A population of bacteria is growing exponentially. At 7:00 a.m. the mass of the population is 12 mg. Five hours later it is 14 mg. (a) What will be the mass of the bacteria after another 5 hours? (b) At 7:00 p.m. what do we expect the mass to be? (c) What was the mass of the population at 8:00 a.m.? Given your answer, by what percent is the mass of the population increasing each hour? By what percent is it increasing each day?

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 hoursAssuming we start our measurement at 7.00 a.m as reference time  t = 0Thus 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 timePart 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.