Exponential and Logarithmic models

Question

The count in a bacteria culture was 500 after 20 minutes and 1300 after 40 minutes. Assuming the count grows exponentially,

What was the initial size of the culture?   

Find the doubling period.   

Find the population after 75 minutes.   

When will the population reach 13000.   

You may enter the exact value or round to 2 decimal places.

Answer 1

The general equation for population growth is P = P_° e^kt where P1 = 500, t1 = 20 min, P2= 1300,

t2=40 min

This forms two equations P1 = P_° e^20k .....eq1, and P2= P_° e^40k....eq2

Then 500 = P_° e^20k ........eq1, P_° = 500 / e^20k

1300 = P_° e^40k ........eq 2 P_° = 1300 / e^40k Equating P_°

500 / e^20k = 1300 / e^40k and (500 /1300) = ( e^20k / e^40k), 0.385 = e^20k . e^(-40k) = e^(-20k)

Take Ln of both sides Ln(0.385) = Ln(e^(-20k)) = -20k, and k = Ln(0.385) / -20 = 0.04773

and P_° = 500 / e^{(20)(0.04773)} = 500 / e ^0.955 = 192.4

or P_° = 1300 / e^{(40)(0.04773)} = 1300 / e^1.909 = 192.7 The difference is due to rounding so lets say P_° = 192 is the initial population of the bacteria

The doubling period is the time it takes the population to double in number that is P = 2 P_° then

2P_° = P_° e^{(0.04773) t} , and 2 = e^{0.04773 t}, take Ln of both sides Ln2 = Ln e^{0.04773 t}, Ln2 = 0.04773 t , and

t = Ln2/0.04773 = 14.52 min

The population after 75 minutes is P = P_° e^0.04773(75) = 192. e^{(0.04773)(75)} = 6886

The population reaches 13,000 is your P in the equation

13000 = 192. e^{(0.04773) t} , solving for t , 13000/192 = e^{(0.04773) t}, 67.70 = e^{(0.04773) t}, taking Ln of both sides

Ln (67.70) = Ln e^{0.04773 t} = 0.04773 t , t = Ln( 67.70) / 0.04773 = 88.3 min

Answer 2

Assume A=Pb^t

ln(A) = ln(P)+tln(b)

ln(500) = ln(P)+20ln(b) eqn1

ln(1300) = ln(P)+40ln(b) eqn2

subtract eqn1 from eqn2

ln(1300)-ln(500) = 20 ln(b)

ln(b) = (1/20)ln(13/5) = ln((13/5)^1/20 ==> b=(13/5)^1/20

A=P(13/5)^t/20

500 = P (13/5)

a)

P = 2500/13

b)

2=(13/5)^t/20

ln(2) = (t/20)ln(13/5)

t = 20 ln(2)/ln(13/5)

c)

A=(2500/13)(13/5)^75/20≈ 6921

d)

13000=(2500/13)(13/5)^t/20

169000/2500=(13/5)^t/20

ln(67.6) = (t/20)ln(13/5)

t=20ln(67.6)/ln(2.6) ≈ 88.2 minutes