Essential Question: What are some types of real-life problems where exponential or

logarithmic equations can be used?

Compound 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 per year compounded t = number of years

◦ We’re only going to be dealing with situations where interest is compounded yearly, so we will use the formula: A = P(1 + r)t

r

n

Example◦ A computer valued at $6500 depreciates at the

rate of 14.3% per year.◦ Write a function that models the value of the

computer. A = P(1 + r)t

Do we know A (the value at the end)? Do we know P (the value at the start)? Do we know r (the rate)? Do we know t (the number of years)?

◦ Find the value of the computer after three years.

NoYes, 6500Yes, 0.143No

A = 6500(1 – 0.143)t = 6500(0.857)t

A = 6500(0.857)3 = $4091.25

depreciates

Continuously Compounding Interest◦ So let’s bring “n” back for just one example:◦ 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.

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

$2.7182825 is the same as the number e to five decimal places (e = 2.71828182…)

So if we’re compounding continuously (instead of yearly), we can instead use 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

Example◦ Suppose you invest $1050 at an annual interest

rate of 5.5% compounded continuously. How much money will you have in the account after 5 years? A = Pert

Do we know A (the value at the end)? Do we know P (the value at the start)? Do we know r (the rate)? Do we know t (the number of years)?

NoYes, 1050Yes, 0.055

Yes, 5

A = 1050 e(0.055 • 5) = $1382.36

Your Turn◦ Suppose you invest $1300 at an annual interest

rate of 4.3% compounded continuously. How much money will you have in the account after 3 years? A = Pert

A = 1300(e)(0.043 • 3) = $1479.00

Assignment◦ Worksheet◦ Round all problems appropriately (if talking about

money: 2 decimal places; if talking about population: nearest integer)

Essential Question: What are some types of real-life problems where exponential or

logarithmic equations can be used?

Radioactive Decay◦ The half-life of a radioactive substance is the time

it takes for half of the material to decay.◦ It’s most often used for things like carbon-14

dating, which determines how old a substance is◦ The function for radioactive decay is

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)xhy P

Example◦ A hospital prepares a 100-mg supply of

technetium-99m, which has a half-life of 6 hours. Write an exponential function to find the amount of technetium-99m.

◦ Do we know y (the value at the end)? Do we know P (the value at the start)? Do we know x (the amount of time)? Do we know h (the half-life)?

◦ Use your function to determine how much technetium-

99m remains after 75 hours.

(0.5)xhy P

NoYes, 100NoYes, 6

6100(0.5)x

y

756100(0.5) 0.017y mg

Example◦ Arsenic-74 is used to locate brain tumors. It has a

half-life of 17.5 days. Write an exponential decay function for a 90-mg sample.

◦ Use the function to find the amount remaining after 6 days.

17.590(0.5)x

y

617.590(0.5) 71y mg

Loudness◦ Logarithms are used to model sound. The intensity

of a sound is a measure of the energy carried by the sound wave. The greater the intensity of a sound, the louder it seems. This apparent loudness L is measured in decibels. You can use the formula

, where I is the intensity of the sound in watts per square

meter (W/m2) I0 is the lowest-intensity sound that the average human

ear can detect. (We will use I0 = 10-12 w/m2)

0

10 logI

LI

Example◦ Suppose you are the supervisor on a road

construction job. Your team is blasting rock to make way for a roadbed. One explosion has an intensity of 1.65 x 10-2 W/m2. What is the loudness of the sound in decibels?

◦ 0

10 logI

LI

2

12

1.65 1010 log 102

10L db

Assignment◦ Worksheet◦ Round all problems appropriately

For half life, round to two decimal places For loudness, round to the nearest dB

Essential Question: What are some types of real-life problems where exponential or

logarithmic equations can be used?

Acidity◦ Scientists use common logarithms to measure

acidity, which increases as the concentration of hydrogen ions in a substance. The pH of a substance equals:

pH = –log[H+],where [H+] is the concentration of hydrogen ions.

Example The pH of lemon juice is 2.3, while the pH of milk is

6.6. Find the concentration of hydrogen ions in each substance. Which substance is more acidic?

Lemon Juice Milk (Your turn) pH = -log[H+]

2.3 = -log[H+] substitutelog[H+] = -2.3 divide by -1[H+] = 10-2.3 convert to exp[H+] = 5.0 x 10-3

pH = -log[H+]6.6 = -log[H+]log[H+] = -6.6[H+] = 10-6.6

[H+] = 2.5 x 10-7

The lemon juice is more acidic as it contains more hydrogen ions

Compound Interest (finding something other than A)◦ Remember our formulas from Monday:

A = P(1 + r)t for compounding annually

A = Pert for compounding continuously

◦ Finding P How much should be invested at 5.5% compounded

annually to yield $3500 at the end of 4 years? 3500 = P(1 + 0.055)4

3500 = P(1.055)4

3500/(1.055)4 = P$2825.26= P

Compound Interest (finding something other than A)◦ Finding r

What interest rate would be required to grow an investment of $1000 to $1407.10 in seven years if that interest is compounded annually?

1407.10 = 1000(1 + r)7

1.4071 = (1 + r)7

1.05 = 1 + r0.05 = r, meaning an interest rate of 5%

Compound Interest (finding something other than A)◦ Finding t

How long will it take to double an investment of $500 at 7% interest, compounded annually?

1000 = 500(1 + 0.07)t

2 = (1.07)t

log1.07 2 = t

10.25 = t, meaning 10.25 years

log 2/log 1.07 = t

Assignment◦ Worksheet◦ Round all problems appropriately

If talking about money: 2 decimal places