Question

A country's population in 1993 was 94 million.

In 1999 it was 99 million. Estimate
the population in 2005 using the exponential
growth formula. Round your answer to the
nearest million.

P = Aekt

Answer 1

Let 1993 correspond to t=0
so 1994 corresponds to t=1
and 1995 corresponds to t=2
and 1996 corresponds to t=3

what does 1999 correspond to in terms of t?
and what does 2005 correspond to in terms of t ?

Answer 2

1999 corresponds to t=5
2005 corresponds to t=11
??

Answer 3

hmmm... I think it should be t=6 and t=12 since 1999-1993=6
and 2005-1993=12

anyways...
we are going to let P be in millions
so we have the following two points given for (t,P)
(0,94) and (6,99)
and we are asked to find P when we have t=12

Answer 4

so we have the equation P=A e^(kt)
we can find A pretty easily using the first point (0,94)

Answer 5

(t,P)=(0,94)
put in 0 for t
and 94 for P
and solve for A.

Answer 6

okk holdon

Answer 7

lol nvm i just dont get this.

Answer 8

\[P=Ae^{kt} \\ (t,P)=(0,94) \\ 94=Ae^{k \cdot 0}\]
k*0 is...

Answer 9

is that the answer?

Answer 10

no

Answer 11

okay what do i have to do ?

Answer 12

We need to find A and k first before we can find what P corresponds to t=12

Answer 13

we are using the point (0,94) to find A

and then we are going to use (6,99) to find k

Answer 14

\[P=Ae^{kt} \\ (t,P)=(0,94) \\ 94=Ae^{k \cdot 0} \]
can you find A here?

Answer 15

e^(k*0) is a very pretty number

Answer 16

what number is it

Answer 17

assuming a isn't 0
then a^0=1

so e^(k*0)=e^0=?

Answer 18

lmfao thanks for your help but this is so hard and konfusing, so im just not gone take the test ..... :/

Answer 19

any number except 0 to the 0 power is just 1

Answer 20

\[P=Ae^{kt} \\ (t,P)=(0,94) \\ 94=Ae^{k \cdot 0} \\ 94=A(1) \\ 94=A \]

if you decide not to give up try to find k in
\[P=94 e^{kt}\]
using the other point which was (6,99)
replace t with 6 and P with 99 and solve for k

Answer 21

after that you will be already to figure out what happens in 2005