1. Splash Screen

2. 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

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

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

5. 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

6. 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

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

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

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

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

11. 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

12. 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

13. 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

14. 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

15. 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

16. 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

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

18. Concept

19. 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

20. 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.
≈ k Use a calculator.
Answer:
Example 1

21. 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.
≈ k Use a calculator.
Answer: The value of k of Sodium-22 is Thus, the equation for the decay of Sodium-22 is y = ae–0.2666t, where t is given in years.
Example 1

22. 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 =
D. k =
Example 1

23. 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 =
D. k =
Example 1

24. 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

25. 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 –
8.64 ≈ t Use a calculator.
Example 2

26. Carbon Dating
Answer:
Example 2

27. 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 t
Original equation
= 1e (8.6)
a = 1 and t = 8.6
≈ or 10% Use a calculator.
Example 2

28. 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 hours later
Example 2

29. 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 hours later
Example 2

30. 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

31. 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

32. 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 , or about 0.66%.
Example 3

33. 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 =
Divide each side by 1.26. ≈
Property of Equality for Logarithmic Functions ≈
Example 3

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

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

36. 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 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

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

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

39. 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 B C D
Example 3

40. 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 B C D
Example 3

41. 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

42. 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

43. C. POPULATION The population of Alaska can be modeled by y = 670,000e0
C. POPULATION The population of Alaska can be modeled by y = 670,000e t. 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

44. C. POPULATION The population of Alaska can be modeled by y = 670,000e0
C. POPULATION The population of Alaska can be modeled by y = 670,000e t. 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

45. Concept

46. 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 Graph the function.
Answer:
Example 4

47. 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 Graph the function.
Answer:
Example 4

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

49. 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 What is the horizontal asymptote?
Answer: The horizontal asymptote is at f (t) =
Example 4

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

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

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

53. Logistic Growth
D. A city’s population in millions is modeled by , where t is the number of years since 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

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

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

56. B. A city’s population in millions is modeled by where t is the number of years since 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

57. B. A city’s population in millions is modeled by where t is the number of years since 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

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

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

60. D. A city’s population in millions is modeled by where t is the number of years since 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

61. D. A city’s population in millions is modeled by where t is the number of years since 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

62. End of the Lesson