Chapter 5: Exponential and Logarithmic Functions 5.3: Applications of Exponential Functions

1. Chapter 5: Exponential and Logarithmic Functions5.3: 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 • $2.7182825 is the same as the number e to five decimal places (e = 2.71828182…) • 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 Functions5.3: Applications of Exponential FunctionsDay 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 1000 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 • 7.60.2x • After 24 hours, there are 1000 • 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.