Answer
the final increase in bacteria at t=35.86 hour.
Work Step by Step
Given that:
$P(t)=P_{0}e^{(kt)}$
Doubling the time:
$t_{d}=\frac{1}{k}\ln{2}$
We know that the bacteria double in 4 hours:
$k=\frac{1}{4}\ln{2}$
$P_{0}=2000$ and time $p(t)=10^6$
So,
$10^6=2000e^{(\frac{1}{4}\ln{2}\times t)}$
$\frac{10^6}{2000}=e^{(\frac{1}{4}\ln{2}\times t)}$
$\ln(500)=\frac{1}{4}\ln{2}\times t$
$t=35.86$
Thus, the final increase in bacteria happens at t=35.86 hours.
