Problema Solution
A certain type of bacteria doubles in population every 6.5 hours. If there were 100 bacteria to start with, what is the hourly growth rate of the bacteria? How many bacteria will there be after a day and a half? Solve algebraically.
Answer provided by our tutors
P = 100 bacteria is the initial population
t = 1.5 days = 24*1.5 hr = 36 hr
k = constant
A = the number of bacteria after time t
A = P*e^(kt)
A = 100*e^(kt)
in 6.5 hours, the population will be 200
200 = 100e^(6.5k)
2 = e^(6.5k)
e^(6.5k) = 2
k = (ln 2)/6.5
now we have the equation A = 100*e^((ln 2)/6.5)t) that is
A = 100*2^(t/6.5)
for t = 36 hr we have
A = 100 * 2^(36/6.5)
A = 4648 bacteria
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