Answer
(a) $m(t) = (100)~e^{-0.023105~t}$
(b) 9.9 mg
(c) 199 years
Work Step by Step
(a) We can find the value of $k$:
$m(t) = m(0)e^{kt}$
$m(30) = m(0)e^{30k} = \frac{1}{2}m(0)$
$e^{30k} = \frac{1}{2}$
$30k = ln(\frac{1}{2})$
$k = \frac{-ln(2)}{30}$
$k = -0.023105$
Then:
$m(t) = (100)~e^{-0.023105~t}$
(b) We can find $m(100)$:
$m(t) = (100)~e^{-0.023105~t}$
$m(100) = (100)~e^{(-0.023105)~(100)}$
$m(100) = 9.9~mg$
(c) We can find the time $t$ when only 1 mg remains:
$m(t) = (100)~e^{-0.023105~t} = 1$
$e^{-0.023105~t} = 0.01$
$(-0.023105)~t = ln(0.01)$
$t = \frac{ln(0.01)}{-0.023105}$
$t = 199~years$
