differential equations - exponential growth and decay 
The population $P$ of bacteria in an experiment grows according to the equation $\frac{dP}{dt}=kP$, where $k$ is a constant and $t$ is measured in hours. If the population of bacteria doubles every $24$ hours, what is the value of $k$?  

I was given this problem and I'm not sure what to do with it. I know the formula for this kind of equation is $ce^{kx}$. But, how do you plug in the values given?

From
$\dfrac{dP}{dt} = kP, \tag 1$
assuming
$P \ne 0, \tag 2$
we deduce that
$\dfrac{1}{P}\dfrac{dP}{dt} = k; \tag 3$
we integrate 'twixt $t_0$ and $t$, assuming $P$ takes the value $P(t_0)$ at $t = t_0$:
$\ln P(t) - \ln P(t_0) = \displaystyle \int_{t_0}^t \dfrac{1}{P(s)}\dfrac{dP(s)}{ds} \; ds = \int_{t_0}^t k \; ds = k(t - t_0), \tag 4$
or
$\ln \left (\dfrac{P(t)}{P(t_0)} \right )  = k(t - t_0); \tag 5$
we apply the function $\exp(\cdot)$ to this to obtain
$\dfrac{P(t)}{P(t_0)} = e^{k(t - t_0)}, \tag 6$
whence
$P(t) = P(t_0)  e^{k(t - t_0)}. \tag 7$
Now given that $P(t)$ doubles every $24$ hours, starting at any $t_0$ we take
$t = t_0 + 24, \tag 8$
and thus
$2P(t_0) = P(t_0 + 24) =  P(t_0)  e^{24k}, \tag 9$
leading to
$e^{24k} = 2; \tag{10}$
solving for $k$ we conclude
$k = \dfrac{\ln 2}{24}. \tag{11}$
