4.4 Logarithmic Functions
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1 SECTION 4.4 Logarithmic Functions Logarithmic Functions PREPARING FOR THIS SECTION Before getting started, review the following: Solving Inequalities (Appendi, Section A.8, pp ) Polnomial and Rational Inequalities (Section.5, pp. 5) Now work the Are You Prepared? problems on page 96. OBJECTIVES Change Eponential Epressions to Logarithmic Epressions and Logarithmic Epressions to Eponential Epressions Evaluate Logarithmic Epressions Determine the Domain of a Logarithmic Function 4 Graph Logarithmic Functions 5 Solve Logarithmic Equations Recall that a one-to-one function = f has an inverse function that is defined (implicitl) b the equation = f. In particular, the eponential function = f = a, a 7 0, a Z, is one-to-one and hence has an inverse function that is defined implicitl b the equation = a, a 7 0, a Z This inverse function is so important that it is given a name, the logarithmic function. The logarithmic function to the base a, where a 7 0 and a Z, is denoted b = log a (read as is the logarithm to the base a of ) and is defined b = log a if and onl if = a The domain of the logarithmic function = log a is 7 0. A logarithm is merel a name for a certain eponent.
2 88 CHAPTER 4 Eponential and Logarithmic Functions EXAMPLE Relating Logarithms to Eponents (a) If = log then =. For eample, 4 = log is equivalent to 8 = 4, 8. (b) If = log then For eample, - = log 5 a 5, = 5. is equivalent to 5 b 5 = 5-. Change Eponential Epressions to Logarithmic Epressions and Logarithmic Epressions to Eponential Epressions We can use the definition of a logarithm to convert from eponential form to logarithmic form, and vice versa, as the following two eamples illustrate. EXAMPLE Changing Eponential Epressions to Logarithmic Epressions Change each eponential epression to an equivalent epression involving a logarithm. (a). = m (b) e b = 9 (c) a 4 = 4 We use the fact that = log and = a a, a 7 0, a Z, are equivalent. (a) If. = m, then = log. m. (b) If e b = 9, then b = log e 9. (c) If a 4 = 4, then 4 = log a 4. NOW WORK PROBLEM 9. EXAMPLE Changing Logarithmic Epressions to Eponential Epressions Change each logarithmic epression to an equivalent epression involving an eponent. (a) log a 4 = 5 (b) log e b = - (c) log 5 = c (a) If log then a 5 a 4 = 5, = 4. (b) If log then e - e b = -, = b. (c) If log then c 5 = c, = 5. NOW WORK PROBLEM. Evaluate Logarithmic Epressions To find the eact value of a logarithm, we write the logarithm in eponential notation and use the fact that if a u = a v then u = v. EXAMPLE 4 Finding the Eact Value of a Logarithmic Epression Find the eact value of: (a) log 6 (b) log 7
3 SECTION 4.4 Logarithmic Functions 89 (a) = log 6 = 6 Change to eponential form. = 4 6 = 4 = 4 Equate eponents. Therefore, log 6 = 4. (b) = log 7 = 7 Change to eponential form. = - = - 7 = = - Equate eponents. Therefore, log 7 = -. NOW WORK PROBLEM. Determine the Domain of a Logarithmic Function The logarithmic function = log a has been defined as the inverse of the eponential function = a. That is, if f = a, then f - = log a. Based on the discussion given in Section 4. on inverse functions, for a function f and its inverse f -, we have Domain of f - = Range of f and Range of f - = Domain of f Consequentl, it follows that Domain of the logarithmic function = Range of the eponential function = 0, q Range of the logarithmic function = Domain of the eponential function = - q, q In the net bo, we summarize some properties of the logarithmic function: = log a defining equation: = a Domain: q Range: - q 6 6 q The domain of a logarithmic function consists of the positive real numbers, so the argument of a logarithmic function must be greater than zero. EXAMPLE 5 Finding the Domain of a Logarithmic Function Find the domain of each logarithmic function. (a) (c) F = log + h = log > ƒƒ (b) g = log 5 a + - b (a) The domain of F consists of all for which + 7 0, that is, 7 -. Using interval notation, the domain of f is -, q.
4 90 CHAPTER 4 Eponential and Logarithmic Functions (b) The domain of g is restricted to Solving this inequalit, we find that the domain of g consists of all between - and, that is, or, using interval notation, -,. (c) Since ƒƒ 7 0, provided that Z 0, the domain of h consists of all real numbers ecept zero or, using interval notation, - q, 0 or 0, q. NOW WORK PROBLEMS 47 AND 5. 4 Graph Logarithmic Functions Since eponential functions and logarithmic functions are inverses of each other, the graph of the logarithmic function = log a is the reflection about the line = of the graph of the eponential function = a, as shown in Figure 5. Figure 5 a a (, a) (, log (a, ) a a ) (0, ) (0, ) (, a) (, a ) (a, ) (, 0) (a) 0 a ( a, ) log a (, 0) (b) a ( a,) For eample, to graph = log graph =, and reflect it about the line See Figure 6. To graph graph = a =. = log >, and reflect it about b the line =. See Figure 7. Figure 6 (, ) (, ) (0, ) log (, ) Figure 7 ( ) (, ) (0, ) (, ) (, ) (, 0) (, ) (, 0) (, ) log / NOW WORK PROBLEM 6.
5 SECTION 4.4 Logarithmic Functions 9 Properties of the Logarithmic Function f () log a. The domain is the set of positive real numbers; the range is the set of all real numbers.. The -intercept of the graph is. There is no -intercept.. The -ais = 0 is a vertical asmptote of the graph. 4. A logarithmic function is decreasing if 0 6 a 6 and increasing if a The graph of f contains the points, 0, a,, and a a, -b. 6. The graph is smooth and continuous, with no corners or gaps. If the base of a logarithmic function is the number e, then we have the natural logarithm function. This function occurs so frequentl in applications that it is given a special smbol, ln (from the Latin, logarithmus naturalis). That is, = ln if and onl if = e () Since = ln and the eponential function = e are inverse functions, we can obtain the graph of = ln b reflecting the graph of = e about the line =. See Figure 8. Figure 8 5 e (,.7) Y e Y ( 0, ) (, 0.7) 0 (, 0) 0 ( 0.7,) (.7,) In 4 Y In Table 8 displas other points on the graph of f = ln. Notice for 0 that we obtain an error message. Do ou recall wh? Table 8
6 9 CHAPTER 4 Eponential and Logarithmic Functions EXAMPLE 6 Graphing Logarithmic Functions Using Transformations Figure 9 0 Graph f = -ln + b starting with the graph of = ln. Determine the domain, range, and vertical asmptote. The domain consists of all for which or 7 - To obtain the graph of = -ln +, we use the steps illustrated in Figure 9. 0 (,.0) (, 0.69) ( (, 0), 0.69) (, 0.69) (, 0) (, 0.69) (,.0) (, 0.69) (, 0) (0, 0.69) (,.0) (a) In Multipl b ; reflect about -ais (b) In Replace b ; shift left units. (c) In ( ) The range of f = -ln + is the interval - q, q, and the vertical asmptote is = -. [Do ou see wh? The original asmptote = 0 is shifted to the left units.] CHECK: Graph Y = -ln + using a graphing utilit to verif Figure 9(c). NOW WORK PROBLEM 75. If the base of a logarithmic function is the number 0, then we have the common logarithm function. If the base a of the logarithmic function is not indicated, it is understood to be 0. That is, = log if and onl if = 0 Figure 40 Since = log and the eponential function = 0 are inverse functions, we can obtain the graph of = log b reflecting the graph of = 0 about the line =. See Figure Y 0 Y (, ) 0 (0, ) (, 0) 4 ( 0, ) log Y log 4
7 SECTION 4.4 Logarithmic Functions 9 EXAMPLE 7 Graphing Logarithmic Functions Using Transformations Graph f = log -. Determine the domain, range, and vertical asmptote of f. The domain consists of all for which or 7 To obtain the graph of = log -, we use the steps illustrated in Figure 4. Figure 4 0 (, ) (0, ) (, 0) (, 0) (, ) (, 0) Replace b ; horizontal shift right unit Multipl b ; vertical stretch b a factor of. (a) log (b) log ( ) (c) log ( ) The range of f = log - is the interval - q, q, and the vertical asmptote is =. CHECK: Graph Figure 4(c). Y = log - using a graphing utilit to verif NOW WORK PROBLEM Solve Logarithmic Equations Equations that contain logarithms are called logarithmic equations. Care must be taken when solving logarithmic equations algebraicall. Be sure to check each apparent solution in the original equation and discard an that are etraneous. In the epression log a M, remember that a and M are positive and a Z. Some logarithmic equations can be solved b changing from a logarithmic epression to an eponential epression. EXAMPLE 8 Solving a Logarithmic Equation Solve: (a) log 4-7 = (b) log 64 = (a) We can obtain an eact solution b changing the logarithm to eponential form. log 4-7 = 4-7 = Change to eponential form. 4-7 = 9 4 = 6 = 4
8 94 CHAPTER 4 Eponential and Logarithmic Functions CHECK: log 4-7 = log 6-7 = log 9 = = 9 (b) We can obtain an eact solution b changing the logarithm to eponential form. log 64 = = 64 = ;64 = ;8 Change to eponential form. Square Root Method. The base of a logarithm is alwas positive. As a result, we discard solution is 8. -8; the onl CHECK: log 8 64 = 8 = 64 EXAMPLE 9 Using Logarithms to Solve Eponential Equations Solve: e = 5 We can obtain an eact solution b changing the eponential equation to logarithmic form. e = 5 ln 5 = = ln 5 L Change to a logarithmic epression using (). Eact solution. Approimate solution. NOW WORK PROBLEMS 9 AND 0. EXAMPLE 0 Alcohol and Driving The concentration of alcohol in a person s blood is measurable. Recent medical research suggests that the risk R (given as a percent) of having an accident while driving a car can be modeled b the equation R = 6e k where is the variable concentration of alcohol in the blood and k is a constant. (a) Suppose that a concentration of alcohol in the blood of 0.04 results in a 0% risk R = 0 of an accident. Find the constant k in the equation. Graph R = 6e k using this value of k. (b) Using this value of k, what is the risk if the concentration is 0.7? (c) Using the same value of k, what concentration of alcohol corresponds to a risk of 00%? (d) If the law asserts that anone with a risk of having an accident of 0% or more should not have driving privileges, at what concentration of alcohol in the blood should a driver be arrested and charged with a DUI (Driving Under the Influence)?
9 SECTION 4.4 Logarithmic Functions 95 (a) For a concentration of alcohol in the blood of 0.04 and a risk of 0%, we let = 0.04 and R = 0 in the equation and solve for k. R = 6e k 0 = 6e k = e0.04k 0.04k = ln 0 6 = k =.77 R = 0; = 0.04 Divide both sides b 6. Change to a logarithmic epression. Solve for k. Figure See Figure 4 for the graph of R = 6e.77. (b) Using k =.77 and = 0.7 in the equation, we find the risk R to be For a concentration of alcohol in the blood of 0.7, the risk of an accident is about 5.6%. (c) Using k =.77 and R = 00 in the equation, we find the concentration of alcohol in the blood to be 00 6 R = 6e k = 6e = 5.6 R = 6e k 00 = 6e.77 = e = ln 00 6 = 0. =.84 R = 00; k =.77 Divide both sides b 6. Change to a logarithmic epression. Solve for. For a concentration of alcohol in the blood of 0., the risk of an accident is 00%. (d) Using k =.77 and R = 0 in the equation, we find the concentration of alcohol in the blood to be R = 6e k 0 = 6e = e = ln 0 6 =.04 = A driver with a concentration of alcohol in the blood of or more (9.4%) should be arrested and charged with DUI. NOTE is given. Most states use 0.08 or 0.0 as the blood alcohol content at which a DUI citation
10 96 CHAPTER 4 Eponential and Logarithmic Functions Summar Properties of the Logarithmic Function f = log a, a 7 Domain: the interval 0, q; Range: the interval - q, q ( = log means = a a ) -intercept: ; -intercept: none; vertical asmptote: = 0 (-ais); increasing; one-to-one See Figure 4(a) for a tpical graph. f = log a, 0 6 a 6 Domain: the interval 0, q; Range: the interval - q, q ( = log means = a a ) -intercept: ; -intercept: none; vertical asmptote: = 0 (-ais); decreasing; one-to-one See Figure 4(b) for a tpical graph. Figure 4 (a, ) log a 0 (a, ) (, 0) (a) a ( a, ) 0 (, 0) ( a, ) (b) 0 a log a
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