a)n(t)= 8600e^0.1508t
b) around 11,600
c) 4.6 hrs.
A culture starts at 8600 bacteria. After one hour the count is 10,000.
Find a function that models the number of bacteria n(t) after t hours.
The answer is n(t) = 8600e^.1506t
Where does this 0.1506 come from?
Thanks.
n(t) = 8600 e^(kt), where k is an unknown constant and t is the number of hours. We know that n(1), or the population after one hour is 10000. So 10000 = 8600e^(k*1)
10000/8600 = e^k
natural log ln(10000/8600) = k = .1508
The general expression is n(t) = no* e^(t/T), where T is the time for an increase by a factor of e.
In this case,
10000 = 8600 e^(1/T)
That equation can be solved for T.
10000/8600 = 1.16279 = e^(1/T)
Take the natural log of both sides.
0.15082 = 1/T
n(t) = e^(0.15082 t)
The 0.1506 in your version is not quite right.
1 answer
1 answer