5.5 Properties of Logarithms. Work with the Properties of Logarithms. 296 CHAPTER 5 Exponential and Logarithmic Functions

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1 296 CHAPTER 5 Exponential and Logarithmic Functions The Richter Scale Problems 3 and 32 use the following discussion: The Richter scale is one way of converting seismographic readings into numbers that provide an easy reference for measuring the magnitude M of an earthquake. All earthquakes are compared to a zero-level earthquake whose seismographic reading measures 0.00 millimeter at a distance of 00 kilometers from the epicenter. An earthquake whose seismographic reading measures x millimeters has magnitude Mx2, given by Mx2 = log x x 0 where x 0 = 0-3 is the reading of a zero-level earthquake the same distance from its epicenter. In Problems 3 and 32, determine the magnitude of each earthquake. 3. Magnitude of an Earthquake Mexico City in 985: seismographic reading of 25,892 millimeters 00 kilometers from the center 32. Magnitude of an Earthquake San Francisco in 906: seismographic reading of 50,9 millimeters 00 kilometers from the center 33. Alcohol and Driving The concentration of alcohol in a person s bloodstream is measurable. Suppose that the relative risk R of having an accident while driving a car can be modeled by an equation of the form R = e kx where x is the percent of concentration of alcohol in the bloodstream and k is a constant. Explaining Concepts: Discussion and Writing (a) Suppose that a concentration of alcohol in the bloodstream of 0.03 percent results in a relative risk of an accident of.4. Find the constant k in the equation. (b) Using this value of k, what is the relative risk if the concentration is 0.7 percent? (c) Using the same value of k, what concentration of alcohol corresponds to a relative risk of 00? (d) If the law asserts that anyone with a relative risk of having an accident of 5 or more should not have driving privileges, at what concentration of alcohol in the bloodstream should a driver be arrested and charged with a DUI? (e) Compare this situation with that of Example 0. If you were a lawmaker, which situation would you support? Give your reasons. 34. Is there any function of the form y = x a, 0 6 a 6, that increases more slowly than a logarithmic function whose base is greater than? Explain. 35. In the definition of the logarithmic function, the base a is not allowed to equal. Why? 36. Critical Thinking In buying a new car, one consideration might be how well the price of the car holds up over time. Different makes of cars have different depreciation rates. One way to compute a depreciation rate for a car is given here. Suppose that the current prices of a certain automobile are as shown in the table. Age in Years New $38,000 $36,600 $32,400 $28,750 $25,400 $2,200 Use the formula New = Olde Rt 2 to find R, the annual depreciation rate, for a specific time t. When might be the best time to trade in the car? Consult the NADA ( blue ) book and compare two like models that you are interested in. Which has the better depreciation rate? Are You Prepared? Answers. (a) x 3 (b) x 6-2 or x x 6-4 or x 7 3. {3} 5.5 Properties of Logarithms OBJECTIVES Work with the Properties of Logarithms (p. 296) 2 Write a Logarithmic Expression as a Sum or Difference of Logarithms (p. 298) 3 Write a Logarithmic Expression as a Single Logarithm (p. 299) 4 Evaluate Logarithms Whose Base Is Neither 0 Nor e (p. 30) Work with the Properties of Logarithms Logarithms have some very useful properties that can be derived directly from the definition and the laws of exponents.

2 SECTION 5.5 Properties of Logarithms 297 EXAMPLE Establishing Properties of Logarithms (a) Show that log a = 0. (b) Show that log a a =. (a) This fact was established when we graphed y = log a x (see Figure 30 on page 286). To show the result algebraically, let y = log a. Then y = log a a y = a y = a 0 y = 0 log a = 0 Change to an exponential statement. a 0 = since a 7 0, a Z Solve for y. y = log a (b) Let y = log a a. Then y = log a a a y = a a y = a y = log a a = To summarize: Change to an exponential statement. a = a Solve for y. y = log a a log a = 0 log a a = THEOREM Properties of Logarithms In the properties given next, M and a are positive real numbers, a Z, and r is any real number. The number log a M is the exponent to which a must be raised to obtain M. That is, a log a M = M () The logarithm to the base a of a raised to a power equals that power.that is, log a a r = r (2) The proof uses the fact that y = a x and y = log a x are inverses. Proof of Property () For inverse functions, Using fx2 = a x and f - x2 = log a x, we find Now let x = M to obtain a log a M = M, where M 7 0. Proof of Property (2) ff - x22 = x for all x in the domain of f - ff - x22 = a log a x = x for x 7 0 For inverse functions, - fx22 = x for all x in the domain of f f Using fx2 = a x and f - x2 = log a x, we find f - fx22 = log a a x = x for all real numbers x Now let x = r to obtain log a a r = r, where r is any real number.

3 298 CHAPTER 5 Exponential and Logarithmic Functions EXAMPLE 2 Using Properties () and (2) (a) 2 log 2 p = p (b) log = -22 (c) Now Work PROBLEM 5 Other useful properties of logarithms are given next. ln e kt = kt THEOREM Properties of Logarithms In the following properties, M, N, and a are positive real numbers, a Z, and r is any real number. The Log of a Product Equals the Sum of the Logs log a MN2 = log a M + log a N (3) The Log of a Quotient Equals the Difference of the Logs log a a M N b = log a M - log a N (4) The Log of a Power Equals the Product of the Power and the Log log a M r = r log a M (5) a x = e x ln a (6) 2 We shall derive properties (3), (5), and (6) and leave the derivation of property (4) as an exercise (see Problem 09). Proof of Property (3) Let A = log a M and let B = log a N. These expressions are equivalent to the exponential expressions Now Law of Exponents Property (2) of logarithms Proof of Property (5) Let A = log a M. This expression is equivalent to Now log a MN2 = log a a A a B 2 = log a a A + B log a M r = log a a A 2 r = log a a ra Law of Exponents Property (2) of logarithms Proof of Property (6) From property (), with a = e, we have e ln M = M Now let M = a x and apply property (5). e ln a x Now Work PROBLEM 9 a A = M and a B = N = A + B = log a M + log a N a A = M = ra = r log a M = e x ln a = a x Write a Logarithmic Expression as a Sum or Difference of Logarithms Logarithms can be used to transform products into sums, quotients into differences, and powers into factors. Such transformations prove useful in certain types of calculus problems.

4 SECTION 5.5 Properties of Logarithms 299 EXAMPLE 3 Writing a Logarithmic Expression as a Sum of Logarithms Write log a Ax 4 x 2 + B, x 7 0, as a sum of logarithms. Express all powers as factors. log a Ax4x 2 + B = log a x + log a 4x 2 + = log a x + log a x >2 log a M # N2 = loga M + log a N EXAMPLE 4 x 2 = log a x + 2 log ax log a a M N b = log a M - log a N log a M r = r log a M log a M r = r log a M Writing a Logarithmic Expression as a Difference of Logarithms Write as a difference of logarithms. Express all powers as factors. ln x 2 x x 7 ln x = ln x2 - lnx = 2 ln x - 3 lnx - 2 c c WARNING In using properties (3) through (5), be careful about the values that the variable may assume. For example, the domain of the variable for log a x is x 7 0 and for log a x - 2 it is x 7. If we add these functions, the domain is x 7. That is, the equality log a x + log a x - 2 = log a 3xx - 24 is true only for x 7. EXAMPLE 5 EXAMPLE 6 3 Writing a Logarithmic Expression as a Sum and Difference of Logarithms Write as a sum and difference of logarithms. Express all powers as factors. log a 4 x 2 + x 3 x = log a 3x log a 3x 3 x = log a 4x log a x 3 + log a x = log a x >2 - log a x 3 - log a x = 2 log ax log a x - 4 log a x + 2 Now Work PROBLEM 5 log a 4 x 2 + x 3 x x 7 0 Property (4) Property (3) Property (5) Write a Logarithmic Expression as a Single Logarithm Another use of properties (3) through (5) is to write sums and/or differences of logarithms with the same base as a single logarithm. This skill will be needed to solve certain logarithmic equations discussed in the next section. Writing Expressions as a Single Logarithm Write each of the following as a single logarithm. 2 (a) log a log a 3 (b) 3 ln 8 - ln52-2 (c) log a x + log a 9 + log a x log a 5

5 300 CHAPTER 5 Exponential and Logarithmic Functions (a) (b) log a log a 3 = log a 7 + log a 3 4 = log a 7 + log a 8 = loga 7 # 82 = log a ln 8 - ln52-2 = ln 8 2>3 - ln25-2 = ln 4 - ln 24 = lna 4 24 b r log a M = log a M r log a M + log a N = log a M # N2 r log a M = log a M r 8 2/3 = ( 3 8) 2 = 2 2 = 4 log a M - log a N = log a a M N b (c) = lna 6 b = ln - ln 6 = -ln 6 ln = 0 log a x + log a 9 + log a x log a 5 = log a 9x2 + log a x log a 5 = log a 39xx log a 5 = log a B 9xx2 + 2 R 5 WARNING A common error made by some students is to express the logarithm of a sum as the sum of logarithms. log a M + N2 is not equal to log a M + log a N Correct statement log a MN2 = log a M + log a N Property (3) Another common error is to express the difference of logarithms as the quotient of logarithms. log a M - log a N is not equal to log a M log a N Correct statement log a M - log a N = log a a M Property (4) N b A third common error is to express a logarithm raised to a power as the product of the power times the logarithm. log a M2 r is not equal to r log a M Correct statement log a M r = r log a M Property (5) Now Work PROBLEM 57 Two other properties of logarithms that we need to know are consequences of the fact that the logarithmic function y = log a x is a one-to-one function. THEOREM Properties of Logarithms In the following properties, M, N, and a are positive real numbers, a Z. If M = N, then log a M = log a N. If log a M = log a N, then M = N. (7) (8) When property (7) is used, we start with the equation M = N and say take the logarithm of both sides to obtain log a M = log a N. Properties (7) and (8) are useful for solving exponential and logarithmic equations, a topic discussed in the next section.

6 SECTION 5.5 Properties of Logarithms 30 4 Evaluate Logarithms Whose Base Is Neither 0 Nor e Logarithms to the base 0, common logarithms, were used to facilitate arithmetic computations before the widespread use of calculators. (See the Historical Feature at the end of this section.) Natural logarithms, that is, logarithms whose base is the number e, remain very important because they arise frequently in the study of natural phenomena. Common logarithms are usually abbreviated by writing log, with the base understood to be 0, just as natural logarithms are abbreviated by ln, with the base understood to be e. Most calculators have both log and ln keys to calculate the common logarithm and natural logarithm of a number. Let s look at an example to see how to approximate logarithms having a base other than 0 or e. EXAMPLE 7 Approximating a Logarithm Whose Base Is Neither 0 Nor e Approximate log 2 7. Round the answer to four decimal places. Remember, log 2 7 means 2 raised to what exponent equals 7. If we let y = log 2 7, then 2 y = 7. Because 2 2 = 4 and 2 3 = 8, we expect log 2 7 to be between 2 and 3. 2 y = 7 ln 2 y = ln 7 y ln 2 = ln 7 y = ln 7 ln 2 y L Property (7) Property (5) Exact value Approximate value rounded to four decimal places Example 7 shows how to approximate a logarithm whose base is 2 by changing to logarithms involving the base e. In general, we use the Change-of-Base Formula. THEOREM Change-of-Base Formula If a Z, b Z, and M are positive real numbers, then log a M = log b M log b a (9) Proof We derive this formula as follows: Let y = log a M. Then a y = M log b a y = log b M y log b a = log b M y = log b M log b a log a M = log b M log b a Property (7) Property (5) Solve for y. y = log a M Since calculators have keys only for log and ln, in practice, the Change-of- Base Formula uses either b = 0 or b = e. That is, log a M = log M log a and log a M = ln M ln a (0)

7 302 CHAPTER 5 Exponential and Logarithmic Functions EXAMPLE 8 Using the Change-of-Base Formula Approximate: (a) (a) log 5 89 Round answers to four decimal places. log 5 89 = or log 5 89 = log 89 log 5 L ln 89 ln 5 L L L (b) or Now Work PROBLEMS 23 AND 7 log 2 25 = log 2 25 = log 25 log 22 = ln 25 ln 22 = 2 log 5 2 log 2 = log 5 log 2 L ln 5 2 ln 2 = ln 5 ln 2 L COMMENT To graph logarithmic functions when the base is different from e or 0 requires the Change-of-Base Formula. For example, to graph y = log 2 x, we would instead graph y = ln x Try it. ln 2. Now Work PROBLEM 79 (b) log SUMMARY Properties of Logarithms In the list that follows, a, b, M, N, and r are real numbers. Also, a 7 0, a Z, b 7 0, b Z, M 7 0, and N 7 0. Definition Properties of logarithms Change-of-Base Formula y = log means x = a y a x log a = 0; log a a = a log a M = M; log a a r = r log a MN2 = log a M + log a N log a a M N b = log a M - log a N log a M = log b M log b a log a M r = r log a M a x = e x ln a If M = N, then log a M = log a N. If log a M = log a N, then M = N. Historical Feature John Napier (550 67) Logarithms were invented about 590 by John Napier (550 67) and Joost Bürgi ( ), working independently. Napier, whose work had the greater influence, was a Scottish lord, a secretive man whose neighbors were inclined to believe him to be in league with the devil. His approach to logarithms was very different from ours; it was based on the relationship between arithmetic and geometric sequences, discussed in a later chapter, and not on the inverse function relationship of logarithms to exponential functions (described in Section 5.4). Napier s tables, published in 64, listed what would now be called natural logarithms of sines and were rather difficult to use. ALondonprofessor,HenryBriggs,becameinterestedinthetables and visited Napier. In their conversations, they developed the idea of common logarithms, which were published in 67. Their importance for calculation was immediately recognized, and by 650 they were being printed as far away as China. They remained an important calculation tool until the advent of the inexpensive handheld calculator about 972, which has decreased their calculational, but not their theoretical, importance. A side effect of the invention of logarithms was the popularization of the decimal system of notation for real numbers.

8 SECTION 5.5 Properties of Logarithms Assess Your Understanding Concepts and Vocabulary. log a = 8. If log a x = log a 6, then x =. 2. log a a = 3. a logam = 4. log a a r = 5. log a (MN) = + 6. log a a M N b = - 7. log a M r = 9. If log 8 M = log 5 7 then M =. log 5 8, 0. True or False. True or False 2. True or False lnx + 32 lnx ln2x2 = ln2x2 log 2 3x 4 2 = 4 log 2 3x2 ln 8 ln 4 = 2 Skill Building In Problems 3 28, use properties of logarithms to find the exact value of each expression. Do not use a calculator. 3. log log ln e ln e log e ln 8 9. log log log log log log log log log 2 6 # log log 3 8 # log log log log log e log e e log e 2 9 In Problems 29 36, suppose that ln 2 = a and ln 3 = b. Use properties of logarithms to write each logarithm in terms of a and b. 29. ln ln ln ln ln ln ln ln A 4 In Problems 37 56, write each expression as a sum and/or difference of logarithms. Express powers as factors log 5 25x2 38. log 3 x log 2 z log 7 x 5 4. lnex2 42. ln e x 43. ln x e x 44. lnxe x log a u 2 v 3 2 u 7 0, v log 2 a a b 2 b a 7 0, b lnax2 2 - xb 0 6 x lnax4 + x2 B x log 2 x3 x - 3 x log 5 43 x2 + x 2 - x 7 xx logb x + 32 R 2 x logb x3 2x + x - 22 R 2 x lnb x2 - x - 2 x >3 R x 7 2 x >3 54. lnb x 2 - R x ln 5x2 + 3x x x lnb 5x x 4x R 0 6 x 6 In Problems 57 70, write each expression as a single logarithm log 5 u + 4 log 5 v log 3 u - log 3 v 59. log 3 x - log 3 x log 2 a x b + log 2 x 2 6. log 4x log 4 x logx 2 + 3x logx + 2 x 63. lna x - b + lna x + x b - lnx log x2 + 2x - 3 x 2 - log x2 + 7x log 2 23x log 2 a 4-4 x + 2 x b + log 2 4

9 304 CHAPTER 5 Exponential and Logarithmic Functions log 3 3 x + log 3 9x log log a 5x log a2x logx logx log 2 x log 2 x log 2 x log 5 3x log 5 2x log 5 x In Problems 7 78, use the Change-of-Base Formula and a calculator to evaluate each logarithm. Round your answer to three decimal places. 7. log log log > log > log log log p e 78. log p 22 In Problems 79 84, graph each function using a graphing utility and the Change-of-Base Formula. 79. y = log 4 x 80. y = log 5 x 8. y = log 2 x y = log 4 x y = log x - x y = log x + 2 x - 22 Mixed Practice 85. If fx2 = ln x, gx2 = e x, and hx2 = x 2, find: 86. If fx2 = log 2 x, gx2 = 2 x, and hx2 = 4x, find: (a) f g2x2. What is the domain of f g? (a) f g2x2. What is the domain of f g? (b) g f2x2. What is the domain of g f? (b) g f2x2. What is the domain of g f? (c) f g252 (c) f g232 (d) f h2x2. What is the domain of f h? (d) f h2x2. What is the domain of f h? (e) f h2e2 (e) f h282 Applications and Extensions In Problems 87 96, express y as a function of x. The constant C is a positive number. 87. ln y = ln x + ln C 88. ln y = lnx + C2 89. ln y = ln x + lnx ln C 90. ln y = 2 ln x - lnx ln C 9. ln y = 3x + ln C 92. ln y = -2x + ln C 93. lny - 32 = -4x + ln C 94. lny + 42 = 5x + ln C ln y = 2 ln2x lnx ln C ln y = - 2 ln x + 3 lnx ln C 97. Find the value of log 2 3 # log3 4 # log4 5 # log5 6 # log6 7 # log Find the value of log 2 4 # log4 6 # log Find the value of log 2 3 # log3 4 # Á # logn n + 2 # logn Find the value of log 2 2 # log2 4 # Á # log2 2 n. 0. Show that log a Ax + 4x 2 - B + log a Ax - 4x 2 - B = Show that log a A x + 2x - B + log a A x - 2x - B = Show that ln + e 2x 2 = 2x + ln + e -2x 2. fx + h2 - fx2 04. Difference Quotient If fx2 = log show that = log h a a + h a x, >h, x b h Z If fx2 = log a x, show that -fx2 = log >a x. 06. If fx2 = log a x, show that fab2 = fa2 + fb If fx2 = log show that fa a x, x b = -fx If fx2 = log show that fx a a x, 2 = afx Show that log a a M where a, M, and N N b = log a M - log a N, 0. Show that log a a where a and N are positive N b = -log a N, are positive real numbers and a Z. real numbers and a Z. Explaining Concepts: Discussion and Writing. Graph Y = logx 2 2 and Y 2 = 2 logx2 using a graphing utility. Are they equivalent? What might account for any differences in the two functions? 2. Write an example that illustrates why log a x2 r Z r log a x. 3. Write an example that illustrates why log 2 x + y2 Z log 2 x + log 2 y. 4. Does 3 log3(-5) = -5? Why or why not?