WRITE AN EXPOENTIAL GROWTH OR DECAY FUNCTION FOR EACH SITUTATION

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 :

The given is exponential growth function.

Let P be the population, x be the number of years and a be the initial population.

r is the rate of increasing.

P = a(1 + r%)^x

P = 5000(1+3%)^x

P = 5000(1+0.03)^x

P = 5000(1.03)^x

When x = 10

P = 5000(1.03)¹⁰

P = 6719.58

So, 6720 is the population after 10 years.

Problem 2 :

Amy makes an initial investment of $5000. The investment loses 13.5% each year. Find the amount Amy has at the end of 8 years.

Solution :

The given is exponential decay function.

Let A be the amount, x be the number of years and a be the initial investment.

A = a(1 - r%)^x

r is the rate of decreasing.

A = 5000 (1 -13.5%)^x

A = 5000 (1 - 0.135)^x

A = 5000 (0.865)^x

When x = 8

A = 5000 (0.865)⁸

A = 1567.1

A = 1567

So, at the end of 8 years the amount will be $1567.

Problem 3 :

Suppose you deposit $1000 in a college fund that pays 7.2% interest compounded annually. Find the account balance after 5 years.

Solution :

The given is exponential growth function.

Let A be the amount, x be the number of years and a be the amount deposited initially.

A = a(1 + r%)^x

r is the rate of increasing.

A = 1000(1 + 7.2%)^x

A = 1000(1 + 0.072)^x

A = 1000(1.072)^x

When x = 5

A = 1000(1.072)⁵

A = 1415.7

So, after 5 years he will receive $1416.

Problem 4 :

Suppose you deposit $1000 in a college fund that pays 7.2% interest. Find the account balance after 5 years.

Suppose the account above paid interest quarterly instead of annually. Find the account balance after 5 years.

Solution :

The given is exponential growth function.

Let A be the amount, x be the number of years and a be the amount deposited initially.

A = a(1 + (r/4)%)^4x

r is the rate of increasing.

A = 1000(1 + 0.018)^4x

When x = 5

A = 1000(1.018)²⁰

A = 1428.74

So, after 5 years he will receive $1429.

Problem 5 :

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 :

P = a(1 + r%)^x

a = 475000, r = 3.75%

For what value of x, P will be > 1000, 000

P = 475000(1+3.75%)^x

1000, 000 > 475000(1+3.75%)^x

(1000, 000/475000) > (1+3.75%)^x

2.11 > (1+3.75%)^x

To solve for x, take log on both sides.

log (2.11) = log (1+3.75%)^x

log (2.11) = x log (1.0375)

x = log (2.11) / log (1.0375)

x = 0.324/0.015

x = 21.6

x = 22

After 22 years the population will be more than 1 million.

Problem 6 :

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

• When will the population of Leavetown drop below 50,000 (to the nearest year)?
• What will the population of Leavetown be 100 years from now?

Solution :

P = a(1 + r%)^x

a = 123,000, r = 2.375%

For what value of x, P will be < 50,000

a(1 + r%)^x < 50,000

123,000(1 - 2.375%)^x < 50,000

123,000(0.97)^x < 50,000

(0.97)^x < 50,000/123000

(0.97)^x < 0.41

x log(0.97) = log (0.41)

x = log (0.41) / log(0.97)

x = 37

After 37 years the population will drop below 50000.

Population after 100 years :

= 123,000(1 - 2.375%)^x

= 123,000(0.97)¹⁰⁰

= 11118

Problem 7 :

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 :

a = 87,000,000, growth rate = 2.4%

P = 100,000,000

P = (1 + r%)^x

100,000,000 = 87,000,000(1 + 2.4%)^x

100,000,000/87,000,000 = (1 + 2.4%)^x

1.15 = (1 + 2.4%)^x

log (1.15) = x log(1.024)

x = 0.061/0.010

x = 6.1

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

Problem 8 :

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?

b) 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 :

a)  a = 6250, r = 3.75%

P = a(1 + r%)^x

P = 6250(1 + 3.75%)^x

x = 1915-1890 ==> 25

P = 6250(1 + 0.0375)²⁵

P = 6250(1.0375)²⁵

P = 15689

b) x = 1940-1890 ==> 50

P = 6250(1.0375)⁵⁰

P = 39381

c) 

x = 2003-1890 ==> 113

P = 6250(1 + 0.0375)¹¹³

P = 6250(1.0375)²⁵

P = 400439

d) P = 1,000,000

1,000,000 = 6250(1 + 0.0375)^x

1,000,000/6250 = (1 + 0.0375)^x

160 = (1.0375)^x

log 160 = x log (1.0375)

x = log 160 /  log (1.0375)

x = 2.204/0.015

x = 147

Problem 9 :

A radioactive element decays at a rate of 5% annually. There are 40 grams of the substance present.

a) How much of the substance remains after 30 years 

b) When will the amount of the substance drop to below 20 grams ?

Solution :

P = a(1 - r%)^x

a = 40, r = 5%

P = 40(1 - 5%)^x

P = 40(0.95)^x

a) When x = 30

P = 40(0.95)³⁰

P = 8.58

b) P = 20, x = ?

20 = 40(0.95)^x

1/2 = (0.95)^x

0.5 = (0.95)^x

log 0.5 = x log 0.95

-0.30/(-0.022) = x

x = 13.6

x = 14