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?

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

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

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

5.3: 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.

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

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

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

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.

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?

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

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.

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

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

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?

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.