Answer
(a) $m(t)=10\cdot 2^{-t/30}$
(b) $m(t)=10\cdot e^{-0.0231t}$
(c) $1.6g$
(d) $70$ years
Work Step by Step
(a) Given $m_0=10,h=30$ (half-life in equation), we have $m(t)=10\cdot 2^{-t/30}$
(b) For this case, $m_0/2=m_0e^{-30r}$, we have $r=ln2/30=0.0231$ and the equation becomes $m(t)=10\cdot e^{-0.0231t}$
(c) Let $t=80$, we have $m(80)=10\times 2^{-80/30}\approx1.6g$
(d) In this case, $2=10e^{-0.0231t}$ and $t=-ln(2/10)/0.0231\approx70$ years
