P(t)=P0e^.019t-220t?

Question

The grownth of the population in a particular country can be modeled by P(t)=P(Sub 0)e^.019t-220t where t represents time in years since 1998, and P(t) represents the population of the country. If after 8 years the population P(t) rose to 5,930,000. Find the initial population P(Sub 0)

Answer 1

If I understand you question correctly.

Given:
P(t) = P(Sò)e^0∙019t - 220t
2006yr - 1998yr = 8yr (time difference).
Now, fill in the information.
5,930,000 = P(Sò)e^0∙019(8) - 220(8)
5,930,000 = P(Sò)e^0∙019(8) - 1760
P(Sò)e^0∙019(8) = 5,930,000 - 1760
P(Sò)e^0∙019(8) = 5,928,240
P(Sò)e^0∙152 = 5,928,240
P(Sò) 1∙164160236 = 5,928,240
P(Sò) = 5,928,240 / 1∙164160236
P(Sò) = 5,092,288∙686

Answer 2

Do you mean:

P(t) = P_0e^(0.019t-220t) OR

P(t) = P_0e^0.019t - 220t

?

Well, you will end up having two equations that you need to solve simultaneously. Since you do not provide a value for P_0, P(0) which is what you are calculating will have this in the answer.

1. P(0) = P_0
2. 5,930,000 = P_0e^(0.019*8 - 220*8)

Answer 3

I'm going to call the initial population N to avoid the whole subscript mess.

P(8) = Ne^(.019*8) - 220(8) = 5,930,000
N(1.164) - 1760 = 5,930,000
1.164N = 5,931,760
N = 5,096,013.746 which rounds to 5,096,014 to the nearest whole person.

.019 is a pretty small growth factor for the population of a country... Are you sure it isn't 0.19? If it is 0.19, you could do the same calculations and arrive at an initial population of 1,297,346 (rounded to the nearest whole person).

Answer 4

P(t) = (P0 * e^.019t) - (220t )

t = 0 @ 1998


P(8) = 5930000 = (P0 * e^.019(8)) - (220(8) )
5930000 = (P0 * e^.152) - (1760)
5931760 = (P0 * e^.152)
P0 = (5931760 / e^.152)
= 5095312 people

Answer 5

Holy crap you guys are good! I never knew about such an equation!