1) bacteria size = b(initial) e^(r t)
initial size = 300
final size = 1600.
Substituting the values into the main equation:
1600 = 300 e^(r 35 - 20)
solving for 'r'.
1600 = 300 e^(r 15)
1600/300 = e^(r 15)
taking natural log on both sides to eliminate the power
ln(16/3) = ln[e^(r 15)]
ln(16/3) = r 15
r = ln(16/3) / 15
= 0.11159843
= 11.16%
To find the initial bacteria size,
Given: t = 20
300 = b e^(0.1116 20)
Solving for b:
300 = b e^2.232
b = 300/e^2.232
b = 32
2) The doubling period = 70% / r (in percent)
= 70% / 11.16%per min
= 6.272 minutes
3) Population after 65 minutes = 32 e^(0.1116 65)
X = 32 e^7.254
X = 631
4) When will the population reach 1400
14000 = 32 e^(0.1116 t)
solving for t
ln(14000/32) = ln(e^(.1116 t)
ln(437.5) =0.1116 * t
t = ln(437.5) /0.1116 = 54.5 minutes
