# Math Help - Population question

1. ## Population question

The population of a city at time t years after 2002 is expressed as
P(t)=12300(e^0.009t).

a) Estimate the population of the city in 2012.

For this one I made t=10 because 2012-2002=10

b) Estimate the year the population will be doubled its population in 2002

I tried to find the initial population by making t=0... also tried t=1 d n both cases the answer was bigger than the answer for a).

What am I doing wrong?

2. You could be working at a higher level than I, but here's an attempt at a start of an answer:

$P(t)=12300(e^{0.009t})$
So you substituted in:
$P(0)=12300(e^{(0.009)(0)})=12300$
Then, for the population to be doubled in 'x' years:
$P(x)=12300(e^{0.009x})=24600$
Then $e^{0.009x}=2$

By taking logs of both sides, I think you can then get an answer. This will need confirmation, though.

3. I think what you need to do is finding "t" when you double the population of the year 2002, this means:
Year 2002 is the year "1", t=1
So,

P(t)=12300(e^0.009(1)) = 12411.20

Then, you double that P(t) and solve for t:

(2*12411.20)=12300(e^0.009t) =

24822.40=12300(e^0.009t)=

Ln(24822.4/12300) = 0.009t

t = 78 which means the population will be double on year 2080

Hope this help!
Best,
August80

4. Originally Posted by August80
I think what you need to do is finding "t" when you double the population of the year 2002, this means:
Year 2002 is the year "1", t=1
No, the problem said "t years after 2002". 2002 was not 1 year after 2002!
t= 0 for 2002.

So,

P(t)=12300(e^0.009(1)) = 12411.20

Then, you double that P(t) and solve for t:

(2*12411.20)=12300(e^0.009t) =

24822.40=12300(e^0.009t)=

Ln(24822.4/12300) = 0.009t

t = 78 which means the population will be double on year 2080

Hope this help!
Best,
August80