Respuesta :
If the growth rate of bacteria at any time is proportional to the number of bacteria present at t then the population after 20 weeks will be 403.42[tex]x_{0}[/tex] in which [tex]x_{0}[/tex] is the initial population.
Given that the growth rate of bacteria at any time t is proportional to the number present at t and triples in 1 week.
We are required to find the number of bacteria present after 10 weeks.
let the number of bacteria present at t is x.
So,
dx/dt∝x
dx/dt=kx
1/x dx=k dt
Now integrate both sides.
[tex]\int\limits {1/x} \, dx[/tex]=[tex]\int\limits{k} \, dt[/tex]
log x=kt+log c----------1
Put t=0
log [tex]x_{0}[/tex]=0 +log c ([tex]x_{0}[/tex] shows the population in beginning)
Cancelling log from both sides.
c=[tex]x_{0}[/tex]
So put c=[tex]x_{0}[/tex] in 1
log x=kt+log [tex]x_{0}[/tex]
log x=log [tex]e^{kt}[/tex]+log [tex]x_{0}[/tex]
log x=log [tex]e^{kt}x_{0}[/tex]
x=[tex]e^{kt}x_{0}[/tex]
We have been given that the population triples in a week so we have to put the value of x=2[tex]x_{0}[/tex] and t=1 to get the value of k.
2[tex]x_{0}[/tex]=[tex]e^{k} x_{0}[/tex]
2=[tex]e^{k}[/tex]
log 2=k
We have to now put the value of t=20 and k=log 2 ,to get the population after 20 weeks.
x=[tex]e^{20log 2}[/tex][tex]x_{0}[/tex]
x=[tex]e^{0.30*2}[/tex][tex]x_{0}[/tex]
x=[tex]e^{6}x_{0}[/tex]
x=403.42[tex]x_{0}[/tex]
If the growth rate of bacteria at any time is proportional to the number of bacteria present at t then the population after 20 weeks will be 403.42[tex]x_{0}[/tex] in which [tex]x_{0}[/tex] is the initial population.
Learn more about growth rate at https://brainly.com/question/25849702
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The given question is incomplete as the question incudes the following:
Calculate the population after 20 weeks.
