1. Exponential Functions

2. Definition of the Exponential Function The exponential function f with base b is defined by f (x) = b x or y = b x Where b is a positive constant other than and x is any real number. The exponential function f with base b is defined by f (x) = b x or y = b x Where b is a positive constant other than and x is any real number. / Here are some examples of exponential functions. f (x) = 2 x g(x) = 10 x h(x) = 3 x+1 Base is 2.Base is 10.Base is 3.

3. Text Example The exponential function f (x) = 13.49(0.967) x – 1 describes the number of O-rings expected to fail, f (x), when the temperature is x°F. On the morning the Challenger was launched, the temperature was 31°F, colder than any previous experience. Find the number of O-rings expected to fail at this temperature. SolutionBecause the temperature was 31°F, substitute 31 for x and evaluate the function at 31. f (x) = 13.49(0.967) x – 1 This is the given function. f (31) = 13.49(0.967) 31 – 1 Substitute 31 for x. Press.967 ^ 31 on a graphing calculator to get.353362693426. Multiply this by 13.49 and subtract 1 to obtain f (31) = 13.49(0.967) 31 – 1=3.77

4. Characteristics of Exponential Functions The domain of f (x) = b x consists of all real numbers. The range of f (x) = b x consists of all positive real numbers. The graphs of all exponential functions pass through the point (0, 1) because f (0) = b 0 = 1. If b > 1, f (x) = b x has a graph that goes up to the right and is an increasing function. If 0 < b < 1, f (x) = b x has a graph that goes down to the right and is a decreasing function. f (x) = b x is a one-to-one function and has an inverse that is a function. The graph of f (x) = b x approaches but does not cross the x-axis. The x- axis is a horizontal asymptote. The domain of f (x) = b x consists of all real numbers. The range of f (x) = b x consists of all positive real numbers. The graphs of all exponential functions pass through the point (0, 1) because f (0) = b 0 = 1. If b > 1, f (x) = b x has a graph that goes up to the right and is an increasing function. If 0 < b < 1, f (x) = b x has a graph that goes down to the right and is a decreasing function. f (x) = b x is a one-to-one function and has an inverse that is a function. The graph of f (x) = b x approaches but does not cross the x-axis. The x- axis is a horizontal asymptote. f (x) = b x b > 1 f (x) = b x 0 < b < 1

5. Transformations Involving Exponential Functions Shifts the graph of f (x) = b x upward c units if c > 0. Shifts the graph of f (x) = b x downward c units if c < 0. g(x) = -b x + cVertical translation Reflects the graph of f (x) = b x about the x-axis. Reflects the graph of f (x) = b x about the y-axis. g(x) = -b x g(x) = b -x Reflecting Multiplying y-coordintates of f (x) = b x by c, Stretches the graph of f (x) = b x if c > 1. Shrinks the graph of f (x) = b x if 0 < c < 1. g(x) = c b x Vertical stretching or shrinking Shifts the graph of f (x) = b x to the left c units if c > 0. Shifts the graph of f (x) = b x to the right c units if c < 0. g(x) = b x+c Horizontal translation DescriptionEquationTransformation

6. Text Example Use the graph of f (x) = 3 x to obtain the graph of g(x) = 3 x+1. SolutionExamine the table below. Note that the function g(x) = 3 x+1 has the general form g(x) = b x+c, where c = 1. Because c > 0, we graph g(x) = 3 x+1 by shifting the graph of f (x) = 3 x one unit to the left. We construct a table showing some of the coordinates for f and g to build their graphs. 3 2+1 = 3 3 = 273 2 = 92 3 1+1 = 3 2 = 93 1 = 31 3 0+1 = 3 1 = 33 0 = 10 3 -1+1 = 3 0 = 13 -1 = 1/3 3 -2+1 = 3 -1 = 1/33 -2 = 1/9-2 g(x) = 3 x+1 f (x) = 3 x x g(x) = 3 x+1 (0, 1) (-1, 1) 123456 -5-4-3-2

7. The Natural Base e An irrational number, symbolized by the letter e, appears as the base in many applied exponential functions. This irrational number is approximately equal to 2.72. More accurately, The number e is called the natural base. The function f (x) = e x is called the natural exponential function. f (x) = e x f (x) = 2 x f (x) = 3 x (0, 1) (1, 2) 1 2 3 4 (1, e) (1, 3)

8. Formulas for Compound Interest After t years, the balance, A, in an account with principal P and annual interest rate r (in decimal form) is given by the following formulas: 1.For n compoundings per year: A = P(1 + r/n) nt 2.For continuous compounding: A = Pe rt.

9. Example Use A= Pe rt to solve the following problem: Find the accumulated value of an investment of $2000 for 8 years at an interest rate of 7% if the money is compounded continuously Solution: A= Pe rt A = 2000e (.07)(8) A = 2000 e (.56) A = 2000 * 1.75 A = $3500

10. Exponential Functions

11. Logarithmic Functions

12. Definition of a Logarithmic Function For x > 0 and b > 0, b = 1, y = log b x is equivalent to b y = x. The function f (x) = log b x is the logarithmic function with base b.

13. Location of Base and Exponent in Exponential and Logarithmic Forms Logarithmic form: y = log b x Exponential Form: b y = x. Exponent Base

14. Text Example Write each equation in its equivalent exponential form. a. 2 = log 5 xb. 3 = log b 64c. log 3 7 = y SolutionWith the fact that y = log b x means b y = x, c. log 3 7 = y or y = log 3 7 means 3 y = 7. a. 2 = log 5 x means 5 2 = x. Logarithms are exponents. b. 3 = log b 64 means b 3 = 64. Logarithms are exponents.

15. Evaluate a. log 2 16b. log 3 9 c. log 25 5 Solution log 25 5 = 1/2 because 25 1/2 = 5.25 to what power is 5?c. log 25 5 log 3 9 = 2 because 3 2 = 9.3 to what power is 9?b. log 3 9 log 2 16 = 4 because 2 4 = 16.2 to what power is 16?a. log 2 16 Logarithmic Expression Evaluated Question Needed for Evaluation Logarithmic Expression Text Example

16. Basic Logarithmic Properties Involving One Log b b = 1because 1 is the exponent to which b must be raised to obtain b. (b 1 = b). Log b 1 = 0because 0 is the exponent to which b must be raised to obtain 1. (b 0 = 1).

17. Inverse Properties of Logarithms For x > 0 and b  1, log b b x = xThe logarithm with base b of b raised to a power equals that power. b log b x = xb raised to the logarithm with base b of a number equals that number.

18. Graph f (x) = 2 x and g(x) = log 2 x in the same rectangular coordinate system. SolutionWe first set up a table of coordinates for f (x) = 2 x. Reversing these coordinates gives the coordinates for the inverse function, g(x) = log 2 x. 4 2 8211/21/4f (x) = 2 x 310-2x 2 4 310-2g(x) = log 2 x 8211/21/4x Reverse coordinates. Text Example

19. Graph f (x) = 2 x and g(x) = log 2 x in the same rectangular coordinate system. Solution We now plot the ordered pairs in both tables, connecting them with smooth curves. The graph of the inverse can also be drawn by reflecting the graph of f (x) = 2 x over the line y = x. -2 6 2345 5 4 3 2 -2 6 f (x) = 2 x f (x) = log 2 x y = x Text Example cont.

20. Characteristics of the Graphs of Logarithmic Functions of the Form f(x) = log b x The x-intercept is 1. There is no y-intercept. The y-axis is a vertical asymptote. If b > 1, the function is increasing. If 0 < b < 1, the function is decreasing. The graph is smooth and continuous. It has no sharp corners or edges.

21. Properties of Common Logarithms General PropertiesCommon Logarithms 1. log b 1 = 01. log 1 = 0 2. log b b = 12. log 10 = 1 3. log b b x = 03. log 10 x = x 4. b log b x = x 4. 10 log x = x

22. Examples of Logarithmic Properties log b b = 1 log b 1 = 0 log 4 4 = 1 log 8 1 = 0 3 log 3 6 = 6 log 5 5 3 = 3 2 log 2 7 = 7

23. Properties of Natural Logarithms General PropertiesNatural Logarithms 1. log b 1 = 01. ln 1 = 0 2. log b b = 12. ln e = 1 3. log b b x = 03. ln e x = x 4. b log b x = x 4. e ln x = x

24. Examples of Natural Logarithmic Properties log e e = 1 log e 1 = 0 e log e 6 = 6 log e e 3 = 3

25. Logarithmic Functions

26. Properties of Logarithms

27. The Product Rule Let b, M, and N be positive real numbers with b  1. log b (MN) = log b M + log b N The logarithm of a product is the sum of the logarithms. For example, we can use the product rule to expand ln (4x): ln (4x) = ln 4 + ln x.

28. The Quotient Rule Let b, M and N be positive real numbers with b  1. The logarithm of a quotient is the difference of the logarithms.

29. The Power Rule Let b, M, and N be positive real numbers with b = 1, and let p be any real number. log b M p = p log b M The logarithm of a number with an exponent is the product of the exponent and the logarithm of that number.

30. Text Example Write as a single logarithm: a. log 4 2 + log 4 32 Solution a. log 4 2 + log 4 32 = log 4 (2 32) Use the product rule. = log 4 64 = 3 Although we have a single logarithm, we can simplify since 4 3 = 64.

31. Properties for Expanding Logarithmic Expressions For M > 0 and N > 0:

32. Example Use logarithmic properties to expand the expression as much as possible.

33. Example cont.

35. Properties for Condensing Logarithmic Expressions For M > 0 and N > 0:

36. The Change-of-Base Property For any logarithmic bases a and b, and any positive number M, The logarithm of M with base b is equal to the logarithm of M with any new base divided by the logarithm of b with that new base.

37. Use logarithms to evaluate log 3 7. Solution: or so Example

38. Properties of Logarithms

39. Exponential and Logarithmic Equations

40. Using Natural Logarithms to Solve Exponential Equations 1. Isolate the exponential expression. 2. Take the natural logarithm on both sides of the equation. 3. Simplify using one of the following properties: ln b x = x ln b or ln e x = x. 4. Solve for the variable.

41. Solve: 5 4x – 7 – 3 = 10 SolutionWe begin by adding 3 to both sides to isolate the exponential expression, 5 4x – 7. Then we take the natural logarithm on both sides of the equation. 5 4x – 7 – 3 = 10 This is the given equation. 5 4x – 7 = 13 Add 3 to both sides. ln 5 4x – 7 = ln 13 Take the natural logarithm on both sides. (4x – 7) ln 5 = ln 13 Use the power rule to bring the exponent to the front. 4x ln 5 – 7 ln 5 = ln 13 Use the distributive property on the left side of the equation. Text Example

42. Solve: 5 4x – 7 – 3 = 10 Solution 4x ln 5 = ln 13 + 7 ln 5 Isolate the variable term by adding 7 ln 5 to both sides. Text Example cont. x = (ln 13)/(4 ln 5) + (7 ln 5)/(4 ln 5) Isolate x by dividing both sides by 4 ln 5. The solution set is {(ln 13 + 7 ln 5)/(4 ln 5)} approximately 2.15.

43. Solve: log 4 (x + 3) = 2. SolutionWe first rewrite the equation as an equivalent equation in exponential form using the fact that log b x = c means b c = x. log 4 (x + 3) = 2 means 4 2 = x + 3 Now we solve the equivalent equation for x. 4 2 = x + 3 This is the equivalent equation. 16 = x + 3 Square 4. 13 = x Subtract 3 from both sides. Check log 4 (x + 3) = 2 This is the logarithmic equation. log 4 (13 + 3) = 2 Substitute 13 for x. log 4 16 = 2 2 = 2 This true statement indicates that the solution set is {13}. ? ? Text Example

44. Example Solve 3 x+2 -7 = 27 Solution: 3 x+2 = 34 ln 3 x+2 = ln 34 (x+2) ln 3 = ln 34 x+2 = (ln 34)/(ln 3) x+2 = 3.21 x = 1.21

45. Example Solve log 2 (3x-1) = 18 Solution: 2 18 = 3x-1 262,144 = 3x - 1 262,145 = 3x 262,145 / 3 = x x = 87,381.67

46. Exponential and Logarithmic Equations

47. Modeling with Exponential and Logarithmic Functions

48. The mathematical model for exponential growth or decay is given by f (t) = A 0 e kt or A = A 0 e kt. If k > 0, the function models the amount or size of a growing entity. A 0 is the original amount or size of the growing entity at time t = 0. A is the amount at time t, and k is a constant representing the growth rate. If k < 0, the function models the amount or size of a decaying entity. A 0 is the original amount or size of the decaying entity at time t = 0. A is the amount at time t, and k is a constant representing the decay rate. The mathematical model for exponential growth or decay is given by f (t) = A 0 e kt or A = A 0 e kt. If k > 0, the function models the amount or size of a growing entity. A 0 is the original amount or size of the growing entity at time t = 0. A is the amount at time t, and k is a constant representing the growth rate. If k < 0, the function models the amount or size of a decaying entity. A 0 is the original amount or size of the decaying entity at time t = 0. A is the amount at time t, and k is a constant representing the decay rate. decreasing A0A0 x y increasing y = A 0 e kt k > 0 x y y = A 0 e kt k < 0 A0A0 Exponential Growth and Decay Models

49. The graph below shows the growth of the Mexico City metropolitan area from 1970 through 2000. In 1970, the population of Mexico City was 9.4 million. By 1990, it had grown to 20.2 million. Find the exponential growth function that models the data. By what year will the population reach 40 million? 20 15 10 5 25 30 1970198019902000 Population (millions) Year Example

50. Solution a. We use the exponential growth model A = A 0 e kt in which t is the number of years since 1970. This means that 1970 corresponds to t = 0. At that time there were 9.4 million inhabitants, so we substitute 9.4 for A 0 in the growth model. A = 9.4 e kt We are given that there were 20.2 million inhabitants in 1990. Because 1990 is 20 years after 1970, when t = 20 the value of A is 20.2. Substituting these numbers into the growth model will enable us to find k, the growth rate. We know that k > 0 because the problem involves growth. A = 9.4 e kt Use the growth model with A 0 = 9.4. 20.2 = 9.4 e k20 When t = 20, A = 20.2. Substitute these values. Example cont.

51. Solution We substitute 0.038 for k in the growth model to obtain the exponential growth function for Mexico City. It is A = 9.4 e 0.038t where t is measured in years since 1970. Example cont. 20.2/ 9.4 = e k20 Isolate the exponential factor by dividing both sides by 9.4. ln(20.2/ 9.4) = lne k20 Take the natural logarithm on both sides. 0.038 = k Divide both sides by 20 and solve for k. 20.2/ 9.4 = 20k Simplify the right side by using ln e x = x.

52. Solution b. To find the year in which the population will grow to 40 million, we substitute 40 in for A in the model from part (a) and solve for t. Because 38 is the number of years after 1970, the model indicates that the population of Mexico City will reach 40 million by 2008 (1970 + 38). A = 9.4 e 0.038t This is the model from part (a). 40 = 9.4 e 0.038t Substitute 40 for A. Example cont. ln(40/9.4) = lne 0.038t Take the natural logarithm on both sides. ln(40/9.4)/0.038 =t Solve for t by dividing both sides by 0.038 ln(40/9.4) =0.038t Simplify the right side by using ln e x = x. 40/9.4 = e 0.038t Divide both sides by 9.4.

53. Use the fact that after 5715 years a given amount of carbon-14 will have decayed to half the original amount to find the exponential decay model for carbon-14. In 1947, earthenware jars containing what are known as the Dead Sea Scrolls were found by an Arab Bedouin herdsman. Analysis indicated that the scroll wrappings contained 76% of their original carbon-14. Estimate the age of the Dead Sea Scrolls. Solution We begin with the exponential decay model A = A 0 e kt. We know that k < 0 because the problem involves the decay of carbon-14. After 5715 years (t = 5715), the amount of carbon-14 present, A, is half of the original amount A 0. Thus we can substitute A 0 /2 for A in the exponential decay model. This will enable us to find k, the decay rate. Text Example

54. Substituting for k in the decay model, the model for carbon-14 is A = A 0 e –0.000121t. k = ln(1/2)/5715=-0.000121 Solve for k. 1/2= e kt5715 Divide both sides of the equation by A 0. Solution A 0 /2= A 0 e k5715 After 5715 years, A = A 0 /2 ln(1/2) = ln e k5715 Take the natural logarithm on both sides. ln(1/2) = 5715k ln e x = x. Text Example cont.

55. Solution The Dead Sea Scrolls are approximately 2268 years old plus the number of years between 1947 and the current year. A = A 0 e -0.000121t This is the decay model for carbon-14. 0.76A 0 = A 0 e -0.000121t A =.76A 0 since 76% of the initial amount remains. 0.76 = e -0.000121t Divide both sides of the equation by A 0. ln 0.76 = ln e -0.000121t Take the natural logarithm on both sides. ln 0.76 = -0.000121t ln e x = x. Text Example cont. t=ln(0.76)/(-0.000121) Solver for t.

56. Logistic Growth Model The mathematical model for limited logistic growth is given by Where a, b, and c are constants, with c > 0 and b > 0.

57. Newton’s Law of Cooling The temperature, T, of a heated object at time t is given by T = C + (T 0 - C)e kt Where C is the constant temperature of the surrounding medium, T 0 is the initial temperature of the heated object, and k is a negative constant that is associated with the cooling object.

58. Expressing an Exponential Model in Base e. y = ab x is equivalent to y = ae (lnb) x

59. Example The value of houses in your neighborhood follows a pattern of exponential growth. In the year 2000, you purchased a house in this neighborhood. The value of your house, in thousands of dollars, t years after 2000 is given by the exponential growth model V = 125e.07t When will your house be worth $200,000?

60. Example Solution: V = 125e.07t 200 = 125e.07t 1.6 = e.07t ln1.6 = ln e.07t ln 1.6 =.07t ln 1.6 /.07 = t 6.71 = t

61. Modeling with Exponential and Logarithmic Functions