Answer
after about $12.6$ days
Work Step by Step
Solve for t when $m(t)=5$.
$15e^{-0.087t}=5 \qquad$ ... $/\div 15$
$e^{-0.087t}=\displaystyle \frac{1}{3} \qquad$ ... apply $\ln$() to both sides
$-0.087t=\displaystyle \ln(\frac{1}{3})\qquad$ ...$\displaystyle \ln(\frac{1}{3})=\ln(3^{-1})=-\ln 3$
$-0.087t=-\ln 3 \qquad$ ... $/\div(-0.087)$
$t=\displaystyle \frac{\ln 3}{0.087}\approx 12.628$.
So
5 grams remain after about $12.6$ days.
