Answer with Step-by-step explanation:
The exponential growth function is given by
[tex]N(t)=N_oe^{mt}[/tex]
where
[tex]N(t)[/tex] is the population of the bacteria at any time 't'
[tex]N_o[/tex] is the population of the bacteria at any time 't = 0'
'm' is a constant and 't' is time after 7.00 a.m in hours
Assuming we start our measurement at 7.00 a.m as reference time  t = 0
Thus we get[tex]N(0)=N_oe^{m\times 0}\\\\12=N_o[/tex]
Now since it is given after 5 hours the population becomes 14 mg thus from the above relation we get
[tex]12\times e^{m\times 5}=14\\\\e^{5m}=\frac{14}{12}\\\\m=\frac{1}{5}\cdot ln(\frac{14}{12})\\\\m=0.031[/tex]
Thus the population of bacteria at any time 't' is given by
[tex]N(t)=12e^{0.031t}[/tex]
Part a)
Population of bacteria after another 5 hours equals the population after 10 hours from start
[tex]N(10)=12e^{0.031\times 10}=16.361mg[/tex]
Part b)
Population of bacteria at 7:00 p.m is mass after 12 hours
[tex]N(1)=12e^{0.031\times 12}=17.41mg[/tex]
Part c)
Population of bacteria at 8:00 p.m is mass after 1 hour
[tex]N(1)=12e^{0.031\times 1}=12.3378mg[/tex]
Part d)
Differentiating the relation of population with respect to time we get
[tex]N'(t)=\frac{d(12\cdot e^{0.031t})}{dt}\\\\N'(t)=12\times 0.031=0.372e^{0.031t}[/tex]
Thus we can see that the percentage increase varies with time initially the percentage increase is 37.2% but this percentage increase increases with increase in time
Part 4)
Since there are 24 hours in 1 day thus the percentage increase in the population is
[tex]\frac{N(24)-N_o}{N_o}\times 100\\\\=\frac{25.25-12}{12}\times 100=110.42[/tex]
Thus there is an increase of 110.42% in the population each day.
