Now, we know that the culture initially contains 2000 bacteria and that it doubles every half hour. Assuming that the time t is in hours, this would mean that
[tex]p(0.5)=4000=2000e^{0.5k}[/tex]Solving for k,
[tex]\begin{gathered} 4000=2000e^{0.5k} \\ \rightarrow2=e^{0.5k} \\ \rightarrow\log (2)=\log (e^{0.5k}) \\ \rightarrow\log (2)=0.5k\log (e) \\ \rightarrow k=\frac{\log (2)}{0.5} \\ \\ \Rightarrow k=2\log (2) \end{gathered}[/tex]This way, we would have that:
[tex]p(t)=2000e^{2\log (2)\cdot t}[/tex]A)
20 minutes is 1/3 of an hour. This way,
[tex]\begin{gathered} p(\frac{1}{3})=2000e^{2\log (2)\cdot\frac{1}{3}} \\ \\ \Rightarrow p(\frac{1}{3})=3174.80 \end{gathered}[/tex]Thereby, the bacterial population after 20 minutes is 3,174.80
B)
[tex]\begin{gathered} p(6)=2000e^{2\log (2)\cdot6} \\ \Rightarrow p(6)=8192000 \end{gathered}[/tex]Thereby, the bacterial population after 6 hours is 8,192,000
