Answer:
About 53022 bacteria present.
Step-by-step explanation:
We can use the continuous growth formula:
[tex]\displaystyle P = A e^{kt}[/tex]
Where k is some constant.
With an initial population of 5000, it grew to 6500 after two hours.
In other words, P = 6500 when A = 5000 and t = 2:
[tex]\displaystyle (6500) = (5000)e^{k(2)}[/tex]
Solve for k:
[tex]\displaystyle \begin{aligned} 6500 & = 5000e^{2k} \\ \\ e^{2k} & = \frac{6500}{5000} = \frac{13}{10} \\ \\ \ln\left(e^{2k}\right) & = \ln\left(\frac{13}{10}\right) \\ \\ 2k & = \ln\frac{13}{10} \\ \\ k & = \frac{1}{2}\ln\frac{13}{10}\end{aligned}[/tex]
Therefore, our equation is:
[tex]\displaystyle P = \bigg{5000e}^{\dfrac{1}{2}\ln\dfrac{13}{10}t[/tex]
After 18 hours, t = 18. Hence:
[tex]\displaystyle \begin{aligned} P(18) & = \bigg{5000e}^{\dfrac{1}{2}\ln\dfrac{13}{10}(18)} \\ \\ & \approx 53022\end{aligned}[/tex]
Therefore, after 18 hours, there will be about 53022 bacteria present.
