Respuesta :
Answer:
a). k = 2.0794
b). [tex]A_{t}=A_{0}(2^{3t} )[/tex]
c). 13107200 cells
d). rate of growth after 6 hours = 27255112
e). 4.76 hours
Step-by-step explanation:
From the formula of bacterial population,
[tex]A_{t}=A_{0}e^{kt}[/tex]
Where [tex]A_{t}[/tex] = Bacterial population after time t
[tex]A_{0}[/tex] = Initial population
k = relative growth factor
t = duration or time
a). For [tex]A_{t}=2\times 50=100[/tex] cells
and [tex]A_{0}=50[/tex] cells
Time t = 20 minutes = [tex]\frac{1}{3}[/tex] hours
Now we plug in these values in the formula,
100 = [tex]50(e^{\frac{k}{3}})[/tex]
[tex]e^{\frac{k}{3}}=2[/tex]
By taking natural log on both the sides of the equation,
[tex]\frac{k}{3}(lne)=ln(2)[/tex]
k = 3ln(2) = 2.0794
b). To get the expression we will plug in the value of k in the formula.
[tex]A_{t}=A_{0}e^{kt}[/tex]
Since k = 3ln(2)
[tex]A_{t}=A_{0}e^{3t(ln2)}[/tex]
Let y = [tex]e^{3t(ln2)}[/tex]
By taking natural log on both the sides of the equation,
lny = [tex]ln(e^{3t(ln2)} )[/tex]
lny = 3t(ln2)ln(e)
ln(y) = 3t(ln2)
ln(y) = [tex]ln(2)^{3t}[/tex]
y = [tex]2^{3t}[/tex]
Now our expression will be [tex]A_{t}=A_{0}(2^{3t} )[/tex]
c). Number of cells after t = 6 hours
[tex]A_{6}=50(2^{3\times 6} )[/tex]
[tex]A_{6}=50(2^{18})[/tex]
[tex]A_{6}=50\times (262144)=13107200[/tex]
d). We can get the rate of growth by finding derivative of the expression
[tex]\frac{d}{dt}(A_{t})=\frac{d}{dt}[A_{0}(e^{kt}})][/tex]
= [tex]A_{0}[ke^{kt}][/tex]
Now [tex]\frac{d}{dt}(A_{6})=k.A_{6}[/tex]
= 2.0794×13107200
= 27255112
e) From the given formula,
[tex]A_{t}=A_{0}(2^{3t} )[/tex]
[tex]1000000=50(2^{3t} )[/tex]
[tex]2^{3t}=20000[/tex]
3t(ln2) = ln(20000)
t = [tex]\frac{ln(20000)}{3(ln2)}[/tex]
t = [tex]\frac{9.90349}{2.07944}=4.76[/tex] hours
