Answer
(a) $m(t)=22\cdot 2^{-t/1600}$
(b) $m(t)=22\cdot e^{-0.000433t}$
(c) $3.9mg$
(d) $463.4$ years
Work Step by Step
(a) Given $m_0=22, m(1600)=m_0/2$, we have $m_0/2=m_0\cdot 2^{-1600/h}$ so $h=1600$ and the model function becomes $m(t)=22\cdot 2^{-t/1600}$
(b) In this model, $m_0/2=m_0e^{-1600r}$. thus $r=-ln(1/2)/1600=0.000433$ and the
model function becomes $m(t)=22\cdot e^{-0.000433t}$
(c) Let $t=4000$ and use model from (a), we have $m(4000)=22\times2^{-4000/1600}\approx3.9mg$
(d) For this case, $18=22\times2^{-t/1600}$ and we can get $t=-1600\times log_2(18/22)\approx463.4$ years
