$ \dot{x}=-\lambda x(t)\Rightarrow x(t)=e^{-\lambda t}x(0) $ Plutonium: half-life = 80 mil years $ \frac{\cancel{x(0)}}{2}=e^{-8\text{e}7\lambda}\cancel{x(0)}\Longrightarrow\lambda=\frac{-\ln({0.5})}{8\text{e}7} $ Polonium: half-life = 138 days (0.35 years) $ \frac{\cancel{x(0)}}{2}=e^{-0.35\lambda}\cancel{x(0)}\Longrightarrow\lambda=\frac{-\ln({0.5})}{0.35} $