Chapter 5: Exponential and Logarithmic Chapter 5: Exponential and Logarithmic FunctionsFunctions5.3: Applications of Exponential 5.3: Applications of Exponential FunctionsFunctions

Essential Question: How do you find a growth factor and a decay factor?

5.3: Applications of Exponential 5.3: Applications of Exponential FunctionsFunctionsCompounding 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 5.3: Applications of Exponential FunctionsFunctionsCompounding 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

r

n

5.3: Applications of Exponential 5.3: Applications of Exponential FunctionsFunctionsDifferent 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

1(3)0.06414000(1 ) $4818.20A

4(3)0.06444000(1 ) $4839.32A

12(3)0.064124000(1 ) $4844.21A

365(3)0.0643654000(1 ) $4846.60A

5.3: Applications of Exponential 5.3: Applications of Exponential FunctionsFunctionsContinuous 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 5.3: Applications of Exponential FunctionsFunctions 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

1111(1 ) $2.00A

4(1)141(1 ) $2.4414A

2(1)121(1 ) 2.25A

12(1)1121(1 ) $2.6130A

365(1)13651(1 ) $2.71457A

8760(1)187601(1 ) $2.718127A

525600(1)15256001(1 ) $2.7182792A

31536000(1)1315360001(1 ) $2.7182825A

5.3: Applications of Exponential 5.3: Applications of Exponential FunctionsFunctions$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 5.3: Applications of Exponential FunctionsFunctionsContinuous 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

0.05(3)4000

$4647.34

rtA Pe

e

5.3: Applications of Exponential 5.3: Applications of Exponential FunctionsFunctionsAssignment

◦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 Chapter 5: Exponential and Logarithmic FunctionsFunctions5.3: Applications of Exponential 5.3: Applications of Exponential FunctionsFunctionsDay 2Day 2

Essential Question: How do you find a growth factor and a decay factor?

5.3: Applications of Exponential 5.3: Applications of Exponential FunctionsFunctionsExponential 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 5.3: Applications of Exponential FunctionsFunctionsExponential 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 5.3: Applications of Exponential FunctionsFunctionsBacteria 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

15 0.25 7.6 7.6 7.6a

5.3: Applications of Exponential 5.3: Applications of Exponential FunctionsFunctionsRadioactive 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

( ) (0.5)xhf x P

5.3: Applications of Exponential 5.3: Applications of Exponential FunctionsFunctionsRadioactive 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?

( ) (0.5)xhf x P

1620( ) 100(0.5)x

f x

8001620( ) 100(0.5) 71.01f x mg 16001620( ) 100(0.5) 50.43f x mg

32001620( ) 100(0.5) 25.43f x mg

5.3: Applications of Exponential 5.3: Applications of Exponential FunctionsFunctionsAssignment

◦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.