In a certain culture of bacteria, the rate of increase is proportional to the number of bacteria present at that instant. It is found that there are 10,000 bacteria at the end of 3 hours and 40,000 bacteria at the end of 5 hours, then the number of bacteria present in the beginning are

Let \(x\) be the number of bacteria present at time \(t\).
\(\begin{aligned}
& \therefore \quad \frac{\mathrm{d} x}{\mathrm{dt}} \propto x \\
& \therefore \quad \frac{\mathrm{d} x}{\mathrm{dt}}=\mathrm{k} x \\
& \therefore \quad \frac{\mathrm{d} x}{x}=\mathrm{kdt} \\
&
\end{aligned}\)
Integrating on both sides, we get
\(\log x=\mathrm{kt}+\mathrm{c}... (i)\)
When \(\mathrm{t}=3, x=10,000\)
Equation (i) becomes
\(\log (10,000)=3 \mathrm{k}+\mathrm{c}... (ii)\)
When \(\mathrm{t}=5, x=40,000\)
Equation (i) becomes
\(\log (40,000)=5 \mathrm{k}+\mathrm{c}... (iii)\)
Subtracting (ii) from (iii), we get
\(\mathrm{k}=\log 2\)
From equation (ii),
\(\log (10,000)=3 \log 2+c\)
\(\therefore \quad c=\log (1250)\)
Now, Initially \(\mathrm{t}=0\)
From (i),
\(\log x=\mathrm{k} \times 0+\log (1250)\)
\(\therefore \quad \log x=\log 1250\)
\(\therefore \quad x=1250\)