Assume that a colony of bacteria grows at a rate proportional to the number of bacteria present. If the number of bacteria doubles in $5$ hours (h), how long will it take for the number of bacteria to triple?

Solution Let $N(t)$ be the number of bacteria present at time $t$. Then the assumption that this colony of bacteria grows at a rate proportional to the number present can be modeled by $$\dfrac{dN}{dt}=kN$$

where $k$ is a positive constant of proportionality. To find $k,$ we write the differential equation as $\dfrac{dN}{N}=kdt$ and integrate both sides. This differential equation is of form (4), and its solution is given by (5). So, we have $$N(t) =N_{0}e^{kt}$$

where $N_{0}$ is the initial number of bacteria in the colony. Since the number of bacteria doubles to $2N_{0}$ in 5h, $$\begin{eqnarray*} N(5) =N_{0}\,e^{5k} &=& 2N_{0} \\ e^{5k} &=&2 \\ k &=&\dfrac{1}{5}\ln 2 \end{eqnarray*}$$

The time $t$ required for this colony to triple obeys the equation $$\begin{eqnarray*} N(t) &=&3N_{0} \\ N_{0}e^{kt} &=&3N_{0} \\ e^{kt} &=&3 \\ t &=&\dfrac{1}{k}\ln 3 \underset{\underset{{\color{#0066A7}{\hbox{\( k=\tfrac{1}{5}\ln 2\)}}}}{\color{#0066A7}{\uparrow }}}{=}5\dfrac{\ln 3}{\ln 2}\approx 8 \end{eqnarray*}$$

The number of bacteria will triple in about 8h.