1. Chapter 5: Exponential and Logarithmic Functions 5
Chapter 5: Exponential and Logarithmic Functions 5.3: Applications of Exponential Functions
Essential Question: How do you find a growth factor and a decay factor?

2. 5.3: Applications of Exponential Functions
Compounding Interest
If you invest $6000 at 8% interest, compounded annually, how much is in the account at the end of 10 years?
After one year, the account balance is
6000(1.08) [“1.08” to get your original back + 8%]
After two years, the account balance is
[6000(1.08)](1.08), or 6000(1.08)2
Because the balance changes by a factor of 1.08 every year, the balance in the account at the end of year x is given by
6000(1.08)x
So the balance after 10 years (to the nearest penny) is
6000(1.08)10 = $12,953.55

3. 5.3: Applications of Exponential Functions
Compounding Interest
A = P(1 + )nt, where
A = Amount at the end of compounding
P = Principal (starting) amount
r = Interest rate (as a decimal)
n = number of times compounding per year
t = number of years
Why didn’t you use n in the previous problem?
Interest was compounded yearly, so n = 1

4. 5.3: Applications of Exponential Functions
Different Compounding Periods
Determine the amount that a $4000 investment over three years at an annual rate of 6.4% for each compounding period.
Annually
Quarterly
Monthly
Daily
Notice that the more frequently interest is compounded, the larger the final amount

5. 5.3: Applications of Exponential Functions
Continuous Compounding
Suppose you invest $1 for one year at 100% annual interest, compounded n times per year. Find the maximum value of the investment in one year.
Observe what happens to the final amount as n grows larger and larger.

6. 5.3: Applications of Exponential Functions
Compounding Continuously
Annually
Semiannually
Quarterly
Monthly
Daily
Hourly: 365 • 24 = 8760 periods
Every minute: 8760 • 60 = 525,600 periods
Every second: 525,600 • 60 = 31,536,000 periods

7. 5.3: Applications of Exponential Functions
$ is the same as the number e to five decimal places (e = …)
So if we’re compounding continuously (instead of some fixed period), we have the equation
A = Pert, where
A = Amount at the end of compounding
P = Principal (starting) amount
r = Interest rate (as a decimal)
t = number of years

8. 5.3: Applications of Exponential Functions
Continuous Compounding
If you invest $4000 at 5% annual interest compounded continuously, how much is in the account at the end of 3 years?
Use the equation A = Pert

9. 5.3: Applications of Exponential Functions
Assignment
Page 353
Problems 1 – 31, odd problems
Show your work
That means: if you don’t show the equations you’re putting into the calculator, you don’t get credit.

10. Chapter 5: Exponential and Logarithmic Functions 5
Chapter 5: Exponential and Logarithmic Functions 5.3: Applications of Exponential Functions Day 2
Essential Question: How do you find a growth factor and a decay factor?

11. 5.3: Applications of Exponential Functions
Exponential Growth
Population Growth
The world population in 1950 was about 2.5 billion people and has been increasing at approximately 1.85% per year. Write the function that gives the world population in year x, where x = 0 corresponds to 1950.
This is similar to word problems from the last chapter.
Think:
In year 0 (1950), the population is 2.5 billion
In year 1 (1951), the population is 2.5(1.0185)
In year 2: population is 2.5(1.0185)(1.0185) = 2.5(1.0185)2
If that pattern continues, the population in year x is f(x) = 2.5(1.0185)x

12. 5.3: Applications of Exponential Functions
Exponential growth can be described by a function of the form:
f(x) = Pax, where:
P is the initial quantity when x = 0
a > 1 is the factor by which the quantity changes when x increases by 1.
Exponential decay works exactly the same, except “a” (the multiplying factor) is between 0 and 1.

13. 5.3: Applications of Exponential Functions
Bacteria Growth
At the beginning of an experiment, a culture contains bacteria. Five hours later, there are 7600 bacteria. Assuming that the bacteria grow exponentially, how many bacteria will there be after 24 hours?
Because there are 1000 bacteria after 5 hours, f(5) = 7600 and 1000a5 = 7600
So we need to figure out a (the acceleration)
1000a5 /1000 = 7600 / 1000
a5 = 7.6
The function is: f(x) = 1000 • x
After 24 hours, there are • 7.6(0.2)(24) ≈ 16,900,721 bacteria

14. 5.3: Applications of Exponential Functions
Radioactive Decay
The amount of a radioactive substance that remains is given by the function
where
P = the initial amount of the substance
x = 0 corresponds to the time since decay began
h = the half-life of the substance

15. 5.3: Applications of Exponential Functions
Radioactive Decay
Example:
The half-life of radium is 1620 years. Find the rule of the function that gives the amount remaining from an initial quantity of 100 milligrams of radium after x years.
How much radium is left after 800 years?
After 1600 years?
After 3200 years?

16. 5.3: Applications of Exponential Functions
Assignment
Page 355
Problems:
39 – 51, odds
Skip #47
Skip any part c
Show your work
That means: if you don’t show the equations you’re putting into the calculator, you don’t get credit.