1. Statement of the problem. Suppose a sample of a substance decayed to 77.8% of its original amount after 300 days, and we are asked to find the half-life and the time to reach one-third of the original amount. 2. Set up the exponential decay model. We use the continuous exponential model $A(t)=A_0 e^{kt}$. After 300 days the fraction remaining is $0.778$, so $\frac{A(300)}{A_0}=0.778=e^{300k}$. 3. Solve for the decay constant $k$. Take natural logarithms of both sides to obtain $300k=\ln(0.778)$. Therefore $k=\frac{\ln(0.778)}{300}$. Evaluate numerically: $\ln(0.778)\approx -0.2513144282809063$. So $k\approx\frac{-0.2513144282809063}{300}\approx -0.0008377147609363543$. 4. (a) Find the half-life. By definition $0.5=e^{kT_{1/2}}$ so $T_{1/2}=\frac{\ln(0.5)}{k}$. Substitute the value of $k$ to get $T_{1/2}\approx\frac{\ln(0.5)}{-0.0008377147609363543}$. Compute numerically to two decimal places to obtain $T_{1/2}\approx 827.68$ days. 5. (b) Time to decay to one-third of the original amount. Set $\frac{A(t)}{A_0}=\frac{1}{3}=e^{kt}$ so $t=\frac{\ln(1/3)}{k}$. Using the computed $k$ gives $t\approx\frac{\ln(1/3)}{-0.0008377147609363543}$. Note that $\frac{\ln(1/3)}{\ln(1/2)}=\frac{\ln 3}{\ln 2}\approx 1.584962500721156$ so $t\approx 1.584962500721156\times T_{1/2}$. Therefore $t\approx 1311.84$ days. Final answers. (a) Half-life $\approx 827.68$ days. (b) Time to one-third $\approx 1311.84$ days.