HOW TO FIND THE GROWTH OR DECAY FACTOR OF AN EXPONENTIAL FUNCTION

Problem 1 :

You deposit $500 into a savings account that earns 6% interest each year and you do not make any deposits or withdrawals.

1) What is the initial amount?

2) Write an equation to model this situation. Use 𝑦 for the total value of the account and 𝑡 for the number of years.

3) What is the growth factor?

4) What is the value of this account in 25 years? 

Solution :

Initial deposit = $500

Interest rate = 6%

1) Initial amount = $500

2) Growth function y = a(1 + r%)^t

y = 500(1 + 6%)^t

Where y is the amount and t is the number of years.

3) Growth factor = (1 + 6%)

= (100 + 6) %

= 106%

Growth factor = 1.06

4)  y = 500(1 + 6%)^t

When t = 25

 y = 500(1 + 6%)²⁵

= 500(106%)²⁵

= 500(1.06)²⁵

= 2145.93

= 2146

Approximately $2146.

Problem 2 :

You are given a car worth $9500 on your 16th birthday. The value of the car declines by 15% per year.

1) What is the initial amount?

2) Write an equation to model this situation. Use 𝑦 for the total value of the car and 𝑡 for the number of years after your 16th birthday.

3) What is the decay factor?

4) What is the value of the car in 5 years? Round to two decimal places, as needed

Solution :

Initial deposit = $9500

Interest rate = 15%

1) Initial amount = $9500

2) Growth function y = a(1 + r%)^t

y = 9500(1 + 15%)^t

Where y is the amount and t is the number of years.

3) Growth factor = (1 + 15)%

= (100 + 15)%

= 115%

Growth factor is = 1.15

4)  y = 9500(1 + 15%)^t

When t = 5

 y = 9500(1 + 15%)⁵

= 9500(115%)⁵

= 9500(1.15)⁵

= 19107.8

= 19108

Approximately $19108

Problem 3 :

A new sports car sells for $35,000. The value of the car decreases by 18% annually.

a)  Which of the following choices models the yearly value of the car since its purchase?

b)  Find the decay factor ?

a)  y = 35000(1.18)^x      b)  y = 35000(0.82)^x

c)  y = 35000(0.18)^x      b)  y = 35000(-0.18)^x

Solution :

a)  Initial value = 35000

Decays 18% annually.

The required function will be in the form

y = a(1 - r%)^x

y = 35000(1 - 18%)^x

y = 35000[(100 - 18)%]^x

y = 35000[82%]^x

y = 35000(0.82)^x

So, option b is correct.

b) Decay factor is 0.82

Problem 4 :

At the end of last year, the population of Jason’s hometown was approximately 75,000 people. The population is growing at the rate of 2.4% each year. Which equation models the growth?

a)  y = 75000(1.24)^x      b)  y = 75000(1.024)^x

c)  y = 75000(0.76)^x      b)  y = 75000(0.976)^x

Solution :

Initial value = 75000

Grows 2.4% each year.

The required function will be in the form

y = a(1 - r%)^x

y = 75000(1 + 2.4%)^x

y = 75000[(100 + 2.4)%]^x

y = 75000(102.4%)^x

y = 75000(1.024)^x

So, option b is correct.

Problem 5 :

Select which exponential function below models this situation: Lupe bought a gold chain for $350, and the value of the chain increases by 4% per year. Use 𝑦 for the value of the chain and 𝑡 for the time in years

a) 𝑦 = 350^4x            b) 𝑦 = 350(4)^x

c) 𝑦 = 350(1.4)^x         d) 𝑦 = 350(1.04)^x

Solution :

Initial value = $350

Increase at the rate of 4%

y = a(1 + r%)^x

y = 350(1 + 4%)^x

y = 350(104%)^x

y = 350(1.04)^x

So, option d is correct.

Problem 6 :

Sam buys a car for $55,000. The value of the car depreciates at a rate of 15% per year.

a) What is the initial value?

b) Write an exponential function to model the value of the car (𝑦) after 𝑡 years.

c) What is the decay factor?

d) Find the value of the car after 4 years.

Solution :

a)  Initial value = 55000

b)  Decays 15% annually.

The required function will be in the form

y = a(1 - r%)^x

y = 55000(1 - 15%)^x

y = 55000[(100 - 15)%]^x

y = 55000[85%]^x

y = 55000(0.85)^x.

c)  Decay factor is 0.85

d)  When x = 4

y = 55000(0.85)^4.

= 55000(0.522)

= 28710.4

Value of car is approximately $28710.