Answer:
Explanation:
Let X be the number of bacteria at time t, and Xo be the initial number of bacteria
Bacteria grows exponentially, the exponential growth model is thus:
X = Xo [tex]e^{kt}[/tex]
[e is the exponential sign]
k is the growth constant = growth rate
At t = 20 minutes, X = 100
At t = 40 minutes, X = 2000
Substituting that into the formula
(i) ... 100 = Xo [tex]e^{k20}[/tex]
(ii) ... 2000 = Xo[tex]e^{k40}[/tex]
Divide (ii) by (i)
[tex]\frac{2000}{100} = \frac{Xoe^{40k} } {Xoe^{20k}}[/tex]
20 = [tex]e^{40k-20k} = e^{20k}[/tex]
[Xo cancels Xo. Division of values raised to a power is done by subtracting their powers]
Take the natural log of both sides
㏑20 = 20k
[taking the ㏑ cancels the exponential]
k = [tex]\frac{ln20}{20}[/tex]
We can now substitute k to solve one of the equations
substituting k in (ii): 2000 = Xo [tex]e\frac{ln 20}{20}.^{40}[/tex]
2000 = Xo [tex]e^{0.15*40}[/tex]
2000 = Xo [tex]e^{6}[/tex]
Making Xo the subject of the formula
Xo = [tex]\frac{2000}{e^{6} }[/tex]
Xo is approximately 5 cells.
THE DOUBLING TIME
The doubling time is the time it takes for the population to double, so 5 cells become 10 cells
Since Xo = 5
Given that X = Xo [tex]e^{kt}[/tex]
When X=10,
10 = 5 [tex]e^{0.15t}[/tex]
solving for t:
[tex]e^{0.15t} = \frac{10}{5}[/tex]
Take ln of both sides
0.15t = ln 2
t =[tex]\frac{ln2}{0.15}[/tex]
t =4.62 minutes
POPULATION AFTER 65 MINUTES
X = 5 [tex]e^{0.15*65}[/tex]
X = 85770
WHEN WILL THE POPULATION REACH 12000
12000 = 5 [tex]e^{0.15t}[/tex]
[tex]e^{0.15t}[/tex] =240
take ln of both sides
0.15t = ln 2400
t = 51.89 minutes (approximately 52 minutes)
