Originally Posted by

**skeeter** $\displaystyle \frac{dQ}{dt} = k(Q - 27)$

$\displaystyle \frac{dQ}{Q-27} = k \, dt$

$\displaystyle \ln(Q-27) = kt + C$

$\displaystyle

Q = 27 + Ae^{kt}

$

at $\displaystyle t = 4$, $\displaystyle Q = 67$

$\displaystyle 67 = 27 + Ae^{4k}$

$\displaystyle

40 = Ae^{4k}

$

at $\displaystyle t = 9$, $\displaystyle Q = 47$

$\displaystyle

47 = 27 + Ae^{9k}

$

$\displaystyle

20 = Ae^{9k}

$

$\displaystyle \frac{40}{20} = \frac{Ae^{4k}}{Ae^{9k}}$

$\displaystyle

2 = e^{-5k}

$

solve for k , then solve for A and finish.