Answer
See below
Work Step by Step
Given: $\frac{dS}{dt}=-kS$
and $S(T)=500,000\\ S(15)=100,000$
Integrating both sides:
$\int \frac{dS}{S}=\int -kdt\\
\rightarrow \ln|S|=-kt+c\\
\rightarrow S(t)=c_1e^{-kt}$
Using $S(0)=500,000$, we find: $c_1=500,000$
The solution $S(t)=500,000e^{-kt}$
Given $t=15$, we have: $100,000=500,000e^{-15k}\\
\rightarrow e^{-15}=\frac{1}{5}\\
\rightarrow -15k=\ln \frac{1}{5}\\
\rightarrow k=\frac{\ln 5}{15}$
Thus, we have the solution $S(t)=500,000e^{-\frac{\ln 5}{15}t}$
a) For $t=35 \rightarrow S(35)=500,000e^{-\frac{\ln 5}{15}35} \approx 11696$
b) For $t=30 \rightarrow S(30)=500,000e^{-\frac{\ln 5}{15}30}=500,000\frac{1}{25}=20,000$
c) With $S(t)=1000 \rightarrow 1000=500,000e^{-\frac{\ln 5}{15}t} \\
\rightarrow e^{-\frac{\ln 5}{15}t}=\frac{1}{500}\\
\rightarrow -\frac{\ln 5}{15}t=\ln \frac{1}{500}\\
\rightarrow t=\frac{15}{\ln 5} \ln 500 \approx 57.92\approx 58$
It is approximately 58 days after April 1.
