Answer
Thus, the initial size of the culture is $2032$ and the doubling time of the population is $7.701$
Work Step by Step
Given that:
$P(t)=P_0e^{(kt)}$
Doubling the time:
$t_d=\frac{1}{k} \ln 2$
We know that after 10 hours, there were 5000 bacteria present.
$5000=P_0e^{(10k)}$
And after 12 hours, there were 6000 bacteria present.
$6000=P_0e^{(12k)}$
To find $k$
$\frac{5000}{6000}=\frac{P_0e^{(10k)}}{P_0e^{(12k)}}$
$e^{-2k}=\frac{5}{6}$
$-2k=\ln(\frac{5}{6})$
$k \approx 0.09$
So,
$5000=P_0e^{(10\times 0.09)}$
$P_0 \approx 2032$
And the doubling time of the population:
$t_d=\frac{1}{0.09} \ln 2$
$t_d \approx7.701$
Thus, the initial size of the culture is $2032$ and the doubling time of the population is $7.701$ hours.
