1. **State the problem:** We have a substance that decays to 77.8% of its original amount after 300 days. We want to find: (a) The half-life of the substance. (b) The time it takes to decay to one-third of the original amount. 2. **Set up the decay model:** The decay follows the exponential decay formula: $$ A = A_0 e^{-kt} $$ where $A_0$ is the original amount, $A$ is the amount after time $t$, and $k$ is the decay constant. 3. **Use given data to find $k$:** Given $A = 0.778 A_0$ at $t = 300$ days, $$ 0.778 A_0 = A_0 e^{-300k} $$ Divide both sides by $A_0$: $$ 0.778 = e^{-300k} $$ Take natural logarithm: $$ \ln(0.778) = -300k $$ $$ k = -\frac{\ln(0.778)}{300} $$ Calculate: $$ \ln(0.778) \approx -0.2513 $$ $$ k = \frac{0.2513}{300} \approx 0.0008377 $$ 4. **Find the half-life $t_{1/2}$:** Half-life is the time when $A = \frac{1}{2} A_0$: $$ \frac{1}{2} = e^{-kt_{1/2}} $$ Take natural logarithm: $$ \ln\left(\frac{1}{2}\right) = -kt_{1/2} $$ $$ t_{1/2} = -\frac{\ln(1/2)}{k} = \frac{\ln(2)}{k} $$ Calculate: $$ \ln(2) \approx 0.6931 $$ $$ t_{1/2} = \frac{0.6931}{0.0008377} \approx 827.56 \text{ days} $$ 5. **Find time to decay to one-third:** Set $A = \frac{1}{3} A_0$: $$ \frac{1}{3} = e^{-kt} $$ Take natural logarithm: $$ \ln\left(\frac{1}{3}\right) = -kt $$ $$ t = -\frac{\ln(1/3)}{k} = \frac{\ln(3)}{k} $$ Calculate: $$ \ln(3) \approx 1.0986 $$ $$ t = \frac{1.0986}{0.0008377} \approx 1311.68 \text{ days} $$ **Final answers:** (a) Half-life $\approx 827.56$ days (b) Time to decay to one-third $\approx 1311.68$ days