Answer:
[tex]Q(t) = 100e^{0.0231t}[/tex]
Step-by-step explanation:
The equation for the number of cells after t minutes is given by the following formula:
[tex]Q(t) = Q(0)e^{rt}[/tex]
In which Q(0) is the initial population and r is the growth rate.
Initial population of 100
So [tex]Q(0) = 100[/tex]
Doubles after 30 minutes.
So Q(30) = 200.
We use this to find r
[tex]Q(t) = Q(0)e^{rt}[/tex]
[tex]Q(t) = 100e^{rt}[/tex]
[tex]200 = 100e^{30r}[/tex]
[tex]e^{30r} = 2[/tex]
[tex]\ln{e^{30r}} = \ln{2}[/tex]
[tex]30r = \ln{2}[/tex]
[tex]r = \frac{\ln{2}}{30}[/tex]
[tex]r = 0.0231[/tex]
So
[tex]Q(t) = 100e^{0.0231t}[/tex]
The equation that expresses the growth of the number of cells of this bacterium as a function of time t (in minutes) is Q(t) = 100e^0.023t.
Let the number of bacteria cells present be Q
Given that;
Initial population of the bacteria is 100, this is is an exponential growth problem.
Q(0) = 100
Q(30) = 200
From;
Qt =Qoe^rt
Where r is the growth constant and t is the time taken.
200 = 100e^r(30)
200/100 = e^r(30)
2 = e^r(30)
ln 2 = lne^r(30)
0.693 = 30r
r = 0.693/30
r = 0.023
So we can write a general equation;
Q(t) = 100e^0.023t
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