Question

Below is a table with estimates of would population since 1AD

Answer 1

1)

`P(t)=1633e^(0.0087t)`

Step-by-step explanation:

Since we are taking t = 0 as 1900. The initial population is the population in 1900:

`P_0=1633`

We can start writting the exponential model:

`P(t)=1633e^(kt)`

:

We know that the population in 1950 (t = 50 in this model) is 2525. Then, we can find k by:

`P(50)=2525=1633e^(k\cdot50)`

And solve for k:

`(2525)/(1633)=e^(50k)`

Now, we can apply natural logarithm on both sides:

`\ln((2525)/(1633))=\ln(e^(50k))`

Since natural log and the exponential functions are inverse functions:

`\ln((2525)/(1633))=50k`
`k=(\ln((2525)/(1633)))/(50)`
`k\approx0.0087`

Thus, the exponential model is:

`P(t)=1633e^(0.0087t)` Question 2

Now, we need to use the model we just create to find the predicted population by the model:

At year 1900, t = 0 (that's how the problem asked us to define t)

Then

`P\left(0\right)=1633e^(0.0087\cdot0)=1633e^0=1633`

At year 1920, t = 20

`P(20)=1633e^(0.0087\cdot20)=1633e^(0.174)=1943`

At year 1940, t = 40:

`P(40)=1633e^(0.0087\cdot40)=1633e^(0.348)=2313`

At year 1960, t = 60

`P(60)=1633e^(0.0087\cdot60)=1633e^(0.522)=2752`

At year 1980, t = 80

`P(80)=1633e^(0.0087\cdot80)=1633e^(0.696)=3275`

At year 2000, t = 100

`P(100)=1633e^(0.0087\cdot100)=1633e^(0.87)=3898`

At year 2020, t = 120

`P(120)=1633e^(0.0087\cdot120)=1633e^(1.044)=4639`

At year 2040, t = 140

`P(140)=1633e^(0.0087\cdot140)=1633e^(1.218)=5520`