Answer
$S(t)=100(\displaystyle \frac{\ln t}{\ln 2}+1)$
Work Step by Step
$\displaystyle \frac{dS}{dt}$ is inversely proportional to time t, so there exists a k such that
$\displaystyle \frac{dS}{dt} =\displaystyle \frac{k}{t},\quad (t > 1)$
$S(t)=\displaystyle \int\frac{k}{t}dt=k\int\frac{1}{t}dt$
$S(t)=k\ln|t|+C\qquad (*)$
Since $t > 1$, we can drop the absolute brackets.
To find k and C, we use the given data,
$S(2)=400$, and $S(4)=300:$
$k\ln 2+C=200$
$k\ln 4+C=300$
Subtract the first from the second equation:
... and ... $\ln 4=\ln 2^{2}=2\ln 2$
$k(2\ln 2-\ln 2)=100$
$k(\ln 2)=100$
$k=\displaystyle \frac{100}{\ln 2}$
Back substitute into $k\ln 2+C=200$,
$\displaystyle \frac{100}{\ln 2}\cdot\ln 2+C=200$
$100+C=200$
$C=100$
So, substituting k and C in (*),
$S(t)=\displaystyle \frac{100}{\ln 2}\cdot\ln t+100$
$S(t)=100(\displaystyle \frac{\ln t}{\ln 2}+1)$
