EXPONENTIAL GROWTH OR DECAY FUNCTIONS

Problem 1 :

y = 600(1 + 0.25)^t

(i) Does this function represent exponential growth or exponential decay?

(ii) What is the initial value?

(iii) What is the rate of growth or rate of decay?

Solution :

Part (i) :

Since the given function is in the form of y = a(1 + r)^t, it represents exponential growth.

Part (ii) :

Comparing y = a(1 + r)^t and y = 600(1 + 0.25)^t, we get

a = 600

So, the initial value is 600.

Part (iii) :

From part (ii) above, we have r = 0.25 or 25%.

So, the rate of growth is 25%.

Problem 2 :

y = 64(1 - 0.03)^t

(i) Does this function represent exponential growth or exponential decay?

(ii) What is the initial value?

(iii) What is the rate of growth or rate of decay?

Solution :

Part (i) :

Since the given function is in the form of y = a(1 - r)^t, it represents exponential decay.

Part (ii) :

Comparing y = a(1 - r)^t and y = 64(1 - 0.03)^t, we get

a = 64

So, the initial value is 64.

Part (iii) :

Comparing y = a(1 - r)^t and y = 64(1 - 0.03)^t, we get

r = 0.03 or 3%.

So, the rate of decay is 3%.

Problem 3 :

y = 100(1.25)^t

(i) Does this function represent exponential growth or exponential decay?

(ii) What is the initial value?

(iii) What is the rate of growth or rate of decay?

Solution :

Part (i) :

Comapring y = ab^t and y = 100(1.25)^t, we get

b = 1.25 > 1

So, the given function represents exponential growth.

Part (ii) :

Comapring y = ab^t and y = 100(1.25)^t, we get

a = 100

So, the initial value is 100.

Part (iii) :

Write the given function in the form of y = a(1 + r)^t.

y = 100(1.25)^t ----> y = 100(1 + 0.25)^t

Comparing y = a(1 + r)^t and y = 100(1 + 0.25)^t, we get

r = 0.25 or 25%

So, the rate of growth is 25%.

Problem 4 :

y = 1500(0.65)^t

(i) Does this function represent exponential growth or exponential decay?

(ii) What is the initial value?

(iii) What is the rate of growth or rate of decay?

Solution :

Part (i) :

Comapring y = ab^t and y = 1500(0.65)^t, we get

b = 0.65 < 1

So, the given function represents exponential decay.

Part (ii) :

Comapring y = ab^t and y = 1500(0.65)^t, we get

a = 1500

So, the initial value is 1500.

Part (iii) :

Write the given function in the form of y = a(1 - r)^t.

y = 1500(0.65)^t ----> y = 100(1 - 0.35)^t

Comparing y = a(1 - r)^t and y = 100(1 - 0.35)^t, we get

r = 0.35 or 35%

So, the rate of decay is 35%.

Problem 5 :

The first day of a concert had an attendance of 500. The attendance y increases by 5% each day.

(i) Write an exponential growth function to represent this situation.

(ii) How many people will attend on the 10th day? Round your answer to the nearest person.

Solution :

Part (i) :

The exponential growth function to represent this situation :

y = 500(1 + 0.05)^t

or

y = 500(1.05)^t

Part (ii) :

To know the number of people attending the concert on the 10th day, substitute t = 10.

y = 500(1.05)¹⁰

y ≈ 814

About 814 people will attend on the 10th day.

Problem 6 :

A company earns a profit of $25,000 this year. Assume that the profit of the company decreases by 6% per year in the upcoming years.

(i) Write an exponential growth function to represent this situation.

(ii) Find the profit of the company in 8 years from now. Round your answer to the nearest dollar.

Solution :

Part (i) :

The exponential decay function to represent this situation :

y = 25000(1 - 0.06)^t

or

y = 250000(0.94)^t

Part (ii) :

To know the profit in 8 years from now, t = 8.

y = 25000(0.94)⁸

y ≈ 15239

The profit of the company in 8 years from now is about $15239.

Problem 7 :

Ken bought $2000 worth of stocks in 2012. Assume that the value of stocks increases by 5% per year in the upcoming years.

(i) Write an exponential growth function to represent this situation.

(ii) What will the stocks be worth in 2017? Round your answer to the nearest cent.

Solution :

Part (i) :

The exponential decay function to represent this situation :

y = 2000(1 + 0.05)^t

or

y = 2000(1.05)^t

Part (ii) :

Time period from 2012 to 2017 :

= 2017 - 2012

= 5 years

To know the worth of the stocks in 2017, substitute t = 5 into the exponential decay function found in part (i)..

y = 2000(1.05)⁵

y ≈ 2553

The worth of the stocks in 2017 is about $2553.

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