Chapter 5 Exponential and Logarithmic Functions

1. Chapter 5 Exponential and Logarithmic Functions

2. Exponential Functions and Models 5.3 Distinguish between linear and exponential growth Recognize exponential growth and decay Calculate compound interest Use the natural exponential functions in application Model data with exponential functions

3. Linear versus Exponential Growth A Linear Function adds a fixed amount to the previous value of y for each unit increase in x For example, in f(x) = 10,000 + 500x, 500 is added to y for each increase of 1 in x. An Exponential Function multiplies a fixed amount by the previous value of y for each unit increase in x. For example, in f(x) = 10,000(1.05)x , y is multiplied by 1.05 for each increase of 1 in x.

4. Exponential Function A function represented by f(x) = Cax, a > 0, a ≠ 1, and C > 0, is an exponential function with base a and coefficient C.

5. Linear versus Exponential Growth For large values of x, an exponential function with a > 1 grows faster than any linearfunction.

6. For each table, find either a linear or an exponential function that models the data. (a) (b) Example: Recognize linear and exponential data

7. Solution (a) For each unit increase in x, the y-values increase by 1.5, so the data are linear. Becausey= –3when x= 0, it follows that the data can be modeled by f(x)= 1.5x– 3. Example: Recognize linear and exponential data

8. (b) For each unit increase in x, the y-values are multiplied by 1/4. This is an exponential function given by f(x)= Caxwith C = f(0)= 16 and a = 1/4, so f(x)= 16(1/4)x. Example: Recognize linear and exponential data

9. Exponential Growth If an exponential function is written asf(x) = Cax, a > 1, then f(x) experiences exponential growth. • The y-values increase bya factor of a for each unit increase in x. We say that the growth factor is a.

10. Exponential Decay If an exponential function is written asf(x) = Cax, 0 < a < 1, then f(x) experiences exponential decay. • The y-values decrease bya factor of a for each unit increase in x. We say that the decay factor is a.

11. Properties of Exponential Functions An exponential functionf, defined by f(x) = Cax, a > 0, a ≠ 1, and C > 0, has the following properties. • 1. The domain of f, is (–∞, ∞) and the range of f is (0, ∞).

12. Properties of Exponential Functions 2. The graph of f, is continuous with no breaks. The x-axis is a horizontal asymptote.There are no x-intercepts and the y-intercept is C. 3. If a > 1, f is increasing on its domain; if0 < a < 1, f is decreasing on its domain. 4. f is one-to-one and therefore has an inverse.

13. Caution Don’t confuse f(x) = 2x with f(x) = x2 f(x) = 2xis an exponential function. f(x) = x2 is a polynomial function, specifically a quadratic function. The functions and consequently their graphs are very different. f(x) = x2 f(x) = 2x

14. Graphs of Exponential Functions The graph of y = axis increasing when a > 1.

15. Graphs of Exponential Functions The graph of y = axis decreasing when0 < a < 1.

16. Graphs of Exponential Functions The graph of y = a–xis a reflection of y = ax across the y-axis.

17. Compound Interest Formula If a principal P dollars is deposited in an account paying an annual rate of interest r (expressed in decimal form), compounded (paid) n times per year, then after t years the account will contain A dollars, where

18. Suppose $1000 is deposited by a 20-year-old worker in an Individual Retirement Account(IRA) that pays an annual interest rate of 12%. Describe the effect on the balance after 45years at age 65 if interest were compounded quarterly rather than annually. Example: Comparing compound interest

19. Solution Compounded Annually: Let P = 1000, r = 0.12, n = 1, and t = 45 Example: Comparing compound interest

20. Compounded Quarterly: Let P = 1000, r = 0.12, n = 4, and t = 45 Example: Comparing compound interest

21. Continuous Compounding Compounding that is done more frequently, by letting n become large without bound, is called continuous compounding. The exponential expression reaches a limitof approximately 2.718281828 as

22. Euler’s Number This value is so important in mathematics that it has been given its own symbol, e, sometimes called Euler’s number. The number e has many of the same characteristics as π. Its decimal expansion never terminates or repeats in a pattern. It is an irrational number.

23. Value of e To eleven decimal places, e 2.71828182846.

24. The Natural Exponential Function The function f, represented by f(x) = ex is the natural exponential function.

25. Continuously Compounded Interest If a principal of P dollars is deposited in an account paying an annual rate of interest r (expressed in decimal form), compounded continuously, then after t years the account will contain A dollars, where A = Pert.

26. The principal in an IRA is $1000 and the interest rate is 12%, compounded continuously. How much money will there be after 45 years? Solution Let P = 1000, r = 0.12, and t = 45. A = 1000e(0.12)45 ≈ $221,406.42 This ismore than the $204,503.36 that resulted from compounding quarterly. Example: Calculating continuously compounded interest

27. Natural Exponential Growth and Decay If A0 is the initial amount of a quantity A at time t = 0 and if k is a positive constant, then exponential growth and decay of A can be modeled by A(t) = A0ekt Growth (k > 0) A(t) = A0e–kt Decay (k > 0)

28. Natural Exponential Growth and Decay Exponential Growth Exponential Decay

29. E. coli(Escherichia coli) is a type of bacteria that inhabits the intestines of animals. Thesebacteria are capable of rapid growth and can be dangerous to humans – especially children. In one study, E. coli bacteria were found to be capable of doubling in number about every49.5 minutes. Their number N after tminutes could be modeled byN(t)=N0e0.014t. Suppose that N0= 500,000 is the initial number of bacteria per milliliter. Example: Modeling the growth of E. coli bacteria

30. (a)Make a conjecture about the number of bacteria per milliliter after 99 minutes. Verifyyour conjecture. (b)Determine graphically the elapsed time when there were 25 million bacteria per milliliter. Example: Modeling the growth of E. coli bacteria

31. Solution (a)Since the bacteria double every 49.5 minutes, there would be 1,000,000 per milliliterafter 49.5 minutes and 2,000,000 after 99minutes. This is verified by evaluating N(99) = 500,000e0.014(99) ≈ 2,000,000 Example: Modeling the growth of E. coli bacteria

32. (b)Graph Y1 = 500000e^(0.014X) andY2 = 25000000. Their graphs intersect near (279.4, 25,000,000). Thus in a 1-milliliter sample, half a million E. coli Example: Modeling the growth of E. coli bacteria • bacteria could increaseto 25 million in approximately 279 minutes, or 4 hours and 39 minutes.

33. Predicted concentrations of atmospheric carbon dioxide (CO2) in parts per million (ppm) are shown in Table 5.17. (These concentrations assume that current trends continue.) The CO2 levels in the year 2000 were greater than they had been at any time in the previous 160,000 years. Example: Modeling atmospheric CO2 concentrations

34. (a)Let x = 0 correspond to 2000 andx = 200 to 2200 to 2200. Find values for C and a so thatf(x)= Caxmodels these data. (b)Estimate CO2 concentrations for the year 2025. Example: Modeling atmospheric CO2 concentrations

35. Solution (a)LetC equal f(0). Then substitute a data point into f(x)= Caxand find a. When x= 0, C = 364: f(x)= 364axUse the last data point: (200, 987): Example: Modeling atmospheric CO2 concentrations

36. (b) Since 2025 corresponds to x = 25, evaluate f(25). f(25) = 364(1.005)25 f(25) ≈ 412 Concentration of CO2 could reach 412 ppm by 2025. Example: Modeling atmospheric CO2 concentrations

37. Modeling Half-Life If a quantity initially equals C units has a half-life of k, then the amount Aremaining after t time is given by

38. Half-life The time it takes for half of the atoms to decay into a different element is called the half-life of an element undergoing radioactive decay.

39. A fossil contains 5% of the amount of carbon-14 that the organism contained when it wasalive. Graphically estimate its age. Example: Finding the age of a fossil

40. Solution The initial amount of carbon-14 is 100% (or 1), the final amount is 5% (or 0.05), and the half-life is 5700 years, soA = 0.05, C = 1, and k= 5700. Solve this equation for x. Example: Finding the age of a fossil

41. Graph Y1 = 0.05 and Y2 = 0.5^(X/5700). Example: Finding the age of a fossil • Their graphsintersect near (24635, 0.05), so the fossil is about 24,635 years old.