Question

In​ 2000, the population of a country was approximately 6.31 million and by 2069 it is projected to grow to 12 million. Use the exponential growth model Upper A equals Upper A 0 e Superscript kt​, in which t is the number of years after 2000 and Upper A 0 is in​ millions, to find an exponential growth function that models the data. b. By which year will the population be 7 ​million?

Answer 1

Answer:

Step-by-step explanation:

This is your exponential growth function for population:

$A=A_{0}e^{kt}$   and these are your initial conditions with the year 2000 being t = 0

(0, 6.31) and (69, 12)

We will use those values to find the equation that models this population growth.  In the coordinates, the first number is the time in years, t; the second number is the population after a certain time t goes by.  In other words, the second number represents the A in our model.  Using those values from the first set of coordinates will help us solve for A₀:

$6.31=A_{0}e^{k0}$  which is basically e raised to the power of 0 which is equal to 1, so we get from that first set of coordinates that A₀ = 6.31

Now we will use that along with the numbers in the second coordinate pair to find the value for k:

$12=6.31e^{69k}$

Begin by dividing both sides by 6.31 to get

$1.901743265=e^{69k}$  and take the natural log of both sides since natural logs and e's undo each other:

$ln(1.901743265)=ln(e^{69k})$  Simplifying both sides give us:

.6427709734 = 69k so

k = .0093155

Now we can finally write the equation that models this population as

$A=6.31e^{.0093155t}$ and we can answer the question about which year, x, will the population be 7 million, A.

$7=6.31e^{.0093155t}$

Begin by dividing both sides by 6.31 to get

$1.109350238=e^{.0093155t}$ and again take the natural log of both sides:

$ln(1.109350238)=ln(e^{.0093155t})$ and simplify to

.1037744728=.0093155t so

t ≈ 11

That means that in the year 2011 the population will be 7 million