Question

Population Growth The population P (in thousands) of Oriando,Florida from 1980 through 2009 can be modeled by p = 788e0.0248t

where t = 0 corresponds to 1980.
(a) What was the population of Phoenix in 2009?
(b) In what year will Phoenix have a population of 2,000,000?

Answer 1

(a). 1,617,599

(b). In year 2018.

Explanation:

We have been given that population P (in thousands) of Phoenix, Arizona from 1980 through 2009 can be modeled by
`p=788e^(0.0248t)`, where,
`t = 0` corresponds to 1980.

(a). To find the population of Phoenix in 2009, we will substitute
`t=2009-1980=29` in our given equation as:

`p=788e^(0.0248*29)`

`p=788e^(0.7192)`

`p=788(2.0527903213187)`

`p=1617.5987731991356`

Since our given population is in thousands, so we will multiply 1617.5987731991356 by 1,000 as:

`1617.5987731991356* 1,000=1,617,598.7731991356\approx 1,617,599`

Therefore, the population of Phoenix would be 1,617,599 in 2009.

(b). To find the time, when Phoenix have a population of 2,000,000, we can set an equation as:

`2,000=788e^(0.0248t)`

We will substitute 2,000 as our given equation represents population in thousands.

Let us solve for t.

`(2,000)/(788)=(788e^(0.0248t))/(788)`

`2.538071065989=e^(0.0248t)`

Now, we will use natural log to solve for t as:

`\text{ln}(2.538071065989)=\text{ln}(e^(0.0248t))`

`\text{ln}(2.538071065989)=0.0248t*\text{ln}(e)`

`0.9314043696842032297=0.0248t*1`

`(0.9314043696842032297)/(0.0248)=(0.0248t)/(0.0248)`

`37.5566278098469=t`

`t\approx 38`

Now, we need to find 38 years after 1980 that is
`1980+38=2018`.

Therefore, in 2018 Phoenix will have a population of 2,000,000.