POPULATION GROWTH AND DECAY WORD PROBLEMS

Write an exponential function to model each situation. Find each amount at the end of the specified time. Round your answers to the nearest whole number.

Problem 1 :

A town with a population of 5,000 grows 3% per year. Find the population at the end of 10 years.

Solution :

y = a(1 + r)^t

Here a = initial population, r = increasing rate and t = number of years

Initial population = 5000

Increasing rate = 3% and

number of years = 10

y = 5000(1 + 3%)¹⁰

y = 5000(1 + 0.03)¹⁰

y = 5000(1.03)¹⁰

y = 5000(1.344)

y = 6720

Problem 2 :

The population of Boomtown is 475,000 and is increasing at a rate of 3.75% each year. When will the population exceed 1 million people (to the nearest year)?

Solution :

Initial population = 475,000

Increasing rate = 3.75%

y = a(1 + r)^t

475000(1 + 3.75%)^t  > 1000000

475000(1 + 0.0375)^t  > 1000000

Dividing by 475000 on both sides.

(1 + 0.0375)^t  > 2.105

(1.0375)^t  > 2.105

To solve use equal sign, we get

t log(1.0375) = log(2.105)

t(0.015) = 0.323

t = 0.323/0.015

t = 21.53

So, it will take 21 years tp reach the population of 1 million.

Problem 3 :

The population of Leave town is 123,000 and is decreasing at a rate of 2.375% each year.

• When will the population of Leave town drop below 50,000 (to the nearest year)?

• What will the population of Leave town be 100 years from now?

Solution :

(i)  y = a(1 - r)^t

Initial population = 123000

When will the population become below 50000.

123000(1 - 2.375%)^t < 50000

Divide by 123000 on both sides.

(1 - 0.02375)^t < 0.4065

(0.97625)^t < 0.4065

Take log on both sides, we get

t log (0.97625) < log (0.4065)

t(-0.0104) < -0.3909

t < 0.3909/0.0104

t < 37.58

So, it will take 38 years.

(ii)  y = a(1 - r)^t

After 100 years the population will be.

y = 123000(1 - 2.375%)¹⁰⁰ 

y = 123000(0.97625)¹⁰⁰ 

y = 123000(0.09038)

y = 11118

Problem 4 :

Problem The 1989 population of Mexico was estimated at 87,000,000. The annual growth rate is 2.4%. When will the population reach 100,000,000 (to the nearest year)?

Solution :

Population is increasing, so we will use the formula

y = a(1 + r)^t

Initial population = 87000000

Growth rate = 2.4%

After how many year the population will become 100,000,000.

87000000(1 + 2.4%)^t = 100,000,000.

(1 + 0.024)^t = 100,000,000/87000000

(1 + 0.024)^t = 100,000,000/87000000

(1.024)^t = 1.149

Taking log on both sides, we get

log (1.024)^t = log (1.149)

t log(1.024) = log(1.149)

t = log(1.149)/log(1.024)

t = 0.0603/0.0102

t = 5.911

So, the population will become 100,000,000 after 6 years.

Problem 5 :

The population of Small town in the year 1890 was 6,250. Since then, it has increased at a rate of 3.75% each year.

a) What was the population of Small town in the year 1915?

a) In 1940?

c) What will the population of Small town be in the year 2003?

d) When will the population reach 1,000,000 (to the nearest year)?

Solution :

Population is increasing :

y = a(1 + r)^t

(a) Initial population a = 6250, increasing rate = 3.75%

Population will be at 1915 :

Difference in years = 1915 - 1890 ==> 25 years

y = 6250(1 + 3.75%)²⁵

y = 6250(1 + 0.0375)²⁵

y = 6250(1.0375)²⁵

y = 6250(2.510)

y =15688

(b) In 1940 :

Difference between 1940 and 1890

= 1940-1890

= 50 years

y = 6250(1 + 3.75%)⁵⁰

y = 6250(1.0375)⁵⁰

y = 6250(6.3)

y = 39380

So, at 1914 the population will be 39380.

c) Population at 2003 :

= 2003 - 1890

= 113

y = 6250(1.0375)¹¹³

y = 6250(64.07)

y = 400438.

(d) After how many the population will be become 1,000,000

1,000,000 = 6250(1 + 3.75%)^t

Dividing by 6250 on both sides.

160 = (1.0375)^t

log 160 = t log (1.0375)

2.204 = t(0.015)

t = 2.204/0.015

t = 146.9

t = 147 

After 147 years, the population will be 1,000,000.