If \(A(t)\) is the amount of oil in the well (let's say in millions of barrels) at time \(t[tex] (years), then
[tex]\dfrac{\mathrm dA}{\mathrm dt}=kA\)
for some negative value of \(k\). You're told that the well started off with 3 million barrels, so \(A(0)=3\), and that after 6 years, there are 1.5 million barrels, so \(A(6)=1.5\).
Solve the differential equation above:
\(\dfrac{\mathrm dA}{\mathrm dt}=kA\implies\dfrac{\mathrm dA}A=k\,\mathrm dt\)
Integrate both sides.
\(\displaystyle\int\frac{\mathrm dA}A=k\int\mathrm dt\)
\(\ln|A|=kt+C\)
\(A=e^{kt+C}\)
\(A=Ce^{kt}\)
Given that \(A(0)=3\) and \(A(6)=1.5\), you have
\(\begin{cases}3=Ce^{0k}\\1.5=Ce^{6k}\end{cases}\implies C=3,\,k=-\dfrac{\ln2}6\)
So the amount of oil in the well is given by the function
\(A(t)=3e^{-\frac{\ln2}6t}=3\times2^{-t/6}\)
