Answer
a) $y(t)$ = $100\left(2^{-\frac{t}{30}}\right)$
b) $\approx$ $9.92$ mg
c) $\approx$ $199.3$ years
Work Step by Step
a)
If $y(t)$ is the mass (in mg) remaining after $t$ years then
$y(t)$ = $y(0)e^{kt}$ = $100e^{kt}$
$y(30)$ = $100e^{30k}$ = $\frac{1}{2}(100)$
$e^{30k}$ = $\frac{1}{2}$
$k$ = $-\frac{\ln2}{30}$
$y(t)$ = $100e^{-\frac{(\ln2)t}{30}}$
$y(t)$ = $100\left(2^{-\frac{t}{30}}\right)$
b)
$y(100)$ = $100\left(2^{-\frac{100}{30}}\right)$ $\approx$ $9.92$ mg
c)
$100\left(2^{-\frac{t}{30}}\right)$ = $1$
$-\frac{(\ln2)t}{30}$ = $\ln\left(\frac{1}{100}\right)$
$t$ $\approx$ $199.3$ years