1. Exponential Functions Define an exponential function. Graph exponential functions. Use transformations on exponential functions. Define simple interest. Develop a compound interest formula. Understand the number e. SECTION 4.1 1 2 3 4 5 6

2. EXPONENTIAL FUNCTION A function f of the form is called an exponential function with base a. Its domain is (–∞, ∞).

3. EXAMPLE 1 Evaluating Exponential Functions d. Let F(x) = 4 x. Find F(3.2).

4. EXAMPLE 1 Evaluating Exponential Functions Solution d. F(3.2) = 4 3.2 ≈ 84.44850629

5. RULES OF EXPONENTS Let a, b, x, and y be real numbers with a > 0 and b > 0. Then

6. EXAMPLE 2 Graphing an Exponential Function with Base a > 1 – Exponential Growth Graph the exponential function Solution Make a table of values. Plot the points and draw a smooth curve.

7. EXAMPLE 2 Graphing an Exponential Function with Base a > 1 Solution continued This graph is typical for exponential functions when a > 1.

8. EXAMPLE 3 Graphing an Exponential Function with Base 0 < a < 1 – Exponential Decay Sketch the graph of Solution Make a table of values. Plot the points and draw a smooth curve.

9. EXAMPLE 3 Graphing an Exponential Function with Base 0 < a < 1 Solution continued As x increases in the positive direction, y decreases towards 0.

10. PROPERTIES OF EXPONENTIAL FUNCTIONS Let f (x) = a x, a > 0, a ≠ 1. 1.The domain of f (x) = a x is (–∞, ∞). 2.The range of f (x) = a x is (0, ∞); the entire graph lies above the x-axis. 3. For a > 1, Exponential Growth (i) f is an increasing function, so the graph rises to the right. (ii) as x → ∞, y → ∞. (iii)as x → –∞, y → 0.

11. 4. For 0 < a < 1, - Exponential Decay (i) f is a decreasing function, so the graph falls to the right. (ii)as x → – ∞, y → ∞. (iii)as x → ∞, y → 0. 5.The graph of f (x) = a x has no x-intercepts, so it never crosses the x-axis. No value of x will cause f (x) = a x to equal 0. 6.The graph of is a smooth and continuous curve, and it passes through the points 7.The x-axis is a horizontal asymptote for every exponential function of the form f (x) = a x.

12. TRANSFORMATIONS ON EXPONENTIAL FUNCTION f (x) = a x TransformationEquationEffect on Equation Horizontal Shift y = a x+b Shift the graph of y = a x, |b| units (i) left if b > 0. (ii) right if b < 0. Vertical Shift y = a x + b Shift the graph of y = a x, |b| units (i) up if b > 0. (ii) down if b < 0.

13. TRANSFORMATIONS ON EXPONENTIAL FUNCTION f (x) = a x TransformationEquationEffect on Equation Stretching or Compressing (Vertically) y = ca x Multiply the y coordinates by c. The graph of y = a x is vertically (i) stretched if c > 1. (ii) compressed if 0 < c < 1.

14. TRANSFORMATIONS ON EXPONENTIAL FUNCTION f (x) = a x TransformationEquationEffect on Equation Reflectiony = –a x Reflect the graph of y = a x in the x-axis. Reflect the graph of y = a x in the y-axis. y = a –x

15. EXAMPLE 6 Sketching Graphs Use transformations to sketch the graph of each function. State the domain and range of each function and the horizontal asymptote of its graph.

16. EXAMPLE 6 Sketching Graphs Solution Domain: (–∞, ∞) Range: (–4, ∞) Horizontal Asymptote: y = –4

17. EXAMPLE 6 Sketching Graphs Solution continued Domain: (–∞, ∞) Range: (0, ∞) Horizontal Asymptote: y = 0

18. EXAMPLE 6 Sketching Graphs Solution continued Domain: (–∞, ∞) Range: (–∞, 0) Horizontal Asymptote: y = 0

19. EXAMPLE 6 Sketching Graphs Solution continued Domain: (–∞, ∞) Range: (–∞, 2) Horizontal Asymptote: y = 2

20. General Exponential Growth/Decay Model Original amount Rate of decay (r 0) Number of time periods Amount after t time periods

21. Compound interest is the interest paid on both the principal and the accrued (previously earned) interest. It is an application of exponential growth. Interest that is compounded annually is paid once a year. For interest compounded annually, the amount A in the account after t years is given by COMPOUND INTEREST – Growth Amount after t time periods Original amount Rate of decay (r 0) Number of time periods

22. EXAMPLE 2 Calculating Compound Interest Juanita deposits $8000 in a bank at the interest rate of 6% compounded annually for five years. a.How much money will she have in her account after five years? b.How much interest will she receive?

23. EXAMPLE 2 Calculating Compound Interest Solution a. Here P = $8000, r = 0.06, and t = 5. b. Interest = A  P = $10,705.80  $8000 = $2705.80.

24. COMPOUND INTEREST FORMULA A = amount after t years P = principal r = annual interest rate (expressed as a decimal) n = number of times interest is compounded each year t = number of years

25. EXAMPLE 3 Using Different Compounding Periods to Compare Future Values If $100 is deposited in a bank that pays 5% annual interest, find the future value A after one year if the interest is compounded (i)annually. (ii)semiannually. (iii)quarterly. (iv)monthly. (v)daily.

26. EXAMPLE 3 Using Different Compounding Periods to Compare Future Values (i) Annual Compounding: Solution In the following computations, P = 100, r = 0.05 and t = 1. Only n, the number of times interest is compounded each year, changes. Since t = 1, nt = n(1) = n.

27. EXAMPLE 3 Using Different Compounding Periods to Compare Future Values (iii) Quarterly Compounding: (ii) Semiannual Compounding:

28. EXAMPLE 3 Using Different Compounding Periods to Compare Future Values (iv) Monthly Compounding: (v) Daily Compounding:

29. EXAMPLE 8 Bacterial Growth A technician to the French microbiologist Louis Pasteur noticed that a certain culture of bacteria in milk doubles every hour. If the bacteria count B(t) is modeled by the equation a.the initial number of bacteria, b.the number of bacteria after 10 hours; and c.the time when the number of bacteria will be 32,000. with t in hours, find

30. EXAMPLE 8 Bacterial Growth a. Initial size c. Find t when B(t) = 32,000 After 4 hours, the number of bacteria will be 32,000. Solution

31. THE VALUE OF e The value of e to 15 places is e = 2.718281828459045. gets closer and closer to a fixed number. This irrational number is denoted by e and is sometimes called the Euler number. As h gets larger and larger,

32. CONTINUOUS COMPOUND FORMULA A = amount after t years P = principal r = annual rate (expressed as a decimal) t = number of years

33. EXAMPLE 4 Calculating Continuous Compound Interest Find the amount when a principal of $8300 is invested at a 7.5% annual rate of interest compounded continuously for eight years and three months. Solution P = $8300 and r = 0.075. Convert eight years and three months to 8.25 years.

34. EXAMPLE 5 Calculating the Amount of Repaying a Loan How much money did the government owe DeHaven’s descendants for 213 years on a $450,000 loan at the interest rate of 6%? Solution a. With simple interest,

35. EXAMPLE 5 Calculating the Amount of Repaying a Loan Solution continued b. With interest compounded yearly, c. With interest compounded quarterly,

36. EXAMPLE 5 Calculating the Amount of Repaying a Loan Solution continued d. With interest compounded continuously, Notice the dramatic difference between quarterly and continuous compounding and the dramatic difference between simple interest and compound interest.

37. THE NATURAL EXPONENTIAL FUNCTION with base e is so prevalent in the sciences that it is often referred to as the exponential function or the natural exponential function. The exponential function

38. EXAMPLE 6 Sketching a Graph Use transformations to sketch the graph of Solution Start with the graph of y = e x.

39. EXAMPLE 6 Sketching a Graph Use transformations to sketch the graph of Solution coninued Shift the graph of y = e x one unit right.

40. EXAMPLE 6 Sketching a Graph Use transformations to sketch the graph of Solution continued Shift the graph of y = e x – 1 two units up.

41. MODEL FOR EXPONENTIAL GROWTH OR DECAY A(t) = amount at time t A 0 = A(0), the initial amount k = relative rate of growth (k > 0) or decay (k < 0) t = time

42. EXAMPLE 7 Modeling Exponential Growth and Decay In the year 2000, the human population of the world was approximately 6 billion and the annual rate of growth was about 2.1%. Using the model on the previous slide, estimate the population of the world in the following years. a.2030 b.1990

43. EXAMPLE 7 a. The year 2000 corresponds to t = 0. So A 0 = 6 (billion), k = 0.021, and 2030 corresponds to t = 30. Solution The model predicts that if the rate of growth is 2.1% per year, over 11.26 billion people will be in the world in 2030. Modeling Exponential Growth and Decay

44. EXAMPLE 7 b. The year 1990 corresponds to t =  10. Solution The model predicts that the world had over 4.86 billion people in 1990. (The actual population in 1990 was 5.28 billion.) Modeling Exponential Growth and Decay

45. Logarithmic Functions Define logarithmic functions. Inverse Functions Evaluate logarithms. Rules of Logarithms Find the domains of logarithmic functions. Graph logarithmic functions. Use logarithms to evaluate exponential equations. SECTION 4.3 1 2 3 4 5 6 7

46. DEFINITION OF THE LOGARITHMIC FUNCTION For x > 0, a > 0, and a ≠ 1, The function f (x) = log a x, is called the logarithmic function with base a. The logarithmic function is the inverse function of the exponential function.

47. Inverse Functions Certain pairs of one-to-one functions “undo” one another. For example, if then

48. Inverse Functions Starting with 10, we “applied” function  and then “applied” function g to the result, which returned the number 10.

49. Inverse Functions As further examples, check that

50. Inverse Functions In particular, for this pair of functions, In fact, for any value of x, or Because of this property, g is called the inverse of .

51. Inverse Function Let  be a one-to-one function. Then g is the inverse function of  if for every x in the domain of g, and for every x in the domain of .

52. EXAMPLE 1 Converting from Exponential to Logarithmic Form Write each exponential equation in logarithmic form. Solution

53. EXAMPLE 2 Converting from Logarithmic Form to Exponential Form Write each logarithmic equation in exponential form. Solution

54. EXAMPLE 3 Evaluating Logarithms Find the value of each of the following logarithms. Solution

55. EXAMPLE 3 Evaluating Logarithms Solution continued

56. EXAMPLE 4 Using the Definition of Logarithm Solve each equation. Solution

57. EXAMPLE 4 Using the Definition of Logarithm Solution continued

58. EXAMPLE 4 Using the Definition of Logarithm Solution continued

59. Rules of Logarithms with Base a If M, N, and a are positive real numbers with a ≠ 1, and x is any real number, then 1. log a (a) = 12. log a (1) = 0 3. log a (a x ) = x4. 5. log a (MN) = log a (M) + log a (N) 6. log a (M/N) = log a (M) – log a (N) 7. log a (M x ) = x · log a (M)8. log a (1/N) = – log a (N) Rules of Logarithms These relationships are used to solve exponential or logarithmic equations

60. COMMON LOGARITHMS 1.log 10 = 1 2.log 1 = 0 3.log 10 x = x The logarithm with base 10 is called the common logarithm and is denoted by omitting the base: log x = log 10 x. Thus, y = log x if and only if x = 10 y. Applying the basic properties of logarithms

61. NATURAL LOGARITHMS 1.ln e = 1 2.ln 1 = 0 3.log e x = x The logarithm with base e is called the natural logarithm and is denoted by ln x. That is, ln x = log e x. Thus, y = ln x if and only if x = e y. Applying the basic properties of logarithms

62. DOMAIN OF LOGARITHMIC FUNCTION Domain of y = log a x is (0, ∞) Range of y = log a x is (–∞, ∞) Logarithms of 0 and negative numbers are not defined.

63. EXAMPLE 5 Finding the Domain Find the domain of Solution Domain of a logarithmic function must be positive, that is, The domain of f is (–∞, 2).

64. EXAMPLE 6 Sketching a Graph Sketch the graph of y = log 3 x. Solution by plotting points (Method 1) Make a table of values.

65. EXAMPLE 6 Sketching a Graph Solution continued Plot the ordered pairs and connect with a smooth curve to obtain the graph of y = log 3 x.

66. 66 EXAMPLE 6 Sketching a Graph Solution by using the inverse function (Method 2) Graph y = f (x) = 3 x. Reflect the graph of y = 3 x in the line y = x to obtain the graph of y = f –1 (x) = log 3 x.

67. GRAPHS OF LOGARITHMIC FUNCTIONS

68. PROPERTIES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS Exponential Function f (x) = a x Logarithmic Function f (x) = log a x Domain (0, ∞) Range (–∞, ∞) 1.Domain (–∞, ∞) Range (0, ∞) x-intercept is 1 No y-intercept 2. y-intercept is 1 No x-intercept 3. x-axis (y = 0) is the horizontal asymptote y-axis (x = 0) is the vertical asymptote

69. PROPERTIES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS Exponential Function f (x) = a x Logarithmic Function f (x) = log a x The graph is a continuous smooth curve that passes through the points (1, 0), and (a, 1). 4.The graph is a continuous smooth curve that passes through the points (0, 1), and (1, a).

70. PROPERTIES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS Exponential Function f (x) = a x Logarithmic Function f (x) = log a x Is one-to-one, that is, log a u = log a v if and only if u = v. 5. Is one-to-one, that is, a u = a v if and only if u = v. Increasing if a > 1 Decreasing if 0 < a < 1 6.Increasing if a > 1 Decreasing if 0 < a < 1

71. EXAMPLE 7 Using Transformations Start with the graph of f (x) = log 3 x and use transformations to sketch the graph of each function. State the domain and range and the vertical asymptote for the graph of each function.

72. EXAMPLE 7 Using Transformations Solution Shift up 2 Domain (0, ∞) Range (–∞, ∞) Vertical asymptote x = 0

73. EXAMPLE 7 Using Transformations Solution continued Shift right 1 Domain (1, ∞) Range (–∞, ∞) Vertical asymptote x = 1

74. EXAMPLE 7 Using Transformations Solution continued Reflect graph of y = log 3 x in the x-axis Domain (0, ∞) Range (–∞, ∞) Vertical asymptote x = 0

75. EXAMPLE 7 Using Transformations Solution continued Reflect graph of y = log 3 x in the y-axis Domain (∞, 0) Range (–∞, ∞) Vertical asymptote x = 0

76. EXAMPLE 8 Using Transformations to Sketch a Graph Sketch the graph of Solution Start with the graph of f (x) = log x. Step 1: Replacing x with x – 2 shifts the graph two units right.

77. EXAMPLE 8 Using Transformations to Sketch a Graph Solution continued Step 2: Multiplying by  1 reflects the graph Step 3: Adding 2 shifts the graph two units up. in the x-axis.

78. Rules of Logarithms with Base a If M, N, and a are positive real numbers with a ≠ 1, and x is any real number, then 1. log a (a) = 12. log a (1) = 0 3. log a (a x ) = x4. 5. log a (MN) = log a (M) + log a (N) 6. log a (M/N) = log a (M) – log a (N) 7. log a (M x ) = x · log a (M)8. log a (1/N) = – log a (N) Rules of Logarithms

79. EXAMPLE 1 Using Rules of Logarithms to Evaluate Expressions Given that log 5 z = 3 and log 5 y = 2, evaluate each expression. Solution

80. EXAMPLE 1 Using Rules of Logarithms to Evaluate Expressions Solution continued

81. EXAMPLE 1 Using Rules of Logarithms to Evaluate Expressions Solution continued

82. EXAMPLE 2 Writing Expressions In Expanded Form Write each expression in expanded form. Solution

83. EXAMPLE 2 Writing Expressions In Expanded Form Solution continued

84. EXAMPLE 3 Writing Expressions in Condensed Form Write each expression in condensed form.

85. EXAMPLE 3 Writing Expressions in Condensed Form Solution

86. EXAMPLE 3 Writing Expressions in Condensed Form Solution continued

87. EXAMPLE 3 Writing Expressions in Condensed Form Solution continued

88. CHANGE-OF-BASE FORMULA Let a, b, and x be positive real numbers with a ≠ 1 and b ≠ 1. Then log b x can be converted to a different base as follows:

89. EXAMPLE 4 Using a Change of Base to Compute Logarithms Compute log 5 13 by changing to a. common logarithms and b. natural logarithms. Solution

90. EXAMPLE 9 Evaluating the Natural Logarithm Evaluate each expression. Solution Use a calculator.

91. EXAMPLE 10 Doubling Your Money a.How long will it take to double your money if it earns 6.5% compounded continuously? b. At what rate of return, compounded continuously, would your money double in 5 years? Solution a.If P is the original amount invested, A = 2P. It will take 11 years to double your money.

92. EXAMPLE 10 Doubling Your Money Solution continued b. Your investment will double in 5 years at the rate of 13.86%.

93. Solving Exponential Or Logarithmic Equations To solve an exponential or logarithmic equation, change the given equation into one of the following forms, where a and b are real numbers, a > 0 and a ≠ 1, and follow the guidelines. 1.a x = b Solve by taking logarithms on both sides. 2. Log a x = b Solve by changing to exponential form a b = x.

94. SOLVING AN EXPONENTIAL EQUATION Solve 7 x = 12. Give the solution to the nearest thousandth. Solution While any appropriate base b can be used, the best practical base is base 10 or base e. We choose base e (natural) logarithms here.

95. SOLVING AN EXPONENTIAL EQUATION Solve 7 x = 12. Give the solution to the nearest thousandth. Solution Property of logarithms Power of logarithms Divide by In 7. Use a calculator. The solution set is {1.277}.

96. SOLVING AN EXPONENTIAL EQUATION Solve 3 2x – 1 =.4 x+2. Give the solution to the nearest thousandth. Solution Take natural logarithms on both sides. Property power Distributive property

97. SOLVING AN EXPONENTIAL EQUATION Solve 3 2x – 1 =.4 x+2. Give the solution to the nearest thousandth. Solution Write the terms with x on one side Factor out x. Divide by 2 In 3 – In.4. Power property

98. SOLVING AN EXPONENTIAL EQUATION Solve 3 2x – 1 =.4 x+2. Give the solution to the nearest thousandth. Solution Apply the exponents. Product property; Quotient property This is approximate. This is exact. The solution set is { –.236}.

99. SOLVING BASE e EXPONENTIAL EQUATIONS Solve the equation. Give solutions to the nearest thousandth. Solution a. Take natural logarithms on both sides. In = x 2

100. SOLVING BASE e EXPONENTIAL EQUATIONS Solution a. Square root property Remember both roots. Use a calculator. The solution set is {  2.302}. Solve the equation. Give solutions to the nearest thousandth.

101. SOLVING BASE e EXPONENTIAL EQUATIONS Solution b. Take natural logarithms on both sides. Power property Solve the equation. Give solutions to the nearest thousandth. Divide by e;

102. SOLVING BASE e EXPONENTIAL EQUATIONS Solution b. In e = 1 Multiply by – ½ The solution set is {–.549}. Solve the equation. Give solutions to the nearest thousandth.

103. SOLVING A LOGARITHMIC EQUATION Solve log(x + 6) – log(x + 2) = log x. Solution Quotient property Property of logarithms

104. SOLVING A LOGARITHMIC EQUATION Solve log(x + 6) – log(x + 2) = log x. Solution Distributive property Standard form Factor. Zero-factor property The proposed negative solution (x = – 3) is not in the domain of the log x in the original equation, so the only valid solution is the positive number 2, giving the solution set {2}.

105. SOLVING A LOGARITHMIC EQUATION Solve log(3x + 2) + log(x – 1 ) = 1. Give the exact value(s) of the solution(s). Solution Substitute. Product property Property of logarithms

106. SOLVING A LOGARITMIC EQUATION Solve log(3x + 2) + log(x – 1 ) = 1. Give the exact value(s) of the solution(s). Solution Multiply. Subtract 10. Quadratic formula

107. SOLVING A LOGARITMIC EQUATION Solve log(3x + 2) + log(x – 1 ) = 1. Give the exact value(s) of the solution(s). Solution The number is negative, so x – 1 is negative. Therefore, log(x – 1) is not defined and this proposed solution must be discarded. Since > 1, both 3x + 2 and x – 1 are positive and the solution set is

108. NEWTON’S LAW OF COOLING Newton’s Law of Cooling states that where T is the temperature of the object at time t, T s is the surrounding temperature, and T 0 is the value of T at t = 0.

109. EXAMPLE 11 McDonald’s Hot Coffee The local McDonald’s franchise has discovered that when coffee is poured from a coffeemaker whose contents are 180ºF into a noninsulated pot, after 1 minute, the coffee cools to 165ºF if the room temperature is 72ºF. How long should the employees wait before pouring the coffee from this noninsulated pot into cups to deliver it to customers at 125ºF?

110. EXAMPLE 11 McDonald’s Hot Coffee Use Newton’s Law of Cooling with T 0 = 180 and T s = 72 to obtain Solution We have T = 165 and t = 1.

111. EXAMPLE 11 McDonald’s Hot Coffee Substitute this value for k. Solution continued Solve for t when T = 125. The employee should wait about 5 minutes.

112. GROWTH AND DECAY MODEL A is the quantity after time t. A 0 is the initial (original) quantity (when t = 0). r is the growth or decay rate per period. t is the time elapsed from t = 0.

113. EXAMPLE 12 Chemical Toxins in a Lake A chemical spill deposits 60,000 cubic meters of soluble toxic waste into a large lake. If 20% of the waste is removed every year, how many years will it take to reduce the toxin to 1000 cubic meters? Solution In the equation A = A 0 e rt, we need to find A 0, r, and the time when A = 1000.

114. EXAMPLE 12 Chemical Toxins in a Lake 1.Find A 0. Initially (t = 0), we are given A 0 = 60,000. So Solution continued 2.Find r. When t = 1 year, the amount of toxin will be 80% of its initial value, or

115. EXAMPLE 12 Chemical Toxins in a Lake Solution continued 2.continued So

116. EXAMPLE 12 Chemical Toxins in a Lake Solution continued 3. Find t when A = 1000. It will take approximately 18 years to reduce toxin to 1000 m 3.