In an experiment of culture of bacteria in a laboratory, the rate of increase of bacteria is proportional to the number of bacteria present at that time.
(i) If in one hour the number of bacteria gets doubled, then what is the number of bacteria at the end of 4 hours?
(ii) If the number of bacteria is 24,000 at the end of 3 hours, find the number of bacteria in the beginning.
Let the original number of bacteria at time $t = 0$ be $= N_0$
Let the number of bacteria at any instant of time $t$ be $= N$
Given: $\text{Rate of increase of bacteria} \propto \text{ Number present at that time}$
i.e. $\dfrac{dN}{dt} \propto N$
i.e. $\dfrac{dN}{dt} = k N$ $\hspace{2em}$ where $k$ is the constant of proportionality
i.e. $\dfrac{dN}{N} = k \; dt$
i.e. $\displaystyle \int \dfrac{dN}{N} = k \int dt$
i.e. $\log \left|N\right| = k \; t + c$ $\;\;\; \cdots \; (1)$ $\hspace{2em}$ where $c$ is the constant of integration
Applying the initial condition, when $t = 0$, $N = N_0$, equation $(1)$ becomes
$c = \log \left|N_0\right|$ $\;\;\; \cdots \; (2)$
$\therefore$ $\;$ In view of equation $(2)$, equation $(1)$ can be written as
$\log \left|N\right| = k \; t + \log \left|N_0\right|$
i.e. $\log \left|N\right| - \log \left|N_0\right| = k \; t$
i.e. $\log \left|\dfrac{N}{N_0}\right| = k \; t$
i.e. $\dfrac{N}{N_0} = e^{k t}$
i.e. $N = N_0 \; e^{kt}$ $\;\;\; \cdots \; (3)$
Now, in time $t = 1$ $\;$ hour, the number of bacteria gets doubled.
i.e. when $t = 1$, $N = 2 \; N_0$
$\therefore$ $\;$ In view of this condition equation $(3)$ becomes
$2 \; N_0 = N_0 \; e^k$
i.e. $e^{k} = 2$ $\implies$ $k = \log_e 2$ $\;\;\; \cdots \; (4)$
$\therefore$ $\;$ In view of equation $(4)$, equation $(3)$ becomes
$N = N_0 \; e^{t \; \log_e 2}$ $\;\;\; \cdots \; (5)$
Given: $t = 4$ hours
$\therefore$ $\;$ We have from equation $(5)$
$\begin{aligned}
N & = N_0 \; e^{4 \; \log_e 2} \\\\
& = N_0 \; e^{\log_e \left(2^4\right)} \\\\
& = N_0 \; e^{\log_e 16} \\\\
& = 16 N_0
\end{aligned}$
i.e. Number of bacteria at the end of 4 hours $=$ 16 times the original number
Given: In time $t = 3$ hours, number of bacteria $= N = 24,000$
$\therefore$ $\;$ We have from equation $(5)$
$\begin{aligned}
24000 & = N_0 \; e^{\log_e 2} \\\\
& = N_0 \; e^{\log_e \left(2^3\right)} \\\\
& = N_0 \; e^{\log_e 8} \\\\
& = 8 N_0
\end{aligned}$
$\therefore$ $N_0 = \dfrac{24000}{8} = 3000$
$\therefore$ $\;$ Original number of bacteria $= 3000$
\end{enumerate}