1. Chapter 6 Exponential and Logarithmic Functions and Applications Section 6.3

2. Section 6.3 The Natural Exponential and Logarithmic Functions The Number e Natural Exponential Function: Characteristics, and Graphs o Transformations Exponential Growth and Decay Natural Logarithmic Functions o Basic Properties of Natural Logarithms o Characteristics and Graphs o Transformations Applications

3. The Number e The number e, which is approximately 2.71828182845, is a famous and very important irrational number. The base e, is called the natural base and it occurs frequently in many fields and important applications. Example: Use your calculator to evaluate e 3. Round your answer to 4 decimal places. We can use the e key or the e x key. Using the e key: ` / ^ 3 e The calculator output is 20.08553692  20.0855 Using the e x key: ` L 3 ) e  20.0855

4. Use your calculator to evaluate the following. Round your answers to 4 decimal places. a. e – 0.3 b. 1/ e c. 10 e 2 d. e 0.03  4 a. Using the e key: ` / ^ _ 0.3 e  0.7408 b. Using the e key: 1 / ` / e  0.3679 c. Using the e x key: 10 ` L 2 ) e  73.8906 d. Using the e x key: ` L 0.03 * 4 ) e  1.1275 Note: If we decide to use the e key for example (d) instead, we need to enclose "0.03  4" within a set of parentheses. Just typing ` / ^ 0.03 * 4 e will yield 4.1218, which is incorrect.

5. The Natural Exponential Function The natural exponential function is the exponential function with base e : f(x) = e x The domain is (– ∞, ∞ ) and the range is (0, ∞ ). The y-intercept is (0, 1). Horizontal asymptote at y = 0 (that is, the x-axis). The function is one-to-one. The graph of y = e x has the following shape:

6. For each function, state the transformations applied to y = e x. Determine the horizontal asymptote, the y-intercept, and the domain and range for each function. a. y = e x + 1 + 4 b. y = e x – 2 – 3 a. Horizontal shift, left 1 unit Vertical shift, up 4 units Horizontal asymptote: y = 4 y-intercept: f(0) = e [(0) + 1] + 4  6.72 Domain: (– ∞, ∞ ) Range: (4, ∞ ) b. Horizontal shift, right 2 units Vertical shift, down 3 units Horizontal asymptote: y = –3 y-intercept: f(0) = e [(0) – 2] – 3  –2.86 Domain: (– ∞, ∞ ) Range: (–3, ∞ )

7. For each function, state the transformations applied to y = e x. Determine the horizontal asymptote, the y-intercept, and the domain and range for each function. a. y = – e x b. y = e – x a. Reflection about the x-axis Horizontal asymptote: y = 0 y-intercept: f(0) = – e (0) = –(1) = –1 Domain: (– ∞, ∞ ) Range: (– ∞, 0 ) b. Reflection about the y-axis Horizontal asymptote: y = 0 y-intercept: f(0) = 1 Domain: (– ∞, ∞ ) Range: (0, ∞ )

8. Exponential Growth and Decay Recall from Section 6.1 the basic exponential function P(t) = P 0 b t. We can also use a natural exponential function to model exponential growth and decay. P(t) = P 0 e kt If k > 0, it models exponential growth. If k < 0, it models exponential decay. t = time P 0 = P(0) = initial amount, or value of P at time 0; P > 0 k = continuous growth or decay rate (expressed as a decimal) e k = growth or decay factor Note: e k is equivalent to the base b in P(t) = P 0 b t. Substituting e k for b, P(t) = P 0 ( e k ) t = P 0 e kt.

9. Given the natural exponential function P(t) = 1075 e 0.0265t, identify the initial value, the continuous growth or decay rate, and the growth or decay factor. Initial value: P 0 = P(0) = 1075 Continuous growth rate: k = 0.0265, thus 2.65% Growth factor: The function is equivalent to P(t) =1075( e 0.0265 ) t. e 0.0265 rounded to 4 decimal places is 1.0269. So, P(t) = 1075( e 0.0265 ) t is equivalent to P(t) = 1075(1.0269) t, and the growth factor is 1.0269.

10. For the natural exponential function N(t) = 483 e – 0.037t, identify the initial value, the continuous growth or decay rate, and the growth or decay factor. Initial value: N 0 = N(0) = 483 Continuous decay rate: k = –0.037, thus –3.7% Decay factor: The function is equivalent to N(t) = 483( e – 0.037 ) t. The decay factor is e – 0.037 or 0.9637, rounded to 4 decimal places.

11. Ronald bought a sport utility vehicle (SUV) in 2009, which started losing its value as soon as he drove off the lot. The standard for vehicle depreciation is that, on average, they lose about 15-20% of their value each year. Ronald's SUV's value in dollars can be modeled by the function V(t) = 21305 e –0.173t, where t represents years after 2009. Sources: www.aaa.com; www.safecarguide.com; www.bankrate.com a. Find and interpret V(0). V 0 = V(0) = 21305. Ronald’s SUV initial value, before it was driven away from the car lot, was $21,305. b. What was the value of Ronald's SUV after the first 2 years? Round your answer to the nearest whole number. V(2) = 21305 e –0.173(2) = 15074 After 2 years, Ronald's SUV value decreased to $15,074. (continued on the next slide)

12. (Contd.) c. Using your graphing calculator, create a table of values for the given function, V(t) = 21305 e –0.173t, for 0  t  13. Find V(5) numerically and algebraically. Round your answer to the nearest whole number. Interpret your answer. From the table, V(5) = 8971. V(5) = 21305 e –0.173(5) = 8971 After 5 years, Ronald's SUV value will decline to $8,971. (continued on the next slide)

13. (Contd.) d. Graph the given function, V(t) = 21305 e –0.173t. Use your table of values from part (c) to help you determine an appropriate viewing window. Use your graph to find when the SUV's value will drop to $5,338. Confirm numerically (with the table). Vehicle Value Years [0, 10, 1] by [0, 25000, 5000] Under normal circumstances, the SUV's value drops to $5,338 after 8 years, or 2017.

14. The Natural Logarithmic Function Recall the inverse of the exponential function y = b x is the logarithmic function y = log b x. Similarly, y = e x and y = log e x are inverse functions. Definition: Let x > 0. The natural logarithmic function with base e is defined as y = log e x. y = log e x is denoted by y = ln x and y = ln x if and only if x = e y

15. Basic Properties of Natural Logarithms The basic properties studied previously for logarithms will apply for natural logarithms. 1. ln 1 = 0 2.ln e = 1 3.ln e x = x 4. Note: Remember that y = ln x is the same as y = log e x.

16. Apply the basic properties to evaluate the following. 1. 2. 3.

17. Graph of the Natural Logarithmic Function The natural logarithmic function is f(x) = ln x. The natural logarithmic function is the inverse of the natural exponential function, thus the graph of y = ln x is the graph of y = e x reflected about the line y = x. The domain is (0, ∞ ) and the range is (– ∞, ∞ ). The x-intercept is (1, 0). Vertical asymptote at x = 0 (that is, the y-axis). The function is one-to-one. The graph of y = ln x has the following shape:

18. For each function, state the transformations applied to y = ln x. Determine the vertical asymptote, domain, and range for each function. a. y = ln x – 4 b. y = ln (x + 4) a. Vertical shift, down 4 units Vertical asymptote: x = 0 Domain: (0, ∞ ) Range: (– ∞, ∞ ) b. Horizontal shift, left 4 units Vertical asymptote: x = –4 Domain: (–4, ∞ ) Range: (– ∞, ∞ )

19. For each function, state the transformations applied to y = ln x. Determine the vertical asymptote, domain, and range for each function. a. y = ln (x – 3) + 4 b. y = – ln x + 4 a. Horizontal shift, right 3 units Vertical shift, up 4 units Vertical asymptote: x = 3 Domain: (3, ∞ ) Range: (– ∞, ∞ ) b. Reflection about the x-axis Vertical shift, up 4 units Vertical asymptote: x = 0 Domain: (0, ∞ ) Range: (– ∞, ∞ )

20. During last winter, a dog from a small populated area was trapped in a 40 ◦ F lake. At 10:00 A. M., a passerby pulled the dog from the water, at which time the dog's temperature was 98 ◦ F (a dog’s average temperature is 101.3 ◦ F). While waiting for help from the nearest town, the dog's temperature had dropped to 94 ◦ F in nine minutes. The dog was taken to a veterinarian for treatment of hypothermia, and had a full recovery. Use the following formula to estimate the number of minutes, t, the dog had been in the cold water. where T w is the constant temperature of the water where the dog was found, T d is the temperature of the dog, and k = 0.0079. Estimate the time when the dog fell into the water. (continued on the next slide)

21. Using T w = 40, T d = 98 (dog’s temperature at 10:00 A. M. ), and k = 0.0079, Since the dog had been in the water for approximately 7 minutes, the estimate time when the dog fell into the water was 9:53 A. M. Note: Using T d = 94,(dog’s temperature at 10:09 A.M. ), t  – 16.0501, which means the dog had been in the water for approximately16 minutes prior to 10:09 A.M., or 9:53 A.M.

22. Using your textbook, practice the problems assigned by your instructor to review the concepts from Section 6.3.