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.

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),
\(\begin{aligned}
& \quad \log x=\mathrm{k} \times 0+\log \left(\frac{10^4}{8}\right) \\
& \quad \Rightarrow \log x=\log \left(\frac{10^4}{8}\right) \\
& \therefore \quad x=\frac{10^4}{8}
\end{aligned}\)