Chapter 8-x Applications of Exponential and Logarithmic Functions Day 1 Essential Question: What are some types of real- life problems where exponential or logarithmic equations can be used?

8-x: Exponential/Logarithmic Applications 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

8-x: Exponential/Logarithmic Applications 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. depreciates No Yes, 6500 Yes, 0.143 No A = 6500(1 – 0.143)t = 6500(0.857)t A = 6500(0.857)3 = $4091.25

8-x: Exponential/Logarithmic Applications 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.

8-x: Exponential/Logarithmic Applications 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

8-x: Exponential/Logarithmic Applications $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

8-x: Exponential/Logarithmic Applications 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)? No Yes, 1050 Yes, 0.055 Yes, 5 A = 1050 e(0.055 • 5) = $1382.36

8-x: Exponential/Logarithmic Applications 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

8-x: Exponential/Logarithmic Applications Assignment Worksheet Round all problems appropriately (if talking about money: 2 decimal places; if talking about population: nearest integer)

Chapter 8-x Applications of Exponential and Logarithmic Functions Day 2 Essential Question: What are some types of real- life problems where exponential or logarithmic equations can be used?

8-x: Exponential/Logarithmic Applications 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

8-x: Exponential/Logarithmic Applications 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. No Yes, 100 No Yes, 6

8-x: Exponential/Logarithmic Applications 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.

8-x: Exponential/Logarithmic Applications 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)

8-x: Exponential/Logarithmic Applications 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?

8-x: Exponential/Logarithmic Applications Assignment Worksheet Round all problems appropriately For half life, round to two decimal places For loudness, round to the nearest dB

Chapter 8-x Applications of Exponential and Logarithmic Functions Day 3 Essential Question: What are some types of real- life problems where exponential or logarithmic equations can be used?

8-x: Exponential/Logarithmic Applications 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.

8-x: Exponential/Logarithmic Applications 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+] substitute log[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

8-x: Exponential/Logarithmic Applications 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

8-x: Exponential/Logarithmic Applications 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 + r 0.05 = r, meaning an interest rate of 5%

8-x: Exponential/Logarithmic Applications 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 log 2/log 1.07 = t 10.25 = t, meaning 10.25 years

8-x: Exponential/Logarithmic Applications Assignment Worksheet Round all problems appropriately If talking about money: 2 decimal places