The general form of an exponential growth or decay function is
y = ab^{t}
Here,
y ---> final value
a ---> initial value
t ---> time period
In the exponential function above, if 0 < b < 1, then it is exponential decay function.
In the exponential function above, if b > 1, then it is exponential growth function.
The above exponential function can also be written as
y = a(1 ± r)^{t}
Here, b is replaced by (1 ± r), where r is the rate of growth or decay.
y = a(1 + r)^{t }----> exponential growth
y = a(1 - r)^{t }----> exponential decay
Usually the rate of growth or decay r is given in percentage. When we use the value of r in the formula, we will convert the percentage to decimal.
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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