Answer :
Answer:
a) The initial size is 506.
b) The population after 75 minutes is 24825.
c) 57 minutes after the start of the experiment will the population reach 11000.
Step-by-step explanation:
Given : Assume that the number of bacteria follows an exponential growth model : [tex]P(t)=P_0e^{kt}[/tex]
The count in the bacteria culture was 800 after 10 minutes and 2000 after 30 minutes.
According to information, we can find value of k
P(t)=2000, [tex]P_0=800[/tex] and t=30-10=20 minute
Substitute value in the model,
[tex]2000=800\times e^{k\times 20}[/tex]
[tex]\frac{2000}{800}=e^{20k}[/tex]
[tex]2.5=e^{20k}[/tex]
Taking ln both side,
[tex]\ln(2.5)=\ln e^{20k}[/tex]
[tex]\ln(2.5)=20k[/tex]
[tex]k=\frac{\ln(2.5)}{20}[/tex]
[tex]k=0.0458[/tex]
The model became [tex]P(t)=P_0e^{0.0458t}[/tex]
a) What was the initial size of the culture?
The count in the bacteria culture was 800 after 10 minutes.
So, [tex]800=P_0e^{0.0458\times 10}[/tex]
[tex]P_0=\frac{800}{e^{0.458}}[/tex]
[tex]P_0=506.03[/tex]
Approximately, the initial size is 506.
b) Find the population after 75 minutes.
So, [tex]P(75)=800\times e^{0.0458\times 75}[/tex]
[tex]P(75)=800\times e^{3.435}[/tex]
[tex]P(75)=24825.13[/tex]
Approximately, the population after 75 minutes is 24825.
(c) How many minutes after the start of the experiment will the population reach 11000?
Here, P(t)=11000, [tex]P_0=800[/tex]
[tex]11000=800\times e^{0.0458\times t}[/tex]
[tex]\frac{11000}{800}=e^{0.0458\times t}[/tex]
[tex]13.75=e^{0.0458t}[/tex]
Taking natural log both side,
[tex]\ln 13.75=\ln e^{0.0458t}[/tex]
[tex]\ln 13.75=0.0458t[/tex]
[tex]t=\frac{\ln 13.75}{0.0458}[/tex]
[tex]t=57.22[/tex]
Approximately, 57 minutes after the start of the experiment will the population reach 11000.