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Five-Minute Check (over Lesson 7–7) CCSS Then/Now New Vocabulary Key Concept: Exponential Growth and Decay Example 1: Exponential Decay Example 2: Real-World Example: Carbon Dating Example 3: Real-World Example: Continuous Exponential Growth Key Concept: Logistic Growth Function Example 4: Logistic Growth Lesson Menu

Write an equivalent exponential function for In 14.2 = x. A. log2 14.2 = x B. 10 log 4.2 = x C. 14.2x = e D. ex = 14.2 5-Minute Check 1

Write an equivalent exponential function for In 14.2 = x. A. log2 14.2 = x B. 10 log 4.2 = x C. 14.2x = e D. ex = 14.2 5-Minute Check 1

Write an equivalent logarithmic function for e6 = y. A. In 6 = y B. In y = 6 C. log6 y = 6 D. 6 In = y 5-Minute Check 2

Write an equivalent logarithmic function for e6 = y. A. In 6 = y B. In y = 6 C. log6 y = 6 D. 6 In = y 5-Minute Check 2

Solve 6 + 4e–x = 12. Round to the nearest ten-thousandth. B. 0 C. 0.4055 D. 1.7918 5-Minute Check 3

Solve 6 + 4e–x = 12. Round to the nearest ten-thousandth. B. 0 C. 0.4055 D. 1.7918 5-Minute Check 3

Solve In (2x + 3) > –3. Round to the nearest ten-thousandth. A. {x | x > 1.9014} B. {x | x > 0.0025} C. {x | x > –1.4751} D. {x | x > –1.9014} 5-Minute Check 4

Solve In (2x + 3) > –3. Round to the nearest ten-thousandth. A. {x | x > 1.9014} B. {x | x > 0.0025} C. {x | x > –1.4751} D. {x | x > –1.9014} 5-Minute Check 4

Write 6 In x – In x2 as a single logarithm. A. 6 In (x2 – x) B. In x4 C. 6 In x D. 6 In x2 5-Minute Check 5

Write 6 In x – In x2 as a single logarithm. A. 6 In (x2 – x) B. In x4 C. 6 In x D. 6 In x2 5-Minute Check 5

You deposit $2000 in an account paying 4% annual interest compounded continuously. Using the formula A = pert, how long will it take for your money to double? A. about 8.7 years B. about 17.3 years C. about 21.4 years D. about 34.6 years 5-Minute Check 6

You deposit $2000 in an account paying 4% annual interest compounded continuously. Using the formula A = pert, how long will it take for your money to double? A. about 8.7 years B. about 17.3 years C. about 21.4 years D. about 34.6 years 5-Minute Check 6

Mathematical Practices Content Standards F.IF.8.b Use the properties of exponents to interpret expressions for exponential functions. F.LE.4 For exponential models, express as a logarithm the solution to abct = d where a, c, and d are numbers and the base b is 2, 10, or e; evaluate the logarithm using technology. Mathematical Practices 1 Make sense of problems and persevere in solving them. CCSS

You used exponential growth and decay formulas. Use logarithms to solve problems involving exponential growth and decay. Use logarithms to solve problems involving logistic growth. Then/Now

rate of continuous growth rate of continuous decay logistic growth model Vocabulary

Concept

Exponential decay formula GEOLOGY The half-life of Sodium-22 is 2.6 years. Determine the value of k and the equation of decay for Sodium-22. If a is the initial amount of the substance, then the amount y that remains after 2.6 years is or 0.5a. Exponential decay formula Replace y with 0.5a and t with 2.6. Divide each side by a. Example 1

In 0.5 = In e–2.6k Property of Equality for Logarithmic Functions Exponential Decay In 0.5 = In e–2.6k Property of Equality for Logarithmic Functions In 0.5 = –2.6k Inverse Property of Exponents and Logarithms Divide each side by –2.6. 0.2666 ≈ k Use a calculator. Answer: Example 1

In 0.5 = In e–2.6k Property of Equality for Logarithmic Functions Exponential Decay In 0.5 = In e–2.6k Property of Equality for Logarithmic Functions In 0.5 = –2.6k Inverse Property of Exponents and Logarithms Divide each side by –2.6. 0.2666 ≈ k Use a calculator. Answer: The value of k of Sodium-22 is 0.2666. Thus, the equation for the decay of Sodium-22 is y = ae–0.2666t, where t is given in years. Example 1

HEALTH The half-life of radioactive iodine used in medical studies is 8 hours. What is the value of k for radioactive iodine? A. k = –0.0866 B. k = –4.1589 C. k = 0.0866 D. k = 4.1589 Example 1

HEALTH The half-life of radioactive iodine used in medical studies is 8 hours. What is the value of k for radioactive iodine? A. k = –0.0866 B. k = –4.1589 C. k = 0.0866 D. k = 4.1589 Example 1

Carbon Dating GEOLOGY A geologist examining a meteorite estimates that it contains only about 10% as much Sodium-22 as it would have contained when it reached the surface of the Earth. How long ago did the meteorite reach the surface of the Earth? Understand The formula for the decay of Sodium-22 is y = ae–kt. You want to find how long ago the meteorite reached Earth. Plan Let a be the initial amount of Sodium-22 in the meteorite. The amount y that remains after t years is 10% of a or 0.10a. Example 2

y = ae–0.2666t Formula for the decay of Sodium-22 Carbon Dating Solve y = ae–0.2666t Formula for the decay of Sodium-22 0.1a = ae–0.2666t Replace y with 0.1a. 0.1 = e–0.2666t Divide each side by a. In 0.1= ln e–0.2666t Property of Equality for Logarithms ln 0.1 = –0.2666t Inverse Property for Exponents and Logarithms Divide each side by –0.2666. 8.64 ≈ t Use a calculator. Example 2

Carbon Dating Answer: Example 2

Answer: It reached the surface of Earth about 8.6 years ago. Carbon Dating Answer: It reached the surface of Earth about 8.6 years ago. Check Use the formula to find the amount of the sample remaining after 8.6 years. Use an original amount of 1. y = ae-0.2666t Original equation = 1e-0.2666(8.6) a = 1 and t = 8.6 ≈ 0.101 or 10% Use a calculator. Example 2

HEALTH The half-life of radioactive iodine used in medical studies is 8 hours. A doctor wants to know when the amount of radioactive iodine in a patient’s body is 20% of the original amount. When will this occur? A. about 0.05 hour later B. about 0.39 hour later C. about 2.58 hours later D. about 18.58 hours later Example 2

HEALTH The half-life of radioactive iodine used in medical studies is 8 hours. A doctor wants to know when the amount of radioactive iodine in a patient’s body is 20% of the original amount. When will this occur? A. about 0.05 hour later B. about 0.39 hour later C. about 2.58 hours later D. about 18.58 hours later Example 2

Formula for continuous exponential growth A. POPULATION In 2007, the population of China was 1.32 billion. In 2000, it was 1.26 billion. Determine the value of k, China’s relative rate of growth. Formula for continuous exponential growth y = 1.32, a = 1.26, and t = 2007 – 2000 or 7 Divide each side by 1.26. Example 3

Property of Equality for Logarithmic Equations Continuous Exponential Growth Property of Equality for Logarithmic Equations ln e x = x Divide each side by 7. Use a calculator. Answer: Example 3

Property of Equality for Logarithmic Equations Continuous Exponential Growth Property of Equality for Logarithmic Equations ln e x = x Divide each side by 7. Use a calculator. Answer: China’s relative rate of growth is about 0.0066, or about 0.66%. Example 3

Formula for continuous exponential growth B. POPULATION In 2007, the population of China was 1.32 billion. In 2000, it was 1.26 billion. When will China’s population reach 1.5 billion? Formula for continuous exponential growth y = 1.5, a = 1.26, and k = 0.0066 Divide each side by 1.26. ≈ Property of Equality for Logarithmic Functions ≈ Example 3

In 1.1905 ≈ 0.0066t ln e x = x Divide each side by 0.0066. Continuous Exponential Growth ln e x = x In 1.1905 ≈ 0.0066t Divide each side by 0.0066. Use a calculator. Answer: Example 3

Answer: China’s population will reach 1.5 billion in 2026. Continuous Exponential Growth ln e x = x In 1.1905 ≈ 0.0066t Divide each side by 0.0066. Use a calculator. Answer: China’s population will reach 1.5 billion in 2026. Example 3

Formula for exponential growth Continuous Exponential Growth C. POPULATION In 2007, the population of China was 1.32 billion. India’s population in 2007 was 1.13 billion and can be modeled by y = 1.13e0.015t. Determine when India’s population will surpass China’s. (Note: t represents years after 2007.) Formula for exponential growth Property of Inequality for Logarithms Example 3

Product Property of Logarithms Continuous Exponential Growth Product Property of Logarithms In 1.32 + In e0.0066t < In 1.13 + In e0.015t ln e x = x In 1.32 + 0.0066t < In 1.13 + 0.015t Subtract (0.0066t + ln 1.13) from each side. In 1.32 – In 1.13 < 0.0084t Divide each side by 0.0084. Use a calculator. Answer: Example 3

Product Property of Logarithms Continuous Exponential Growth Product Property of Logarithms In 1.32 + In e0.0066t < In 1.13 + In e0.015t ln e x = x In 1.32 + 0.0066t < In 1.13 + 0.015t Subtract (0.0066t + ln 1.13) from each side. In 1.32 – In 1.13 < 0.0084t Divide each side by 0.0084. Use a calculator. Answer: India’s population will surpass China’s in 18.5 years, or midway through 2025. Example 3

A. POPULATION In 2006, the population of Alaska was 670,000 A. POPULATION In 2006, the population of Alaska was 670,000. In 2000, it was 620,000. Determine the value of k, Alaska’s relative rate of growth. A. 0.00043 B. 0.01293 C. 0.02459 D. 0.03946 Example 3

A. POPULATION In 2006, the population of Alaska was 670,000 A. POPULATION In 2006, the population of Alaska was 670,000. In 2000, it was 620,000. Determine the value of k, Alaska’s relative rate of growth. A. 0.00043 B. 0.01293 C. 0.02459 D. 0.03946 Example 3

B. POPULATION In 2006, the population of Alaska was 670,000 B. POPULATION In 2006, the population of Alaska was 670,000. In 2000, it was 620,000. When did Alaska’s population reach 700,000? A. in 2009 B. in 2010 C. in 2011 D. in 2013 Example 3

B. POPULATION In 2006, the population of Alaska was 670,000 B. POPULATION In 2006, the population of Alaska was 670,000. In 2000, it was 620,000. When did Alaska’s population reach 700,000? A. in 2009 B. in 2010 C. in 2011 D. in 2013 Example 3

C. POPULATION The population of Alaska can be modeled by y = 670,000e0 C. POPULATION The population of Alaska can be modeled by y = 670,000e0.01293t. Wyoming’s population in 2006 was 515,000 and can be modeled by y = 515,000e0.020t. Determine when Wyoming’s population will surpass Alaska’s. (Note: t represents years after 2006.) A. in 2021. B. in 2023. C. in 2025. D. in 2027. Example 3

C. POPULATION The population of Alaska can be modeled by y = 670,000e0 C. POPULATION The population of Alaska can be modeled by y = 670,000e0.01293t. Wyoming’s population in 2006 was 515,000 and can be modeled by y = 515,000e0.020t. Determine when Wyoming’s population will surpass Alaska’s. (Note: t represents years after 2006.) A. in 2021. B. in 2023. C. in 2025. D. in 2027. Example 3

Concept

A. A city’s population in millions is modeled by Logistic Growth A. A city’s population in millions is modeled by , where t is the number of years since 2000. Graph the function. Answer: Example 4

A. A city’s population in millions is modeled by Logistic Growth A. A city’s population in millions is modeled by , where t is the number of years since 2000. Graph the function. Answer: Example 4

Logistic Growth B. A city’s population in millions is modeled by , where t is the number of years since 2000. What is the horizontal asymptote? Answer: Example 4

Answer: The horizontal asymptote is at f (t) = 1.432. Logistic Growth B. A city’s population in millions is modeled by , where t is the number of years since 2000. What is the horizontal asymptote? Answer: The horizontal asymptote is at f (t) = 1.432. Example 4

Logistic Growth C. A city’s population in millions is modeled by , where t is the number of years since 2000. What will be the maximum population? Answer: Example 4

Logistic Growth C. A city’s population in millions is modeled by , where t is the number of years since 2000. What will be the maximum population? Answer: The population will reach a maximum of a little less than 1,432,000 people. Example 4

Logistic Growth D. A city’s population in millions is modeled by , where t is the number of years since 2000. According to the function, when will the city’s population reach 1 million? Answer: Example 4

Logistic Growth D. A city’s population in millions is modeled by , where t is the number of years since 2000. According to the function, when will the city’s population reach 1 million? Answer: The graph indicates the population will reach 1 million people at t ≈ 3. Replacing f (t) with 1 and solving for t in the equation yields t = 2.78 years. So, the population of the city will reach 1 million people by 2003. Example 4

A. A city’s population in millions is modeled by where t is the number of years since 2000. Graph the function. A. B. C. D. Example 4

A. A city’s population in millions is modeled by where t is the number of years since 2000. Graph the function. A. B. C. D. Example 4

B. A city’s population in millions is modeled by where t is the number of years since 2000. What is the horizontal asymptote? A. f(t) = 2.971 B. f(t) = 1.13 C. f(t) = –0.28 D. f(t) = 1.563 Example 4

B. A city’s population in millions is modeled by where t is the number of years since 2000. What is the horizontal asymptote? A. f(t) = 2.971 B. f(t) = 1.13 C. f(t) = –0.28 D. f(t) = 1.563 Example 4

C. A city’s population in millions is modeled by where t is the number of years since 2000. What will be the maximum population? A. 1 million people B. 1.563 million people C. 1.13 million people D. 0.28 million people Example 4

C. A city’s population in millions is modeled by where t is the number of years since 2000. What will be the maximum population? A. 1 million people B. 1.563 million people C. 1.13 million people D. 0.28 million people Example 4

D. A city’s population in millions is modeled by where t is the number of years since 2000. According to the function, when will the city’s population reach 1.5 million? A. by the year 2008 B. by the year 2010 C. by the year 2012 D. by the year 2014 Example 4

D. A city’s population in millions is modeled by where t is the number of years since 2000. According to the function, when will the city’s population reach 1.5 million? A. by the year 2008 B. by the year 2010 C. by the year 2012 D. by the year 2014 Example 4

End of the Lesson