Question

16)

ne population of the country Mathematica was 40 million in the year 2000. It has grown continuously by
2.7% in the years following. (pt)
a
write an exponential model for the population P, in millions, of the country t years after 2000 where
t 20
b) Using your model, what is the population today?
c) In what year will the population reach 90 million?

Answer 1

Explanation:

We would apply the formula for exponential growth which is expressed as

y = b(1 + r)^ t

Where

y represents the population after t years.

t represents the number of years.

b represents the initial population.

r represents rate of growth.

From the information given,

b = 40 × 10^6

r = 2.7% = 2.7/100 = 0.027

a) Therefore, exponential model for the population P after t years is

P = 40 × 10^6(1 + 0.027)^t

P = 40 × 10^6(1.027)^t

b) t = 2020 - 2000 = 20 years

P = 40 × 10^6(1.027)^20

P = 68150471

c) when P = 90 × 10^6

90 × 10^6 = 40 × 10^6(1.027)^t

90 × 10^6/40 × 10^6 = (1.027)^t

2.25 = (1.027)^t

Taking log of both sides to base 10

Log 2.25 = log1.027^t = tlog1.027

0.352 = t × 0.01157

t = 0.352/0.01157 = 30.4 years

Answer 2

Explanation:

The standard form for an exponential function is

`y=a(b)^x`

We could adjust it a little bit to fit our situation:

`P(t)=a(b)^t` where P(t) is the population after a certain number of years has gone by, a is the starting number (aka the number of people present when you started counting), b is the growth rate, and t is the time in years. Our P(t) is the unknown for part b, a = 40 million, t is 20 for part b, and the growth rate is tricky and earns a bit of an explanation.

b is a growth rate or a rate of decay. If b as a decimal is greater than 1, it makes the exponential function a growth function; if b as a decimal is greater than 0 but less than 1, it makes the function a decay function. We are told in our particular problem that this is a growth problem, not a decay problem. In particular, we are told that the population is growing continuously by 2.7%. If we state 2.7% as a decimal, we would have .027. If we use .027 as the rate, it would be decay because that decimal is less than 1. This is how we need to think about growth (in this case. This does get tricky sometimes, so I'm just going to explain growth in this context so as to not confuse you.). Think about having 100% of the population that is already present, then growing it by another 2.7% every year. That means that we would have 102.7% of the population after t years. 102.7% as a decimal is 1.027, which is greater than 1. That is our growth rate.

Filling in, we have our model:

`P(t)=40(1.027)^t` That's part a.

For part b, we will simply sub in a 20 in place of t:

`P(t)=40(1.027)^(20)` which gives us a population after 20 years of

P(t) = 68.1 million

For part c, we are asked after how many years, t, will the population P(t) be 90 million. We will sub in 90 for P(t) and solve for t this time:

`90=40(1.027)^t`

Begin by dividing both sides by 40 to get:

`2.25=1.027^t`

In order to get the t out of its current exponential position, we will take the natural log of both sides, allowing us (by the power rule of logs) to bring the t down out front. I'm going to do that in one step:

`ln(2.25)=xln(1.027)`

We will divide both sides by ln(1.027) to get x alone:

`x=(ln(2.25))/(ln(1.027))`

Do this on your calculator to get that

t = 30.4 years, or rounded, 30 years.

That means, in this context, that in 2030 the population will hit 90 million if the growth trend continues at 2.7% per year.