Question

In 1995, the population of a town was 33,500. It is decreasing at a rate of 2.5% per decade.

a. Identify whether the function is a growth or decay function.

b. Write the function, f(n), that expresses the population of the town after (n) decades.

c. What is the population in the year 2025 to the nearest hundred?

Answer 1

a. The function is a decay function

b. The function is f(n) = 33500
`(0.975)^(n)`

c. The population in the year 2025 is 31,800 to the nearest hundred

Explanation:

The form exponential growth function is y = a
`(1+r)^(x)` , where

• a is the initial value

• r is the rate of increase in decimal

The form exponential decay function is y = a
`(1-r)^(x)` , where

• a is the initial value

• r is the rate of decrease in decimal

∵ In 1995, the population of a town was 33,500

∴ a = 33,500

∵ It is decreasing at a rate of 2.5% per decade

→ The decrease means, it is a decay function, then use the 2nd form above

∴ f(n) = a
`(1-r)^(n)` , where f(n) is the population in n decades

∴ It is a decay function

a. The function is a decay function

∵ The rate is 2.5% per decade ⇒ 2.5% each 10 years

∴ r = 2.5%

→ Divide it by 100 to change it to decimal

∵ 2.5% = 2.5 ÷ 100 = 0.025

∴ r = 0.025

→ Substitute the values of a and r in the f(n)

∴ f(n) = 33500
`(1-0.025)^(n)`

∴ f(n) = 33500
`(0.975)^(n)`

b. The function is f(n) = 33500
`(0.975)^(n)`

∵ We need to find the population in 2025

→ Find how many decades from 1995 to 2025

∵ There are 20 years from 1995 to 2025

∵ 1 decade = 10 years

∵ n is the number of decads

∴ n = 2 ⇒ (20/10 = 2)

→ Substitute n in the function by 2

∵ f(2) = 33500
`(0.975)^(2)`

∴ f(2) = 31845.9375

→ Round it to the nearest hundred

∴ f(2) = 31800

c. The population in the year 2025 is 31,800 to the nearest hundred