Chapter 6 Exponential and Logarithmic Functions and Applications Section 6.2 Section 6.2 Logarithmic Functions and Their Graphs • Definition of Logarithm • Common Logarithms • Evaluating Logarithms • Basic Logarithmic Properties • Logarithmic Functions: Definition, Characteristics, and Graphs • Transformations of Graphs of Logarithmic Functions • Applications Introduction to Logarithms A logarithm (denoted by “log”) is basically another name for an exponent. 52 = 25 log525 = 2 In this example, the logarithm (or “exponent”) is 2. “log525 = ?” means we are asking the following: What is the exponent to which we must raise base 5 to obtain 25? 52 = 25 is the exponential form, and log525 = 2 is the logarithmic form. Both are equivalent. Basically, both are telling us that the outcome of raising base 5 to the exponent 2 will be 25. Thus, a logarithm is simply the exponent or power to which a base must be raised to produce a given number. General Definition of Logarithms Let b > 0, and b 1, logbx = y if and only if x = by for x > 0 and for every real number y. We read logbx = y as "the logarithm base b of x is y." Examples: log216 = 4: “The logarithm base 2 of 16 is 4” because 2 raised to the exponent 4 is 16, or 24 = 16. log101000 = 3: “The logarithm base 10 of 1000 is 3” because 10 to the power of 3 is 1000, or 103 = 1000. If the logarithm has a base b = 10, we do not need to write the base; we call it a common logarithm. Convert from exponential form to logarithmic form: a. 9 = 32 b. 10–3 = 1/1000 c. 41/2 = 2 d. t = rs a. We know x = by is equivalent to logbx = y. The base in “9 = 32” is 3; the exponent is 2. Hint: When converting from exponential to log form, the base of the exponent becomes the base of the log; the exponent of the exponential form becomes the exponent (“answer”) of the log. Base, 3, of the exponential form becomes base, 3, of the log. Exponent 2 is the exponent (“answer”) of the log. The logarithmic form of 9 = 32 is log39 = 2. (continued on the next slide) (Contd.) Convert from exponential form to logarithmic form: a. 9 = 32 b. 10–3 = 1/1000 c. 41/2 = 2 d. t = rs b. The base in “10–3 = 1/1000” is 10; the exponent is –3. The base, 10, of the exponential form becomes the base, 10, of the log. Exponent –3 is the exponent (“answer”) of the log. The logarithmic form is log10(1/1000) = –3, which we can also write as log(1/1000) = –3. c. The logarithmic form of “41/2 = 2” is log42= 1/2. Note: Remember that 41/2 is equivalent to “the square root of 4.” d. The logarithmic form is logr t = s. Convert from logarithmic form to exponential form: a. log749 = 2 b. log(100) = 2 c. log5(1/25) = –2 a. We know logbx = y is equivalent to x = by. The exponential form of log749 = 2 is 72 = 49. Hint: Start at the base of the log and move counterclockwise: log749 = 2 is equivalent to 72 = 49. (continued on the next slide) (Contd.) Convert from logarithmic form to exponential form: a. log749 = 2 b. log(100) = 2 c. log5(1/25) = –2 b. We know log(100) = 2 is the same as log10(100) = 2. log10(100) = 2 is equivalent to 102 = 100. c. The exponential form of log5(1/25) = –2 is 5–2 = 1/25. Evaluate each logarithm. a. log416 Let log416 = y. The exponential form is 4y = 16. To what exponent y must we raise 4 to get 16? 42 = 16 Therefore, log416 = 2. b. log7 7 Let log7 7 y. The exponential form is 7y 7. 71/2 7 Therefore, log7 7 1/2. Evaluate each logarithm. c. log2(0.25) Let log2(0.25) = y. The exponential form is 2y = 0.25. To what exponent y must we raise 2 to get 0.25? 1 1 22 2 0.25 2 4 Therefore, log2(0.25) = –2. d. log 1 This is equivalent to log10(1). Let log10(1) = y. The exponential form is 10y = 1. We know a0 =1 for all a ≠ 0. Therefore, 100 = 1, and log10(1) = 0. Basic Properties of Logarithms For b > 0, and b 1, 1. logb 1 = 0 2. logb b = 1 3. logb bx = x 4. blog x x b Apply the basic properties to simplify each logarithm. 1. log1/2 (1/2) log1/2 (1/2) = 1 Note that (1/2)1 = 1/2 2. log7 1 log7 1 = 0 Note that (7)0 = 1 3. 5log (13) 5log (13) 13 We know log5(13) = log5(13), changed to exponential form, 5 5 5log (13) 13. 5 4. log8 82x log8 82x = 2x Changing log8 82x = 2x to exponential form, we have (8)2x = 82x. Logarithmic Functions Let x > 0, b > 0, and b 1. The inverse function of the exponential function f(x) = bx is called the logarithmic function with base b, and it is denoted by f –1(x) = logb x. y = logbx is equivalent to by = x Since x > 0, the domain of f –1(x) = logb x is the set of all positive real numbers. Examples: The inverse function of y = 0.3x is y = log0.3 x. The inverse function of y = 10x is y = log10 x, or simply y = log x. If f(x) = 5x, find the following: 1. f(1) f(1) = 5(1) = 5 2. f–1(x) f–1(x) = log5 x 3. f–1(1) f–1(1) = log5 (1) = 0 Note that 5(0) = 1 4. f–1(–25) f–1(–25) = log5 (–25) which is undefined. Note that the definition of logarithmic function states that x > 0. (Recall that if y = log5 x, base 5 raised to any exponent will not yield a negative value.) Graphs of Exponential and Logarithmic Functions We know that the logarithmic function is the inverse of the exponential function, thus the graph of f–1(x) = logb x is the graph of f(x) = bx reflected about the line y = x. The domain of f–1 is the same as the range of f, and the range of f–1 is the same as the domain of f. For any point (a, b) on the graph of f(x) = bx , the point (b, a) lies on the graph of f–1(x) = logb x. Graphs of Logarithmic Functions For the logarithmic function f(x) = logb x, b > 0, b 1: The domain is the set of all positive real numbers, or (0, ∞) and the range is the set all real numbers, or (–∞, ∞). The x-intercept is (1, 0). The y-axis (that is, the line x = 0) is a vertical asymptote. The function is one-to-one. The graph passes though the point (b, 1) (continued on the next slide) Graphs of Logarithmic Functions The graph has one of the following shapes: If b > 1, the function is increasing. If 0 < b < 1, the function is decreasing. Transformations of Graphs of Logarithmic Functions Transformations of graphs of logarithmic functions are similar to other basic graphs we have studied. That is, we will shift the graphs vertically, horizontally, stretch or compress them, and reflect them across the axes. Next, we will see some examples. Transformations of Graphs of Logarithmic Functions (I) Apply transformations to the graph of f(x) = log x to construct the graph of each function. Determine the domain and asymptote for each function. a. f(x) = log x + 4 b. f(x) = log x – 4 a. Vertical shift, up 4 units Vertical asymptote: x = 0 Domain: (0, ∞) b. Vertical shift, down 4 units Vertical asymptote: x = 0 Domain: (0, ∞) Transformations of Graphs of Logarithmic Functions (II) Apply transformations to the graph of f(x) = log x to construct the graph of each function. Determine the domain and asymptote for each function. a. f(x) = log (x + 4) b. f(x) = log (x – 4) a. Horizontal shift, left 4 units Vertical asymptote: x = –4 Domain: (–4, ∞) b. Horizontal shift, right 4 units Vertical asymptote: x = 4 Domain: (4, ∞) Transformations of Graphs of Logarithmic Functions (III) Apply transformations to the graph of f(x) = log x to construct the graph of f(x) = –log x – 5. Determine the domain and asymptote. Reflection about the x-axis Vertical shift, down 5 units Vertical asymptote: x = 0 Domain: (0, ∞) Application: Measuring Earthquake Intensity The Richter Scale is used to measure the magnitude of earthquakes, and is given by the formula I R log I0 where I and I0 are numbers that represent the wave energy of the earthquake. I = intensity of the current earthquake I0 = intensity of a standard earthquake (reference intensity) The Richter scale grows by powers of 10. This means that an increase of 1 point results in an earthquake 10 times greater than the previous level. Magnitude measures the energy released at the source of the earthquake. Intensity measures the strength of shaking produced by the earthquake at a certain location. On January 12, 2010, an earthquake of magnitude 7.0 caused major destruction in the country of Haiti. On December 26, 2004, the ocean floorof Sumatra, Indonesia was the epicenter of an earthquake of magnitude of 9.3. This earthquake caused a disastrous tsunami that washed away whole cities in Indonesia. Sources: www.earthquake.usgs.gov; www.mapsofworld.com a. Find the intensity of the 2004 earthquake. Express its intensity in terms of the intensity of a standard earthquake, that is, I0. Applying the Richter formula, we let R = 9.3 and solve for I in terms of I0. I I 9.3 log R log I0 I0 I 10 9.3 I0 I 10 9.3 I0 The intensity of the Sumatra earthquake was 109.3 times that of a standard earthquake. (continued on the next slide) (Contd.) b. Find the intensity of the 2010 earthquake. Express its intensity in terms of the intensity of a standard earthquake, that is, I0. Applying the Richter formula, we let R = 7 and solve for I in terms of I0. I 7 log I0 I 10 7 I0 I 107 I0 The intensity of the 2010 earthquake was 107 times that of a standard earthquake. (continued on the next slide) (Contd.) c. Compare the intensity of the 2004 earthquake with the one in 2010. We now calculate: I2004 I2010 10 9.3 I0 2.3 10 199.53 7 10 I0 The Indonesian earthquake was approximately 199.53 times more intense than the Haitian earthquake. Application: Acidity or Alkalinity of a Solution (pH values) Potential of hydrogen (pH) is a measure to determine the relative alkalinity, acidity, or neutrality of a solution. It is measured on a scale of 0 to 14. The lower the pH value, the more acidic the solution; the higher the pH value the more alkaline (or base) the solution. A value of 7 represents a neutral solution, such as distilled water. The formula for calculating potential of hydrogen is given by pH = – log [H+] where [H+] represents the hydrogen ion concentration measured in moles of hydrogen per liter. Note: A mole is simply a unit of measurement that provides a consistent method to convert between atoms or molecules and grams. Many dentists recommend their patients to try to limit soda consumption. The threshold pH for enamel dissolution is 5.5. At a pH of less than 5.5, the acid begins to dissolve the hard enamel of teeth. The pH level of a soda is, on average, 3. Find its hydrogen ion concentration. Sources: www.healthenlightenment.com; www.dentalgentlecare.com/diet_soda.htm To find the hydrogen ion concentration, we will use the formula for calculating potential of hydrogen, replacing pH with 3 and solving for [H+]. pH = – log [H+] 3 = – log [H+] –3 = log [H+] 10–3 = [H+] The hydrogen ion concentration of a soda is 10–3 or 0.001. Using your textbook, practice the problems assigned by your instructor to review the concepts from Section 6.2.