{\frac{1}{2}}(0.0002) $$ 15. **Answer (b):** The time until only 0.01 grams remain is $$ t = 4 \log_{\frac{1}{2}}(0.0002) $$ This is the simplified logarithmic expression for the time.