Exponential Growth and Decay Help......

The doubling period of a baterial population is, 10 minutes. At time, t = 80 minutes, the baterial population was 60000. For some constant A, the formula for the population is, p (t) = Ae^kt where, k = ln (2) / 10 . What was the initial population at time, t = 0 ? ____________

Find the size of the baterial population after 3 hours. ___________

p(t) = Ae^(kt)
p(t) = Ae^[t(ln 2)/10]


i'm assuming you forgot brackets around kt because the formula then doubles A at t = 10 minutes.


we know that population = 60000 at t = 80, so we can find A:
p(80) = Ae^[80(ln 2)/10]
60000 = Ae^[80(ln 2)/10]
60000 = Ae^[8(ln 2)]
60000 = Ae^(ln 2⁸)
60000 = A2⁸
60000/2⁸ = A
234.375 = A


now that we have A, we can get population at any time


at t = 0:
p(t) = 234.375e^(kt)
p(0) = 234.375e^(k*0)
p(0) = 234.375e^(0)
p(0) = 234.375


at t = 3 hours = 3*60 minutes = 180 minutes:
p(t) = 234.375e^(kt)
p(t) = 234.375e^(180k)
p(180) = 234.375e^[180(ln 2)/10]
p(180) = 234.375e^[18(ln 2)]
p(180) = 234.375e^(ln 2¹⁸)
p(180) = 234.375(2¹⁸)
p(180) = 61440000