Answer
(a) $7.1~mg~~$ remains after 20 years.
(b) $1~mg~~$ remains after 34.8 years
Work Step by Step
(a) We can find the value of $k$:
$m(t) = m(0)e^{kt}$
$m(5.24) = m(0)e^{5.24k} = 0.5~m(0)$
$e^{5.24k} = 0.5$
$5.24k = ln(0.5)$
$k = \frac{ln(0.5)}{5.24}$
$k = -0.13228$
Then:
$m(t) = m(0)~e^{-0.13228~t}$
We can find the mass after 20 years:
$m(t) = m(0)~e^{-0.13228~t}$
$m(20) = (100)~e^{(-0.13228)(20)}$
$m(20) = 7.1$
$7.1~mg~~$ remains after 20 years.
(b) We can find the time $t$ when only 1 mg remains:
$m(t) = m(0)~e^{-0.13228~t}$
$m(t) = (100)~e^{-0.13228~t} = 1$
$e^{-0.13228~t} = 0.01$
$-0.13228~t = ln(0.01)$
$t = \frac{ln(0.01)}{-0.13228}$
$t = 34.8$
$1~mg~~$ remains after 34.8 years
