In a certain culture of bacteria, the rate of increase is proportional to the number present. If there are…
In a certain culture of bacteria, the rate of increase is proportional to the number present. If there are $10^4$ at the end of 3 hours and $4 \cdot 10^4$ at the end of 5 hours, then there were $\qquad$ in the beginning.
$10^4$
$\frac{10^4}{4^1}$
$410^4$
$\frac{10^4}{8}$
Solution
Let $x$ be the number of bacteria present at time $t$.
$\begin{array}{ll}
\therefore & \frac{\mathrm{d} x}{\mathrm{dt}} \propto x \\
\therefore & \frac{\mathrm{~d} x}{\mathrm{dt}}=\mathrm{k} x \\
\therefore & \frac{\mathrm{~d} x}{x}=\mathrm{kdt}
\end{array}$
Integrating on both sides, we get
$\log x=\mathrm{kt}+\mathrm{c}$...(i)
When $\mathrm{t}=3, x=10^4=10,000$
Equation (i) becomes
$\log (10,000)=3 \mathrm{k}+\mathrm{c}$
...(ii)
When $t=5, x=4.10^4=40,000$
Equation (i) becomes
$\log (40,000)=5 \mathrm{k}+\mathrm{c}$...(iii)
Subtracting (ii) from (iii), we get
$k=\log 2$ From equation (ii),
$\log (10 ; 000)=3 \log 2+c$
$\therefore \quad c=\log \left(\frac{10^4}{8}\right)$ Now, Initially $\mathrm{t}=0$
From (i),
$\begin{aligned}
& \quad \log x=\mathrm{k} \times 0+\log \left(\frac{10^4}{8}\right) \\
& \quad \Rightarrow \log x=\log \left(\frac{10^4}{8}\right) \\
& \therefore \quad x=\frac{10^4}{8}
\end{aligned}$