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?


So you know $P=ce^{kt}$. The population doubles in $24$ hours, or $2P=ce^{k(t+24)}$. Now can you find $k$ by dividing the two equations?

As a side note, you might want to know how that formula was obtained.$$\frac{dP}{dt}=kP\implies\frac{dP}P=k~dt$$Now integrate both sides,$$\int_{P(t=0)}^{P(t=t)}\frac{dP}P=k\int_{t=0}^{t=t}dt$$giving you $P(t)=P(0)e^{kt}$.

  • $\begingroup$ Can you explain further? I don't understand. $\endgroup$ – Burt Aug 1 at 21:43
  • $\begingroup$ What part do you not understand? @burt $\endgroup$ – Shubham Johri Aug 1 at 21:45
  • $\begingroup$ You were clear. I just don't understand how to do the dividing. $\endgroup$ – Burt Aug 1 at 21:48
  • 2
    $\begingroup$ @burt$$\frac{2P}P=\frac{ce^{k(t+24)}}{ce^{kt}}$$ $\endgroup$ – Shubham Johri Aug 1 at 21:50
  • $\begingroup$ Am I missing something here? I am completely lost. $\endgroup$ – Burt Aug 1 at 21:51


$\dfrac{dP}{dt} = kP, \tag 1$


$P \ne 0, \tag 2$

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 $t = t_0$:

$\ln P(t) - \ln P(t_0) = \displaystyle \int_{t_0}^t \dfrac{1}{P(s)}\dfrac{dP(s)}{ds} \; ds = \int_{t_0}^t k \; ds = k(t - t_0), \tag 4$


$\ln \left (\dfrac{P(t)}{P(t_0)} \right ) = k(t - t_0); \tag 5$

we apply the function $\exp(\cdot)$ to this to obtain

$\dfrac{P(t)}{P(t_0)} = e^{k(t - t_0)}, \tag 6$


$P(t) = P(t_0) e^{k(t - t_0)}. \tag 7$

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

$k = \dfrac{\ln 2}{24}. \tag{11}$

  • 1
    $\begingroup$ @burt: an archaic form of the modern word "between"; quite intentinal. Cheers! $\endgroup$ – Robert Lewis Aug 1 at 22:58

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