Question

The bacteria culture grows at a rate proportional to its size. After $2$ hours, there are $600$ bacteria and after $8hours$ the count is $75,000$. Find the initial population and when will the population reach $2,00,000$ ?

Answer 1

Use $P\left(t\right)=P\left(0\right)e^{kt}$ , where $P\left(t\right)$ is the population after t hours.

Then $600=P\left(0\right)e^{2k}$ and $75000=P\left(0\right)e^{8k}$.

Dividing, we get

$125=\frac{75000}{600}=\frac{P\left(0\right)e^{8k}}{P\left(0\right)e^{2k}}=e^{6k}$.

Solving for $k$ gives $\frac{\ln \left(125\right)}{6}=k$.

Plugging that into our equation for the population after 2 hours gives

$600=P\left(0\right)e^{\frac{\ln \left(125\right)}{3}}$ , so the initial population

$P\left(0\right)=\frac{600}{e^{\frac{\ln \left(125\right)}{3}}}$ .

A “cleaner” answer here is $P\left(0\right)=\mathrm{120.......}$ since $e^{\frac{\ln \left(125\right)}{3}}=e^{\frac{\ln \left(5^3\right)}{3}}=5$.