1. 5  Exponential Functions  Logarithmic Functions  Compound Interest  Differentiation of Exponential Functions  Exponential Functions as Mathematical Models Exponential and Logarithmic Functions

2. 5.1 Exponential Functions

3. Exponential Function  The function defined by is called an exponential function with base b and exponent x.  The domain of f is the set of all real numbers.

4. Example  The exponential function with base 2 is the function with domain (– ,  ).  Find the values of f(x) for selected values of x follow:

5. Example  The exponential function with base 2 is the function with domain (– ,  ).  Find the values of f(x) for selected values of x follow:

6. Laws of Exponents  Let a and b be positive numbers and let x and y be real numbers. Then, 1. 1. 2. 2. 3. 3. 4. 4. 5. 5.

7. Examples  Let f(x) = 2 2x – 1. Find the value of x for which f(x) = 16. Solution  We want to solve the equation 2 2x – 1 = 16 = 2 4  But this equation holds if and only if 2x – 1 = 4 giving x =. Example 2, page 331

8. Examples  Sketch the graph of the exponential function f(x) = 2 x. Solution  First, recall that the domain of this function is the set of real numbers.  Next, putting x = 0 gives y = 2 0 = 1, which is the y-intercept. (There is no x-intercept, since there is no value of x for which y = 0) Example 3, page 331

9. Examples  Sketch the graph of the exponential function f(x) = 2 x. Solution  Now, consider a few values for x:  Note that 2 x approaches zero as x decreases without bound: ✦ There is a horizontal asymptote at y = 0.  Furthermore, 2 x increases without bound when x increases without bound.  Thus, the range of f is the interval (0,  ). x – 5 – 4 – 3 – 2 – 1 012345 y1/321/161/81/41/212481632 Example 3, page 331

10. Examples  Sketch the graph of the exponential function f(x) = 2 x. Solution  Finally, sketch the graph: x y – 2 2 – 2 2 42 f(x) = 2 x Example 3, page 331

11. Examples  Sketch the graph of the exponential function f(x) = (1/2) x. Solution  First, recall again that the domain of this function is the set of real numbers.  Next, putting x = 0 gives y = (1/2) 0 = 1, which is the y-intercept. (There is no x-intercept, since there is no value of x for which y = 0) Example 4, page 332

12. Examples  Sketch the graph of the exponential function f(x) = (1/2) x. Solution  Now, consider a few values for x:  Note that (1/2) x increases without bound when x decreases without bound.  Furthermore, (1/2) x approaches zero as x increases without bound: there is a horizontal asymptote at y = 0.  As before, the range of f is the interval (0,  ). x – 5 – 4 – 3 – 2 – 1 012345 y321684211/21/41/81/161/32 Example 4, page 332

13. Examples  Sketch the graph of the exponential function f(x) = (1/2) x. Solution  Finally, sketch the graph: x y – 2 2 – 2 2 42 f(x) = (1/2) x Example 4, page 332

14. Examples  Sketch the graph of the exponential function f(x) = (1/2) x. Solution  Note the symmetry between the two functions: x y – 2 2 4242 f(x) = (1/2) x f(x) = 2 x Example 4, page 332

15. Properties of Exponential Functions  The exponential function y = b x (b > 0, b ≠ 1) has the following properties: 1.Its domain is (– ,  ). 2.Its range is (0,  ). 3.Its graph passes through the point (0, 1) 4.It is continuous on (– ,  ). 5.It is increasing on (– ,  ) if b > 1 and decreasing on (– ,  ) if b 1 and decreasing on (– ,  ) if b < 1.

16. The Base e  Exponential functions to the base e, where e is an irrational number whose value is 2.7182818…, play an important role in both theoretical and applied problems.  It can be shown that

17. Examples  Sketch the graph of the exponential function f(x) = e x. Solution  Since e x > 0 it follows that the graph of y = e x is similar to the graph of y = 2 x.  Consider a few values for x: x – 3 – 2 – 1 0123 y0.050.140.3712.727.3920.09 Example 5, page 333

18. 531Examples  Sketch the graph of the exponential function f(x) = e x. Solution  Sketching the graph: x y – 3 – 11 3 – 3 – 11 3 f(x) = e x Example 5, page 333

19. Examples  Sketch the graph of the exponential function f(x) = e –x. Solution  Since e –x > 0 it follows that 0 0 it follows that 0 < 1/e < 1 and so f(x) = e –x = 1/e x = (1/e) x is an exponential function with base less than 1. f(x) = e –x = 1/e x = (1/e) x is an exponential function with base less than 1.  Therefore, it has a graph similar to that of y = (1/2) x.  Consider a few values for x: x – 3 – 2 – 1 0123 y20.097.392.7210.370.140.05 Example 6, page 333

20. 531531Examples  Sketch the graph of the exponential function f(x) = e –x. Solution  Sketching the graph: x y – 3 – 11 3 f(x) = e –x Example 6, page 333

21. 5.2 Logarithmic Functions

22. Logarithms  We’ve discussed exponential equations of the form y = b x (b > 0, b ≠ 1)  But what about solving the same equation for y?  You may recall that y is called the logarithm of x to the base b, and is denoted log b x. ✦ Logarithm of x to the base b y = log b x if and only if x = b y (x > 0)

23. Examples  Solve log 3 x = 4 for x: Solution  By definition, log 3 x = 4 implies x = 3 4 = 81. Example 2, page 338

24. Examples  Solve log 16 4 = x for x: Solution  log 16 4 = x is equivalent to 4 = 16 x = (4 2 ) x = 4 2x, or 4 1 = 4 2x, from which we deduce that Example 2, page 338

25. Examples  Solve log x 8 = 3 for x: Solution  By definition, we see that log x 8 = 3 is equivalent to Example 2, page 338

26. Logarithmic Notation log x= log 10 x Common logarithm ln x= log e x Natural logarithm

27. Laws of Logarithms  If m and n are positive numbers, then 1. 1. 2. 2. 3. 3. 4. 4. 5. 5.

28. Examples  Given that log 2 ≈ 0.3010, log 3 ≈ 0.4771, and log 5 ≈ 0.6990, use the laws of logarithms to find Example 4, page 339

29. Examples  Given that log 2 ≈ 0.3010, log 3 ≈ 0.4771, and log 5 ≈ 0.6990, use the laws of logarithms to find Example 4, page 339

30. Examples  Given that log 2 ≈ 0.3010, log 3 ≈ 0.4771, and log 5 ≈ 0.6990, use the laws of logarithms to find Example 4, page 339

31. Examples  Given that log 2 ≈ 0.3010, log 3 ≈ 0.4771, and log 5 ≈ 0.6990, use the laws of logarithms to find Example 4, page 339

32. Examples  Expand and simplify the expression: Example 5, page 340

33. Examples  Expand and simplify the expression: Example 5, page 340

34. Examples  Expand and simplify the expression: Example 5, page 340

35. Logarithmic Function  The function defined by is called the logarithmic function with base b.  The domain of f is the set of all positive numbers.

36. Properties of Logarithmic Functions  The logarithmic function y = log b x(b > 0, b ≠ 1) has the following properties: 1.Its domain is (0,  ). 2.Its range is (– ,  ). 3.Its graph passes through the point (1, 0). 4.It is continuous on (0,  ). 5.It is increasing on (0,  ) if b > 1 and decreasing on (0,  ) if b 1 and decreasing on (0,  ) if b < 1.

37. Example  Sketch the graph of the function y = ln x. Solution  We first sketch the graph of y = e x. 1 x y 1 y = e x y = ln x y = x  The required graph is the mirror image of the graph of y = e x with respect to the line y = x: Example 6, page 341

38. Properties Relating Exponential and Logarithmic Functions  Properties relating e x and ln x: e ln x = x(x > 0) ln e x = x (for any real number x)

39. Examples  Solve the equation 2e x + 2 = 5. Solution  Divide both sides of the equation by 2 to obtain:  Take the natural logarithm of each side of the equation and solve: Example 7, page 342

40. Examples  Solve the equation 5 ln x + 3 = 0. Solution  Add – 3 to both sides of the equation and then divide both sides of the equation by 5 to obtain: and so: Example 8, page 343

41. 5.3 Compound Interest

42.  Compound interest is a natural application of the exponential function to business.  Recall that simple interest is interest that is computed only on the original principal.  Thus, if I denotes the interest on a principal P (in dollars) at an interest rate of r per year for t years, then we have I = Prt  The accumulated amount A, the sum of the principal and interest after t years, is given by

43. Compound Interest  Frequently, interest earned is periodically added to the principal and thereafter earns interest itself at the same rate. This is called compound interest.  Suppose $1000 (the principal) is deposited in a bank for a term of 3 years, earning interest at the rate of 8% per year compounded annually.  Using the simple interest formula we see that the accumulated amount after the first year is or $1080.

44. Compound Interest  To find the accumulated amount A 2 at the end of the second year, we use the simple interest formula again, this time with P = A 1, obtaining: or approximately $1166.40.

45. Compound Interest  We can use the simple interest formula yet again to find the accumulated amount A 3 at the end of the third year: or approximately $1259.71.

46. Compound Interest  Note that the accumulated amounts at the end of each year have the following form:  These observations suggest the following general rule: ✦ If P dollars are invested over a term of t years earning interest at the rate of r per year compounded annually, then the accumulated amount is or:

47. Compounding More Than Once a Year  The formula was derived under the assumption that interest was compounded annually.  In practice, however, interest is usually compounded more than once a year.  The interval of time between successive interest calculations is called the conversion period.

48. Compounding More Than Once a Year  If interest at a nominal a rate of r per year is compounded m times a year on a principal of P dollars, then the simple interest rate per conversion period is  For example, the nominal interest rate is 8% per year, and interest is compounded quarterly, then or 2% per period.

49. Compounding More Than Once a Year  To find a general formula for the accumulated amount, we apply repeatedly with the interest rate i = r/m.  We see that the accumulated amount at the end of each period is as follows:

50. Compound Interest Formula where A= Accumulated amount at the end of t years P= Principal r= Nominal interest rate per year m= Number of conversion periods per year t= Term (number of years)  There are n = mt periods in t years, so the accumulated amount at the end of t year is given by

51. Example  Find the accumulated amount after 3 years if $1000 is invested at 8% per year compounded a.Annually b.Semiannually c.Quarterly d.Monthly e.Daily Example 1, page 347

52. ExampleSolution a.Annually. Here, P = 1000, r = 0.08, m = 1, and t = 3, so or $1259.71. Example 1, page 347

53. ExampleSolution b.Semiannually. Here, P = 1000, r = 0.08, m = 2, and t = 3, so or $1265.32. Example 1, page 347

54. ExampleSolution c.Quarterly. Here, P = 1000, r = 0.08, m = 4, and t = 3, so or $1268.24. Example 1, page 347

55. ExampleSolution d.Monthly. Here, P = 1000, r = 0.08, m = 12, and t = 3, so or $1270.24. Example 1, page 347

56. ExampleSolution e.Daily. Here, P = 1000, r = 0.08, m = 365, and t = 3, so or $1271.22. Example 1, page 347

57. Effective Rate of Interest  The last example demonstrates that the interest actually earned on an investment depends on the frequency with which the interest is compounded.  For clarity when comparing interest rates, we can use what is called the effective rate (also called the true rate): ✦ This is the simple interest rate that would produce the same accumulated amount in 1 year as the nominal rate compounded m times a year.  We want to derive a relation between the nominal compounded rate and the effective rate.

58. Effective Rate of Interest  The accumulated amount after 1 year at a simple interest rate r eff per year is  The accumulated amount after 1 year at a nominal interest rate r per year compounded m times a year is  Equating the two expressions gives Since t = 1

59. Effective Rate of Interest Formula  Solving the last equation for r eff we obtain the formula for computing the effective rate of interest: where r eff = Effective rate of interest r= Nominal interest rate per year m= Number of conversion periods per year

60. Example  Find the effective rate of interest corresponding to a nominal rate of 8% per year compounded a.Annually b.Semiannually c.Quarterly d.Monthly e.Daily Example 2, page 350

61. ExampleSolution a.Annually. Let r = 0.08 and m = 1. Then or 8%. Example 2, page 350

62. ExampleSolution b.Semiannually. Let r = 0.08 and m = 2. Then or 8.16%. Example 2, page 350

63. ExampleSolution c.Quarterly. Let r = 0.08 and m = 4. Then or 8.243%. Example 2, page 350

64. ExampleSolution d.Monthly. Let r = 0.08 and m = 12. Then or 8.3%. Example 2, page 350

65. ExampleSolution e.Daily. Let r = 0.08 and m = 365. Then or 8.328%. Example 2, page 350

66. Effective Rate Over Several Years  If the effective rate of interest r eff is known, then the accumulated amount after t years on an investment of P dollars may be more readily computed by using the formula

67. Present Value  Consider the compound interest formula:  The principal P is often referred to as the present value, and the accumulated value A is called the future value, since it is realized at a future date.  On occasion, an investor may wish to determine how much money he should invest now, at a fixed rate of interest, so that he will realize a certain sum at some future date.  This problem may be solved by expressing P in terms of A.

68. Present Value  Present value formula for compound interest

69. Examples  How much money should be deposited in a bank paying a yearly interest rate of 6% compounded monthly so that after 3 years the accumulated amount will be $20,000? Solution  Here, A = 20,000, r = 0.06, m = 12, and t = 3.  Using the present value formula we get Example 3, page 351

70. Examples  Find the present value of $49,158.60 due in 5 years at an interest rate of 10% per year compounded quarterly. Solution  Here, A = 49,158.60, r = 0.1, m = 4, and t = 5.  Using the present value formula we get Example 4, page 350

71. Continuous Compounding of Interest  One question arises on compound interest: ✦ What happens to the accumulated amount over a fixed period of time if the interest is compounded more and more frequently?  We’ve seen that the more often interest is compounded, the larger the accumulated amount.  But does the accumulated amount approach a limit when interest is computed more and more frequently?

72. Continuous Compounding of Interest  Recall that in the compound interest formula the number of conversion periods is m.  So, we should let m get larger and larger (approach infinity) and see what happens to the accumulated amount A.  But first, for clarity, lets rewrite the equation as follows:

73. Continuous Compounding of Interest  Letting m → , we find that  We can substitute u = m/r (note that u →  as m →  ).  Thus

74. Continuous Compounding of Interest  Now, you may recall that  So, we can restate as follows:  Thus, as the frequency with which interest is compounded increases without bound, the accumulated amount approaches Pe rt.

75. Continuous Compounding of Interest  Continuous Compound Interest Formula A = Pe rt where P= Principal r= Annual interest rate compounded continuously. continuously. t= Time in years. A= Accumulated amount at the end of t years. of t years.

76. Examples  Find the accumulated amount after 3 years if $1000 is invested at 8% per year compounded (a) daily, and (b) continuously. Solution a.Using the compound interest formula with P = 1000, r = 0.08, m = 365, and t = 3, we find b.Using the continuous compound interest formula with P = 1000, r = 0.08, and t = 3, we find A = Pe rt = 1000e (0.08)(3) ≈ 1271.25 Note that both solutions are very similar. Example 5, page 352

77. Examples  How long will it take $10,000 to grow to $15, 000 if the investment earns an interest rate of 12% per year compounded quarterly? Solution  Using the compound interest formula with A = 15,000, P = 10,000, r = 0.12, and m = 4, we obtain Example 7, page 354

78. Examples  How long will it take $10,000 to grow to $15, 000 if the investment earns an interest rate of 12% per year compounded quarterly? Solution  Taking logarithms on both sides gives  So, it will take approximately 3.4 years for the investment to grow from $10,000 to $15,000. Example 7, page 354

79. Examples  Find the interest rate needed for an investment of $10,000 to grow to an amount of $18,000 in 5 years if the interest is compounded monthly. Solution  Using the compound interest formula with A = 18,000, P = 10,000, m = 12, and t = 5, we find Example 8, page 355

80. Examples  Find the interest rate needed for an investment of $10,000 to grow to an amount of $18,000 in 5 years if the interest is compounded monthly. Solution  Taking the 60 th root on both sides and solving for r we get Example 8, page 355

81. Examples  Find the interest rate needed for an investment of $10,000 to grow to an amount of $18,000 in 5 years if the interest is compounded monthly. Solution  Converting back into an exponential equation, and  Thus, the interest rate needed is approximately 11.81% per year. Example 8, page 355

82. 5.4 Differentiation of the Exponential Function

83. Rule 1 Derivative of the Exponential Function  The derivative of the exponential function with base e is equal to the function itself:

84. Examples  Find the derivative of the function Solution  Using the product rule gives Example 1, page 362

85. Examples  Find the derivative of the function Solution  Using the general power rule gives Example 1, page 362

86. Rule 2 Chain Rule for Exponential Functions  If f(x) is a differentiable function, then

87. Examples  Find the derivative of the function Solution Example 2, page 363

88. Examples  Find the derivative of the function Solution Example 2, page 363

89. Examples  Find the derivative of the function Solution Example 2, page 363

90. Examples  Find the derivative of the function Solution Example 3, page 363

91. Examples  Find the derivative of the function Solution Example 4, page 364

92. Examples  Find the inflection points of the function Solution  Find the first and second derivatives of f :  Setting f ″ = 0 gives e – x 2 = 0, and 2x 2 – 1 = 0.  Since e – x 2 never equals zero for any real value of x, the only candidates for inflection points of f are  Testing values around these numbers we conclude that they are indeed inflection points. Example 6, page 364

93. Examples  Find the inflection points of the function Solution 1 x y – 1 1 – 1 1 Example 6, page 364

94. 5.5 Differentiation of Logarithmic Functions

95. Rule 3 Derivative of the Natural Logarithm  The derivative of ln x is

96. Examples  Find the derivative of the function Solution Example 1, page 372

97. Examples  Find the derivative of the function Solution Example 1, page 372

98. Rule 4 Chain Rule for Logarithmic Functions  If f(x) is a differentiable function, then

99. Examples  Find the derivative of the function Solution Example 2, page 373

100. Examples  Find the derivative of the function Solution Example 3, page 373

101. Logarithmic Differentiation  We have seen how finding derivatives of logarithmic functions becomes easier when applying the laws of logarithms.  These laws can also be used in a process called logarithmic differentiation to permit the differentiation of functions that would be difficult to differentiate or even not be differentiable through other means.

102. Examples  Use logarithmic differentiation to find the derivative of Solution  Take the natural logarithm of both sides of the equation:  Use the laws of logarithms to rewrite the equation:  Differentiate both sides of the equation: Example 5, page 374

103. Examples  Use logarithmic differentiation to find the derivative of Solution  On the left side, note that y is a function of x, therefore: Example 5, page 374

104. Examples  Use logarithmic differentiation to find the derivative of Solution  Thus, we have: Example 5, page 374

105. Examples  Use logarithmic differentiation to find the derivative of Solution  Finally, solving for y′ we get: Example 5, page 374

106. Logarithmic Differentiation  To find dy/dx by logarithmic differentiation: 1.Take the natural logarithm on both sides of the equation and use the properties of logarithms to write any “complicated expression” as a sum of simpler terms. 2.Differentiate both sides of the equation with respect to x. 3.Solve the resulting equation for dy/dx.

107. Examples  Use logarithmic differentiation to find the derivative of Solution 1.Take the natural logarithm of both sides of the equation and use the laws of logarithms to rewrite the equation: Example 6, page 375

108. Examples  Use logarithmic differentiation to find the derivative of Solution 2.Differentiate both sides of the equation: Example 6, page 375

109. Examples  Use logarithmic differentiation to find the derivative of Solution 3.Solve for dy/dx: Example 6, page 375

110. Examples  Use logarithmic differentiation to find the derivative of Solution 1.Take the natural logarithm of both sides of the equation and use the laws of logarithms to rewrite the equation: Example 7, page 376

111. Examples  Use logarithmic differentiation to find the derivative of Solution 2.Differentiate both sides of the equation: Example 7, page 376

112. Examples  Use logarithmic differentiation to find the derivative of Solution 3.Solve for dy/dx: Example 7, page 376

113. 5.6 Exponential Functions as Mathematical Models 1.Growth of bacteria 2.Radioactive decay 3.Assembly time

114. Applied Example: Growth of Bacteria  Under a laboratory, the number of bacteria in a culture grows according to where Q 0 denotes the number of bacteria initially present in the culture, k is a constant determined by the strain of bacteria under consideration, and t is the elapsed time measured in hours.  Suppose 10,000 bacteria are present initially in the culture and 60,000 present two hours later. a.How many bacteria will there be in the culture at the end of four hours? b.What is the rate of growth of the population after four hours? Applied Example 1, page 380

115. Applied Example: Growth of Bacteria Solution a.We are given that Q(0) = Q 0 = 10,000, so Q(t) = 10,000e kt. At t = 2 there are 60,000 bacteria, so Q(2) = 60,000, thus: Taking the natural logarithm on both sides we get: So, the number of bacteria present at any time t is given by: Applied Example 1, page 380

116. Applied Example: Growth of Bacteria Solution a.At the end of four hours (t = 4), there will be or 360,029 bacteria. Applied Example 1, page 380

117. Applied Example: Growth of Bacteria Solution b.The rate of growth of the bacteria at any time t is given by Using the result from part (a), we find that the rate of bacterial growth at the end of four hours is or approximately 322,550 bacteria per hour. Applied Example 1, page 380

118. Applied Example: Radioactive Decay  Radioactive substances decay exponentially.  For example, the amount of radium present at any time t obeys the law where Q 0 is the initial amount present and k is a suitable positive constant.  The half-life of a radioactive substance is the time required for a given amount to be reduced by one-half.  The half-life of radium is approximately 1600 years.  Suppose initially there are 200 milligrams of pure radium. a.Find the amount left after t years. b.What is the amount after 800 years? Applied Example 2, page 382

119. Applied Example: Radioactive Decay Solution a.The initial amount is 200 milligrams, so Q(0) = Q 0 = 200, so Q(t) = 200e –kt The half-life of radium is 1600 years, so Q(1600) = 100, thus Applied Example 2, page 382

120. Applied Example: Radioactive Decay Solution a.Taking the natural logarithm on both sides yields: Therefore, the amount of radium left after t years is: Applied Example 2, page 382

121. Applied Example: Radioactive Decay Solution b.In particular, the amount of radium left after 800 years is: or approximately 141 milligrams. or approximately 141 milligrams. Applied Example 2, page 382

122. Applied Example: Assembly Time  The Camera Division of Eastman Optical produces a single lens reflex camera.  Eastman’s training department determines that after completing the basic training program, a new, previously inexperienced employee will be able to assemble model F cameras per day, t months after the employee starts work on the assembly line. a.How many model F cameras can a new employee assemble per day after basic training? b.How many model F cameras can an employee with one month of experience assemble per day? c.How many model F cameras can the average experienced employee assemble per day? Applied Example 5, page 384

123. Applied Example: Assembly Time Solution a.The number of model F cameras a new employee can assemble is given by b.The number of model F cameras that an employee with 1, 2, and 6 months of experience can assemble per day is given by or about 32 cameras per day. c.As t increases without bound, Q(t) approaches 50. Hence, the average experienced employee can be expected to assemble 50 model F cameras per day. Applied Example 5, page 384

124. End of Chapter