Question

The formula A=22.9e^0.0183t models the population of Texas, A, in millions, t years after 2005.
a)--
b) when will the population of Texas reach 27 million?

Answer 1

Plug in:
27 = 22.9e^(0.0183t)
Divide by 22.9:
27/22.9 = e^(0.0183t)
Take the natural logarithm:
ln(27/22.9) = ln(e^(0.0183t))
ln(27/22.9) = 0.0183t
Divide by 0.0183:
ln(27/22.9)/0.0183 = t
t ≈ 8.999
Would this be correct?

Answer 2

well substitute it and check..

Answer 3

But wait idk how to find "What was the population of texas in 2005" im stuck there

Answer 4

its extremely close... of you had used t = 9 you get the correct answer

Answer 5

dude but im serious idk how to do this; its the only problem i dont get>.<

Answer 6

it's not 9, but more like 8.999997565

which I guess is close enough

Answer 7

well 2005 let t = 0

so\[A_{0} = 22 900 000\]

Answer 8

well I found

\[22.9 \times e^{0.0183 \times 9} = 27\]

Answer 9

and Ao is the initial population

Answer 10

rounding error maybe

Answer 11

^ Wait for which one?

Answer 12

so 9 years after 2005 is...

Answer 13

thats when the population will reach 27 mill

Answer 14

and the beauty of the general equation

A = P*e^(kt)

is that when t =0, the initial population is P

So in this case, when t = 0 (ie the year 2005), the population is 22.9 million

Answer 15

I get you campbell but i dont get you jim where do you plug in what?

Answer 16

A = 22.9e^(0.0183t)

A = 22.9e^(0.0183*0) ... plug in t = 0

A = 22.9e^(0)

A = 22.9(1)

A = 22.9

Answer 17

So when t = 0, A = 22.9

Which means that in the year 2005, the population is 22.9 million

Answer 18

Thats the population f texas in 2005 right?

Answer 19

yes

Answer 20

thanks!:D

Answer 21

you're welcome

Answer 22

glad to help

Answer 23

Dude you both guys just help me big time! Im really thankful no lie!:D

Answer 24

lol that's great that we were able to help you out