Given:
Count in bacteria culture after 20 minutes = 200
Count in bacteria culture after 40 minutes = 1900
Required: Find the initial size, doubling period, population after 60 minutes and time at which the population will reach 11000
Explanation:
The exponential growth can be modeled by
[tex]A=A_0e^{kt}[/tex]
where A0 is the initial size and k is the growth rate.
Plug the condition A(20) = 200.
[tex]\begin{gathered} 200=A_0e^{k\cdot20} \\ =A_0e^{20k} \end{gathered}[/tex]
Plug the condition A(40) = 1900.
[tex]\begin{gathered} 1900=A_0e^{k\cdot40} \\ =A_0e^{40k} \end{gathered}[/tex]
Divide both of them.
[tex]\begin{gathered} \frac{1900}{200}=\frac{A_0e^{40k}}{A_0e^{20k}}=e^{20k} \\ 20k=\ln(9.5) \\ k=\frac{\ln(9.5)}{20} \end{gathered}[/tex]
Thus,
[tex]A=A_0e^{\frac{\ln(9.5)}{20}t}[/tex]
Substitute A(20) = 200 to find the initial count.
[tex]\begin{gathered} 200=A_0e^{\frac{\ln(9.5)}{20}\cdot20} \\ =9.5A_0 \\ A_0=\frac{200}{9.5}\approx21 \end{gathered}[/tex]
Thus,
[tex]A=21e^{\frac{\ln(9.5)}{20}t}[/tex]
To find the doubling, find t at which A(t) = 42.
[tex]\begin{gathered} 42=21e^{\frac{ln(9.5)}{20}t} \\ 2=e^{\frac{ln(9.5)}{20}t} \\ \frac{\ln(9.5)}{20}t=\ln(2) \\ t=\frac{\ln(2)\cdot20}{\ln(9.5)} \\ =6.158 \end{gathered}[/tex]
The doubling time is 6.158 minutes.
To find the population after 60 minutes, substitute 60 for t into A(t).
[tex]\begin{gathered} A=21e^{\frac{\ln(9.5)}{20}\cdot60} \\ \approx18005 \end{gathered}[/tex]
The count of bacteria after 60 minutes is 18005.
To find the time required for the culture of 11000, find t at which A(t) = 11000.
[tex]\begin{gathered} 21e^{\frac{\ln(9.5)}{20}t}=11000 \\ e^{\frac{\ln(9.5)}{20}t}=523.81 \\ \frac{\ln(9.5)}{20}t=6.261 \\ t=55.622 \end{gathered}[/tex]
The count of bacteria will reach 11000 in 55.622 minutes.
Final Answer:
Initial count = 21
The doubling time is 6.158 minutes.
The count of bacteria after 60 minutes is 18005.
The count of bacteria will reach 11000 in 55.622 minutes.
