Examples of Applications of Exponential Functions
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We have seen in past courses that exponential functions are used to represent growth and decay.
 

Where a = initial amount
           b = growth/decay factor
           x = time
           y = ending amount

Growth
When a > 0 and b > 1,
the function models growth.
(b is called the growth factor)
(a represents the initial amount)
 

Decay
When  a > 0 and 0 < b < 1,
the function models decay.
(b is called the decay factor)
(a represents the initial amount)


  Let's look at examples of these exponential functions at work.

1. Population:  The population of the popular town of Smithville in 2003 was estimated to be 35,000 people with an annual rate of increase (growth) of about 2.4%. 

a.)  What is the growth factor for Smithville? 
                 After one year the population would be 35,000 + 0.024(35000).
                 By factoring we see that this is 35,000(1 + 0.024) or 35,000(1.024).
                 The growth factor is 1.024.  (Remember that the growth factor is greater than 1.)

b.)  Write an equation to model future growth.
                
                  where y is the population;  x is the number of years since 2003

c.)  Use your equation to estimate the population in 2007 to the nearest hundred people.
                                

 

 

2. Money:  Marisa invests $300 at a bank that offers 5% compounded annually.

a.)  What is the growth factor for the investment? 
                 The growth factor is 1.05.  (Remember that the growth factor is greater than 1.)

b.)  Write an equation to model the growth of the investment.
                
                  where y is the money;  x is the number of years since the initial investment

c.)  How many years will it take for the initial investment to double?              
  Algebraic solution:and solve for x.

Answer:  approximately 14.2 years      		Graphical solution:
   

    

 

 

3. Depreciation:  Matt bought a new car at a cost of $25,000.  The car depreciates approximately 15% of its value each year. 

a.)  What is the decay factor for the value of this car? 
                 The decay factor is 1 - 0.15 = 0.85.  (Remember that the decay factor is 0 <  b < 1.)

b.)  Write an equation to model the decay value of this car.
                
                  where y is the value of the car;  x is the number of years since new purchase

c.)  What will the car be worth in 10 years?
                 
                   Answer:  $4,921.86

 

 

Where y0 = initial amount
           k = constant of proportionality
           t = time
           y = ending amount

Growth
models growth since e > 1.

Exponential functions with the base e are often used to describe continuous growth or decay.

  

Decay

 models decay since which is between 0 and 1

 

4. More Money:  Most banks compound interest more than once a year.  When interest is compounded n times per year for t years at an interest rate of r, the principal, P, grows to the amount A given by the formula:

If you allow n to get larger and larger, you will discover that as n increases,
approaches e.

If compounding takes place without interruption (called compounded continuously), this formula becomes:
Compound Quarterly:
Jose invests $500 at a bank offering 10% compounded quarterly.  Find the amount of the investment at the end of 5 years (if untouched).
 Compound Continuously:
Tamika invests $500 at a bank offering 10% compounded continuously.  Find the amount of the investment at the end of 5 years (if untouched).

                        

 

 

5. Half-Life:  Radium-226, a common isotope of radium, has a half-life of 1620 years.  Professor Korbel has a 120 gram sample of radium-226 in his laboratory. 
a.)  Find the constant of proportionality for radium-226.
It should make sense that k is a negative value,
since this is an example of decay.

b.)  How many grams of the 120 gram sample will remain after 100 years?
The "official" equation for radium-226 is now written using the value for k.  Substitute the time and solve.

 

 

6. Bacteria Growth:  A certain strain of bacteria that is growing on your kitchen counter doubles every 5 minutes.  Assuming that you start with only one bacterium, how many bacteria could be present at the end of 8 hours?
8 hours = 480 minutes
at 5 minute intervals gives 96 time intervals.


 

How to use your
TI-83+/84+ graphing calculator with exponentials.
Click calculator.