(a) If [tex]y(t)[/tex] is the mass (in mg) remaining after [tex]t[/tex] years, then [tex]y(t) = y(0) e^{kt}=100e^{kt}.[/tex]
[tex]y(30) = 100 e^{30k} = \frac{1}{2}(100) \implies e^{30k} = \frac{1}{2} \implies k = -(\ln 2) /30 \implies \\ \\
y(t) = 100e^{-(\ln 2)t/30} = 100 \cdot 2^{-t/30}[/tex]
(b) [tex]y(100) = 100 \cdot 2^{-100/30} \approx \text{9.92 mg} [/tex]
(c)
[tex]100 e^{- (\ln 2)t/30} = 1\ \implies\ -(\ln 2) t / 30 = \ln \frac{1}{100}\ \implies\ \\ \\
t = -30 \frac{\ln 0.01}{\ln 2} \approx \text{199.3 years}[/tex]
