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Verify that each function in Exercises 64 - 69 is the inverse of the corresponding function in Exercises 58 - 63. (Match up #58 and #64, and so on.) In Exercises 71 - 74, ﬁnd the inverse of the function from the ‘procedural perspective’ discussed in Example 6.1.5 and graph the function and its inverse on the same set of axes. 71. f (x) = 3x+2 − 4 73. f (x) = −2−x + 1 72. f (x) = log4(x − 1) 74. f (x) = 5 log(x) − 2 6.1 Introduction to Exponential and Logarithmic Functions 431 (Logarithmic Scales) In Exercises 75 - 77, we introduce three widely used measurement scales which involve common logarithms: the Richter scale, the decibel scale and the pH scale. The computations involved in all three scales are nearly identical so pay attention to the subtle diﬀerences. 75. Earthquakes are complicated events and it is not our intent to provide a complete discussion of the science involved in them. Instead, we refer the interested reader to a solid course in Geology12 or the U.S. Geological Survey’s Earthquake Hazards Program found here and present only a simpliﬁed version of the Richter scale. The Richter scale measures the magnitude of an earthquake by comparing the amplitude of the seismic waves of the given earthquake to those of a “magnitude 0 event”, which was chosen to be a seismograph reading of 0.001 millimeters recorded on a seismometer 100 kilometers from the earthquake’s epicenter. Speciﬁcally, the magnitude of an earthquake is given by M (x) = log x 0.001 where x is the seismograph reading in millimeters of the earthquake recorded 100 kilometers from the epicenter. (a) Show that M (0.001) = 0. (b) Compute M (80, 000). (c) Show that an earthquake which registered 6.7 on the Richter scale had a seismograph reading ten times larger than one which measured 5.7. (d) Find two news stories about recent earthquakes which give their magnitudes on the Richter scale. How many times larger was the seismograph reading of the earthquake with larger magnitude? 76. While
the decibel scale can be used in many disciplines,13 we shall restrict our attention to its use in acoustics, speciﬁcally its use in measuring the intensity level of sound.14 The Sound Intensity Level L (measured in decibels) of a sound intensity I (measured in watts per square meter) is given by L(I) = 10 log I 10−12. Like the Richter scale, this scale compares I to baseline: 10−12 W hearing. m2 is the threshold of human (a) Compute L(10−6). 12Rock-solid, perhaps? 13See this webpage for more information. 14As of the writing of this exercise, the Wikipedia page given here states that it may not meet the “general notability guideline” nor does it cite any references or sources. I ﬁnd this odd because it is this very usage of the decibel scale which shows up in every College Algebra book I have read. Perhaps those other books have been wrong all along and we’re just blindly following tradition. 432 Exponential and Logarithmic Functions (b) Damage to your hearing can start with short term exposure to sound levels around 115 decibels. What intensity I is needed to produce this level? (c) Compute L(1). How does this compare with the threshold of pain which is around 140 decibels? 77. The pH of a solution is a measure of its acidity or alkalinity. Speciﬁcally, pH = − log[H+] where [H+] is the hydrogen ion concentration in moles per liter. A solution with a pH less than 7 is an acid, one with a pH greater than 7 is a base (alkaline) and a pH of 7 is regarded as neutral. (a) The hydrogen ion concentration of pure water is [H+] = 10−7. Find its pH. (b) Find the pH of a solution with [H+] = 6.3 × 10−13. (c) The pH of gastric acid (the acid in your stomach) is about 0.7. What is the corresponding hydrogen ion concentration? 78. Show that logb 1 = 0 and logb b = 1 for every b > 0, b = 1. 79. (Crazy bonus question) Without using your calculator, determine which is larger: eπ or πe. 6.1 Introduction
to Exponential and Logarithmic Functions 433 6.1.2 Answers 1. log2(8) = 3 4. log 1 3 (9) = −2 7. ln(1) = 0 10. 3−4 = 1 81 13. 10−1 = 0.1 2. log5 1 125 = −3 5. log 4 25 5 2 = − 1 2 8. 52 = 25 11. 4 3 −1 = 3 4 14. e1 = e 3. log4(32) = 5 2 6. log(0.001) = −3 9. (25) 1 2 = 5 12. 102 = 100 15. e− 1 2 = 1√ e 16. log3(27) = 3 17. log6(216) = 3 18. log2(32) = 5 19. log6 1 36 = −2 20. log8(4) = 2 3 21. log36(216) = 3 2 22. log 1 5 (625) = −4 23. log 1 6 (216) = −3 25. log 1 1000000 = −6 28. log4(8) = 3 2 31. log36 √ 4 36 = 1 4 26. log(0.01) = −2 29. log6(1) = 0 32. 7log7(3) = 3 34. log36 36216 = 216 35. ln(e5) = 5 37. log 3√ 105 = 5 3 40. log eln(100) = 2 38. ln 1√ e = − 1 2 24. log36(36) = 1 27. ln e3 = 3 30. log13 √ 13 = 1 2 33. 36log36(216) = 216 36. log 9√ 1011 = 11 9 39. log5 3log3 5 = 1 41. log2 3− log3(2) = −1 42. ln 426 log(1) = 0 43. (−∞, ∞) 44. (−2, ∞) 45. (5, ∞) 46. (−∞, −6) ∪ (−3, ∞) 47. (−2, −1) ∪ (1, ∞) 48. (−6, −3) ∪ (5, ∞) 49. (4, 7) 52. [1, ∞) 50. (
5, ∞) 51. (−∞, ∞) 53. (−∞, −7) ∪ (1, ∞) 54. (13, ∞) 55. (0, 125) ∪ (125, ∞) 56. No domain 57. (−∞, −3) ∪ 1 2, 2 434 Exponential and Logarithmic Functions 58. Domain of g: (−∞, ∞) Range of g: (−1, ∞) 59. Domain of g: (−∞, ∞) Range of g: (0, ∞) 3−2−1 x 1 2 3 H.A. y = −1 y = g(x) = 2x − 1 60. Domain of g: (−∞, ∞) Range of g: (2, ∞) y 11 10.A. y = 2 −3−2−1 1 2 3 x y = g(x) = 3−3−2−1 1 2 3 x y = g(x) = 1 3 x−1 61. Domain of g: (−∞, ∞) Range of g: (−20, ∞) y 80 70 60 50 40 30 20 10 −3−2 −10 1 2 3 x H.A. y = −20 y = g(x) = 10 x+1 2 − 20 62. Domain of g: (−∞, ∞) Range of g: (−∞, 8) 63. Domain of g: (−∞, ∞) Range of g: (0, ∞) H.A3−2−1 1 2 3 x y = g(x) = 8 − e−x y 80 70 60 50 40 30 20 10 −10 10 20 30 x y = g(x) = 10e−0.1x 6.1 Introduction to Exponential and Logarithmic Functions 435 64. Domain of g: (−1, ∞) Range of g: (−∞, ∞) 65. Domain of g: (0, ∞) Range of g: (−∞, ∞) 1 x −2 −3 V.A. x = −1 y = g(x) = log2(x + 1) y 3 2 1 −1 −2 −(x) = log 1 3 (x) + 1 66. Domain of g: (2, �
�) Range of g: (−∞, ∞) 67. Domain of g: (−20, ∞) Range of g: (−∞, ∞) y 3 2 1 −1 −2 − 10 11 x V.A. x = 2 y = g(x) = − log3(x − 2) y 3 2 1 −10 10 20 30 40 50 60 70 80 90 100 x −2 −3 V.A. x = −20 y = g(x) = 2 log(x + 20) − 1 68. Domain of g: (−∞, 8) Range of g:(−∞, ∞) y 3 2 1 −1 −2 −.A. x = 8 69. Domain of g: (0, ∞) Range of g: (−∞, ∞) y 30 20 10 −10 10 20 30 40 50 60 70 80 x y = g(x) = − ln(8 − x) y = g(x) = −10 ln x 10 436 Exponential and Logarithmic Functions 71. f (x) = 3x+2 − 4 f −1(x) = log3(x + 4) − 2 72. f (x) = log4(x − 1) f −1(x) = 4x + 1 y 6 5 4 3 2 1 −4−3−2−1 −1 1 2 3 4 5 6 x −2 −3 −4 y = f (x) = 3x+2 − 4 y = f −1(x) = log3(x + 4) − 2 73. f (x) = −2−x + 1 f −1(x) = − log2(1 − x) y 2 1 −2 −1 1 2 x −1 −2 y = f (x) = −2−x + 1 y = f −1(x) = − log2(1 − x) y 6 5 4 3 2 1 −2 −1 1 2 3 4 5 6 −1 −2 x y = f (x) = log4(x − 1) y = f −1(x) = 4x + 1 74. f (x) = 5 log(x) − 2 f −1(x) = 10 x+2 5 y 5 4 3 2 1 −4−3−2−1 −1 1 2 3 4 5
x −2 −3 −4 y = f (x) = 5 log(x) − 2 x+2 y = f −1(x) = 10 5 75. 76. 77. (a) M (0.001) = log 0.001 0.001 80,000 0.001 (b) M (80, 000) = log = log(1) = 0. = log(80, 000, 000) ≈ 7.9. (a) L(10−6) = 60 decibels. (b) I = 10−.5 ≈ 0.316 watts per square meter. (c) Since L(1) = 120 decibels and L(100) = 140 decibels, a sound with intensity level 140 decibels has an intensity 100 times greater than a sound with intensity level 120 decibels. (a) The pH of pure water is 7. (b) If [H+] = 6.3 × 10−13 then the solution has a pH of 12.2. (c) [H+] = 10−0.7 ≈.1995 moles per liter. 6.2 Properties of Logarithms 437 6.2 Properties of Logarithms In Section 6.1, we introduced the logarithmic functions as inverses of exponential functions and discussed a few of their functional properties from that perspective. In this section, we explore the algebraic properties of logarithms. Historically, these have played a huge role in the scientiﬁc development of our society since, among other things, they were used to develop analog computing devices called slide rules which enabled scientists and engineers to perform accurate calculations leading to such things as space travel and the moon landing. As we shall see shortly, logs inherit analogs of all of the properties of exponents you learned in Elementary and Intermediate Algebra. We ﬁrst extract two properties from Theorem 6.2 to remind us of the deﬁnition of a logarithm as the inverse of an exponential function. Theorem 6.3. (Inverse Properties of Exponential and Logarithmic Functions) Let b > 0, b = 1. ba = c if and only if logb(c) = a logb (bx) = x for all x and blogb(x) = x for all x > 0 Next, we spell out what it means
for exponential and logarithmic functions to be one-to-one. Theorem 6.4. (One-to-one Properties of Exponential and Logarithmic Functions) Let f (x) = bx and g(x) = logb(x) where b > 0, b = 1. Then f and g are one-to-one and bu = bw if and only if u = w for all real numbers u and w. logb(u) = logb(w) if and only if u = w for all real numbers u > 0, w > 0. We now state the algebraic properties of exponential functions which will serve as a basis for the properties of logarithms. While these properties may look identical to the ones you learned in Elementary and Intermediate Algebra, they apply to real number exponents, not just rational exponents. Note that in the theorem that follows, we are interested in the properties of exponential functions, so the base b is restricted to b > 0, b = 1. An added beneﬁt of this restriction is that it eliminates the pathologies discussed in Section 5.3 when, for example, we simpliﬁed x2/33/2 and obtained \|x\| instead of what we had expected from the arithmetic in the exponents, x1 = x. Theorem 6.5. (Algebraic Properties of Exponential Functions) Let f (x) = bx be an exponential function (b > 0, b = 1) and let u and w be real numbers. Product Rule: f (u + w) = f (u)f (w). In other words, bu+w = bubw Quotient Rule: f (u − w) = f (u) f (w). In other words, bu−w = bu bw Power Rule: (f (u))w = f (uw). In other words, (bu)w = buw While the properties listed in Theorem 6.5 are certainly believable based on similar properties of integer and rational exponents, the full proofs require Calculus. To each of these properties of 438 Exponential and Logarithmic Functions exponential functions corresponds an analogous property of logarithmic functions. We list these below in our next theorem. Theorem 6.6. (Algebraic Properties of Logarithmic Functions) Let g(x) = logb(x
) be a logarithmic function (b > 0, b = 1) and let u > 0 and w > 0 be real numbers. Product Rule: g(uw) = g(u) + g(w). In other words, logb(uw) = logb(u) + logb(w) Quotient Rule: g u w = g(u) − g(w). In other words, logb u w = logb(u) − logb(w) Power Rule: g (uw) = wg(u). In other words, logb (uw) = w logb(u) There are a couple of diﬀerent ways to understand why Theorem 6.6 is true. Consider the product rule: logb(uw) = logb(u) + logb(w). Let a = logb(uw), c = logb(u), and d = logb(w). Then, by deﬁnition, ba = uw, bc = u and bd = w. Hence, ba = uw = bcbd = bc+d, so that ba = bc+d. By the one-to-one property of bx, we have a = c + d. In other words, logb(uw) = logb(u) + logb(w). The remaining properties are proved similarly. From a purely functional approach, we can see the properties in Theorem 6.6 as an example of how inverse functions interchange the roles of inputs in outputs. For instance, the Product Rule for exponential functions given in Theorem 6.5, f (u + w) = f (u)f (w), says that adding inputs results in multiplying outputs. Hence, whatever f −1 is, it must take the products of outputs from f and return them to the sum of their respective inputs. Since the outputs from f are the inputs to f −1 and vice-versa, we have that that f −1 must take products of its inputs to the sum of their respective outputs. This is precisely what the Product Rule for Logarithmic functions states in Theorem 6.6: g(uw) = g(u) + g(w). The reader is encouraged to view the remaining properties listed in Theorem 6.6 similarly. The following examples help build familiarity with these properties. In our �
�rst example, we are asked to ‘expand’ the logarithms. This means that we read the properties in Theorem 6.6 from left to right and rewrite products inside the log as sums outside the log, quotients inside the log as diﬀerences outside the log, and powers inside the log as factors outside the log.1 Example 6.2.1. Expand the following using the properties of logarithms and simplify. Assume when necessary that all quantities represent positive real numbers. 1. log2 8 x 4. log 3 100x2 yz5 Solution. 2. log0.1 10x2 5. log117 x2 − 4 3. ln 2 3 ex 1. To expand log2 8 x, we use the Quotient Rule identifying u = 8 and w = x and simplify. 1Interestingly enough, it is the exact opposite process (which we will practice later) that is most useful in Algebra, the utility of expanding logarithms becomes apparent in Calculus. 6.2 Properties of Logarithms 439 log2 8 x = log2(8) − log2(x) Quotient Rule = 3 − log2(x) = − log2(x) + 3 Since 23 = 8 2. In the expression log0.1 10x2, we have a power (the x2) and a product. In order to use the Product Rule, the entire quantity inside the logarithm must be raised to the same exponent. Since the exponent 2 applies only to the x, we ﬁrst apply the Product Rule with u = 10 and w = x2. Once we get the x2 by itself inside the log, we may apply the Power Rule with u = x and w = 2 and simplify. log0.1 10x2 = log0.1(10) + log0.1 x2 Product Rule = log0.1(10) + 2 log0.1(x) = −1 + 2 log0.1(x) = 2 log0.1(x) − 1 Power Rule Since (0.1)−1 = 10 3. We have a power, quotient and product occurring in ln 3 ex. Since the exponent 2 applies to the entire quantity inside the logarithm, we begin with the Power Rule with u = 3 ex and w = 2. Next, we see the
Quotient Rule is applicable, with u = 3 and w = ex, so we replace is being multiplied by 2, the entire ln 3 ex quantity ln(3) − ln(ex) is multiplied by 2. Finally, we apply the Product Rule with u = e and w = x, and replace ln(ex) with the quantity ln(e) + ln(x), and simplify, keeping in mind that the natural log is log base e. with the quantity ln(3) − ln(ex). Since ln 3 ex 2 2 3 ex ln = 2 ln 3 ex Power Rule = 2 [ln(3) − ln(ex)] Quotient Rule = 2 ln(3) − 2 ln(ex) = 2 ln(3) − 2 [ln(e) + ln(x)] Product Rule = 2 ln(3) − 2 ln(e) − 2 ln(x) = 2 ln(3) − 2 − 2 ln(x) Since e1 = e = −2 ln(x) + 2 ln(3) − 2 4. In Theorem 6.6, there is no mention of how to deal with radicals. However, thinking back to 3 exponent. We begin by using the Power Deﬁnition 5.5, we can rewrite the cube root as a 1 440 Exponential and Logarithmic Functions Rule2, and we keep in mind that the common log is log base 10. log 3 100x2 yz5 = log 1/3 100x2 yz5 100x2 yz5 = 1 3 log log 100x2 − log yz5 3 log yz5 3 log 100x2 − 1 log(100) + log x2 − 1 3 3 log x2 − 1 3 log(100) + 1 3 log(x) − 1 3 log(100) + 2 3 + 2 3 log(x) − 1 3 log(x) − 1 3 log(y) − 5 3 log(y log(y) + log z5 3 log z5 3 log(y) − 1 3 log(z) 3 log(y) − 5 3 log(z) 3 log(z) + 2 3 Power Rule Quotient Rule Product Rule Power Rule Since 102 = 100 5. At ﬁrst it seems as if
we have no means of simplifying log117 x2 − 4, since none of the properties of logs addresses the issue of expanding a diﬀerence inside the logarithm. However, we may factor x2 − 4 = (x + 2)(x − 2) thereby introducing a product which gives us license to use the Product Rule. log117 x2 − 4 = log117 [(x + 2)(x − 2)] Factor = log117(x + 2) + log117(x − 2) Product Rule A couple of remarks about Example 6.2.1 are in order. First, while not explicitly stated in the above example, a general rule of thumb to determine which log property to apply ﬁrst to a complicated problem is ‘reverse order of operations.’ For example, if we were to substitute a number for x into 10x2, we would ﬁrst square the x, then multiply by 10. The last step is the the expression log0.1 multiplication, which tells us the ﬁrst log property to apply is the Product Rule. In a multi-step problem, this rule can give the required guidance on which log property to apply at each step. The reader is encouraged to look through the solutions to Example 6.2.1 to see this rule in action. Second, while we were instructed to assume when necessary that all quantities represented positive real numbers, the authors would be committing a sin of omission if we failed to point out that, for x2 − 4 and g(x) = log117(x + 2) + log117(x − 2) have diﬀerent instance, the functions f (x) = log117 domains, and, hence, are diﬀerent functions. We leave it to the reader to verify the domain of f is (−∞, −2) ∪ (2, ∞) whereas the domain of g is (2, ∞). In general, when using log properties to 2At this point in the text, the reader is encouraged to carefully read through each step and think of which quantity is playing the role of u and which is playing the role of w as we apply each property. 6.2 Properties of Logarithms 441 expand a logarithm, we may very well be restricting the domain as we do so. One last comment before we move to reassembling logs from their various bits and pieces. The authors
are well aware of the propensity for some students to become overexcited and invent their own properties of logs x2 − log117(4), which simply isn’t true, in general. The unwritten3 like log117 property of logarithms is that if it isn’t written in a textbook, it probably isn’t true. x2 − 4 = log117 Example 6.2.2. Use the properties of logarithms to write the following as a single logarithm. 1. log3(x − 1) − log3(x + 1) 2. log(x) + 2 log(y) − log(z) 3. 4 log2(x) + 3 4. − ln(x) − 1 2 Solution. Whereas in Example 6.2.1 we read the properties in Theorem 6.6 from left to right to expand logarithms, in this example we read them from right to left. 1. The diﬀerence of logarithms requires the Quotient Rule: log3(x−1)−log3(x+1) = log3 x−1 x+1. 2. In the expression, log(x) + 2 log(y) − log(z), we have both a sum and diﬀerence of logarithms. However, before we use the product rule to combine log(x) + 2 log(y), we note that we need to somehow deal with the coeﬃcient 2 on log(y). This can be handled using the Power Rule. We can then apply the Product and Quotient Rules as we move from left to right. Putting it all together, we have log(x) + 2 log(y) − log(z) = log(x) + log y2 − log(z) Power Rule = log xy2 − log(z) Product Rule = log xy2 z Quotient Rule 3. We can certainly get started rewriting 4 log2(x) + 3 by applying the Power Rule to 4 log2(x) x4, but in order to use the Product Rule to handle the addition, we need to to obtain log2 rewrite 3 as a logarithm base 2. From Theorem 6.3, we know 3 = log2 23, so we get 4 log2(x) + 3 = log2 = log
2 = log2 = log2 x4 + 3 x4 + log2 x4 + log2(8) 8x4 23 Since 3 = log2 Power Rule 23 Product Rule 3The authors relish the irony involved in writing what follows. 442 Exponential and Logarithmic Functions 4. To get started with − ln(x) − 1 2, we rewrite − ln(x) as (−1) ln(x). We can then use the Power Rule to obtain (−1) ln(x) = ln x−1. In order to use the Quotient Rule, we need to write 1 2 as a natural logarithm. Theorem 6.3 gives us 1 e). We have 2 = ln e1/2 = ln ( √ − ln(x) − 1 2 2 = (−1) ln(x) − 1 = ln x−1 − 1 2 = ln x−1 − ln e1/2 Since 1 = ln x−1 − ln ( x−1 √ e 1 √ x = ln = ln e) √ e Power Rule 2 = ln e1/2 Quotient Rule As we would expect, the rule of thumb for re-assembling logarithms is the opposite of what it was for dismantling them. That is, if we are interested in rewriting an expression as a single logarithm, we apply log properties following the usual order of operations: deal with multiples of logs ﬁrst with the Power Rule, then deal with addition and subtraction using the Product and Quotient Rules, respectively. Additionally, we ﬁnd that using log properties in this fashion can increase the domain of the expression. For example, we leave it to the reader to verify the domain of f (x) = log3(x−1)−log3(x+1) is (1, ∞) but the domain of g(x) = log3 is (−∞, −1)∪(1, ∞). We will need to keep this in mind when we solve equations involving logarithms in Section 6.4 - it is precisely for this reason we will have to check for extraneous solutions. x−1 x+1 The two logarithm buttons commonly found on calculators are the ‘LOG’ and ‘LN�
� buttons which correspond to the common and natural logs, respectively. Suppose we wanted an approximation to log2(7). The answer should be a little less than 3, (Can you explain why?) but how do we coerce the calculator into telling us a more accurate answer? We need the following theorem. Theorem 6.7. (Change of Base Formulas) Let a, b > 0, a, b = 1. ax = bx logb(a) for all real numbers x. loga(x) = logb(x) logb(a) for all real numbers x > 0. The proofs of the Change of Base formulas are a result of the other properties studied in this section. If we start with bx logb(a) and use the Power Rule in the exponent to rewrite x logb(a) as logb (ax) and then apply one of the Inverse Properties in Theorem 6.3, we get bx logb(a) = blogb(ax) = ax, 6.2 Properties of Logarithms 443 as required. To verify the logarithmic form of the property, we also use the Power Rule and an Inverse Property. We note that loga(x) · logb(a) = logb aloga(x) = logb(x), and we get the result by dividing through by logb(a). Of course, the authors can’t help but point out the inverse relationship between these two change of base formulas. To change the base of an exponential expression, we multiply the input by the factor logb(a). To change the base of a logarithmic expression, we divide the output by the factor logb(a). While, in the grand scheme of things, both change of base formulas are really saying the same thing, the logarithmic form is the one usually encountered in Algebra while the exponential form isn’t usually introduced until Calculus.4 What Theorem 6.7 really tells us is that all exponential and logarithmic functions are just scalings of one another. Not only does this explain why their graphs have similar shapes, but it also tells us that we could do all of mathematics with a single base - be it 10, e, 42, or 117. Your Calculus teacher will have more to say about this when the time comes. Example 6.2.3. Use an appropriate change of base formula to convert
the following expressions to ones with the indicated base. Verify your answers using a calculator, as appropriate. 1. 32 to base 10 3. log4(5) to base e Solution. 2. 2x to base e 4. ln(x) to base 10 1. We apply the Change of Base formula with a = 3 and b = 10 to obtain 32 = 102 log(3). Typing the latter in the calculator produces an answer of 9 as required. 2. Here, a = 2 and b = e so we have 2x = ex ln(2). To verify this on our calculator, we can graph f (x) = 2x and g(x) = ex ln(2). Their graphs are indistinguishable which provides evidence that they are the same function. 4The authors feel so strongly about showing students that every property of logarithms comes from and corresponds to a property of exponents that we have broken tradition with the vast majority of other authors in this ﬁeld. This isn’t the ﬁrst time this happened, and it certainly won’t be the last. y = f (x) = 2x and y = g(x) = ex ln(2) 444 Exponential and Logarithmic Functions 3. Applying the change of base with a = 4 and b = e leads us to write log4(5) = ln(5) this in the calculator gives ln(5) By deﬁnition, log4(5) is the exponent we put on 4 to get 5. The calculator conﬁrms this.5 ln(4). Evaluating ln(4) ≈ 1.16. How do we check this really is the value of log4(5)? 4. We write ln(x) = loge(x) = log(x) both graphs appear to be identical. log(e). We graph both f (x) = ln(x) and g(x) = log(x) log(e) and ﬁnd y = f (x) = ln(x) and y = g(x) = log(x) log(e) 5Which means if it is lying to us about the ﬁrst answer it gave us, at least it is being consistent. 6.2 Properties of Logarithms 445 6.2.
1 Exercises In Exercises 1 - 15, expand the given logarithm and simplify. Assume when necessary that all quantities represent positive real numbers. 1. ln(x3y2) 4. log(1.23 × 1037) 7. log√ 2 4x3 10. log3 x2 81y4 13. log √ 100x √ 3 10 y 2. log2 128 x2 + 4 √ z xy 5. ln 8. log 1 3 (9x(y3 − 8)) 11. ln 4 xy ez 4 3√ √ y x2 z 14. log 1 2 3. log5 3 z 25 6. log5 x2 − 25 9. log 1000x3y5 12. log6 4 216 x3y 15. ln √ 3 x √ yz 10 In Exercises 16 - 29, use the properties of logarithms to write the expression as a single logarithm. 16. 4 ln(x) + 2 ln(y) 17. log2(x) + log2(y) − log2(z) 18. log3(x) − 2 log3(y) 19. 1 2 log3(x) − 2 log3(y) − log3(z) 20. 2 ln(x) − 3 ln(y) − 4 ln(z) 22. − 1 3 ln(x) − 1 3 ln(y) + 1 3 ln(z) 24. 3 − log(x) 26. ln(x) + 1 2 21. log(x) − 1 3 log(z) + 1 2 log(y) 23. log5(x) − 3 25. log7(x) + log7(x − 3) − 2 27. log2(x) + log4(x) 28. log2(x) + log4(x − 1) 29. log2(x) + log 1 2 (x − 1) 446 Exponential and Logarithmic Functions In Exercises 30 - 33, use the appropriate change of base formula to convert the given expression to an expression with the indicated base. 30. 7x−1 to base e x 32. to base e 2 3 31. log3(x + 2) to base 10 33. log
(x2 + 1) to base e In Exercises 34 - 39, use the appropriate change of base formula to approximate the logarithm. 34. log3(12) 1 10 37. log4 35. log5(80) 38. log 3 5 (1000) 36. log6(72) 39. log 2 3 (50) 40. Compare and contrast the graphs of y = ln(x2) and y = 2 ln(x). 41. Prove the Quotient Rule and Power Rule for Logarithms. 42. Give numerical examples to show that, in general, (a) logb(x + y) = logb(x) + logb(y) (b) logb(x − y) = logb(x) − logb(y) (c) logb x y = logb(x) logb(y) 43. The Henderson-Hasselbalch Equation: Suppose HA represents a weak acid. Then we have a reversible chemical reaction HA H + + A−. The acid disassociation constant, Ka, is given by Kα = [H +][A−] [HA] = [H +] [A−] [HA], where the square brackets denote the concentrations just as they did in Exercise 77 in Section 6.1. The symbol pKa is deﬁned similarly to pH in that pKa = − log(Ka). Using the deﬁnition of pH from Exercise 77 and the properties of logarithms, derive the Henderson-Hasselbalch Equation which states pH = pKa + log [A−] [HA] 44. Research the history of logarithms including the origin of the word ‘logarithm’ itself. Why is the abbreviation of natural log ‘ln’ and not ‘nl’? 45. There is a scene in the movie ‘Apollo 13’ in which several people at Mission Control use slide rules to verify a computation. Was that scene accurate? Look for other pop culture references to logarithms and slide rules. 6.2 Properties of Logarithms 447 6.2.2 Answers 1. 3 ln(x) + 2 ln(y) 3. 3 log5(z) − 6 5. 1 2 ln(z) − ln(x) − ln
(y) 7. 3 log√ 2(x) + 4 2. 7 − log2(x2 + 4) 4. log(1.23) + 37 6. log5(x − 5) + log5(x + 5) 8. −2 + log 1 3 (x) + log 1 3 (y − 2) + log 1 3 (y2 + 2y + 4) 9. 3 + 3 log(x) + 5 log(y) 10. 2 log3(x) − 4 − 4 log3(y) 11. 1 4 ln(x) + 1 4 ln(y) − 1 4 − 1 4 ln(z) 12. 12 − 12 log6(x) − 4 log6(y) 13. 5 3 + log(x) + 1 2 log(y) 14. −2 + 2 3 log 1 2 (x) − log 1 2 (y) − 1 2 log 1 2 (z) 15. 1 3 ln(x) − ln(10) − 1 17. log2 xy z 2 ln(y) − 1 2 ln(z) 20. ln x2 y3z4 23. log5 x 125 26. ln (x √ e) 29. log2 x x−1 32. 2 3 x = ex ln( 2 3 ) 34. log3(12) ≈ 2.26186 36. log6(72) ≈ 2.38685 38. log 3 5 (1000) ≈ −13.52273 18. log3 x y2 21. log √ x 3√ y z 24. log 1000 x 27. log2 x3/2 16. ln(x4y2) 19. log3 √ x y2z 22. ln z xy 3 25. log7 28. log2 x(x−3) 49 √ x x − 1 30. 7x−1 = e(x−1) ln(7) 31. log3(x + 2) = log(x+2) log(3) 33. log(x2 + 1) = ln(x2+1) ln(10) 35. log5(80) ≈ 2.72271 37. log4 1 10 ≈ −1.66096 39. log 2 3 (50)
≈ −9.64824 448 Exponential and Logarithmic Functions 6.3 Exponential Equations and Inequalities In this section we will develop techniques for solving equations involving exponential functions. Suppose, for instance, we wanted to solve the equation 2x = 128. After a moment’s calculation, we ﬁnd 128 = 27, so we have 2x = 27. The one-to-one property of exponential functions, detailed in Theorem 6.4, tells us that 2x = 27 if and only if x = 7. This means that not only is x = 7 a solution to 2x = 27, it is the only solution. Now suppose we change the problem ever so slightly to 2x = 129. We could use one of the inverse properties of exponentials and logarithms listed in Theorem 6.3 to write 129 = 2log2(129). We’d then have 2x = 2log2(129), which means our solution is x = log2(129). This makes sense because, after all, the deﬁnition of log2(129) is ‘the exponent we put on 2 to get 129.’ Indeed we could have obtained this solution directly by rewriting the equation 2x = 129 in its logarithmic form log2(129) = x. Either way, in order to get a reasonable decimal approximation to this number, we’d use the change of base formula, Theorem 6.7, to give us something more calculator friendly,1 say log2(129) = ln(129) ln(2). Another way to arrive at this answer is as follows 2x = 129 ln (2x) = ln(129) Take the natural log of both sides. x ln(2) = ln(129) Power Rule ln(129) ln(2) x = ‘Taking the natural log’ of both sides is akin to squaring both sides: since f (x) = ln(x) is a function, as long as two quantities are equal, their natural logs are equal.2 Also note that we treat ln(2) as any other non-zero real number and divide it through3 to isolate the variable x. We summarize below the two common ways to solve exponential equations, motivated by our examples. Steps for Solving an Equation involving Exponential Functions 1. Isolate
the exponential function. 2. (a) If convenient, express both sides with a common base and equate the exponents. (b) Otherwise, take the natural log of both sides of the equation and use the Power Rule. Example 6.3.1. Solve the following equations. Check your answer graphically using a calculator. 1. 23x = 161−x 4. 75 = 100 1+3e−2t Solution. 2. 2000 = 1000 · 3−0.1t 3. 9 · 3x = 72x 5. 25x = 5x + 6 6. ex−e−x 2 = 5 1You can use natural logs or common logs. We choose natural logs. (In Calculus, you’ll learn these are the most ‘mathy’ of the logarithms.) 2This is also the ‘if’ part of the statement logb(u) = logb(w) if and only if u = w in Theorem 6.4. 3Please resist the temptation to divide both sides by ‘ln’ instead of ln(2). Just like it wouldn’t make sense to √ √ divide both sides by the square root symbol ‘ ’ when solving x 2 = 5, it makes no sense to divide by ‘ln’. 6.3 Exponential Equations and Inequalities 449 1. Since 16 is a power of 2, we can rewrite 23x = 161−x as 23x = 241−x. Using properties of exponents, we get 23x = 24(1−x). Using the one-to-one property of exponential functions, we get 3x = 4(1−x) which gives x = 4 7. To check graphically, we set f (x) = 23x and g(x) = 161−x and see that they intersect at x = 4 7 ≈ 0.5714. 2. We begin solving 2000 = 1000 · 3−0.1t by dividing both sides by 1000 to isolate the exponential which yields 3−0.1t = 2. Since it is inconvenient to write 2 as a power of 3, we use the natural log to get ln 3−0.1t = ln(2). Using the Power Rule, we get −0.1t ln(3) = ln(2), so we divide both sides by −0.1 ln
(3) to get t = − ln(2) ln(3). On the calculator, we graph f (x) = 2000 and g(x) = 1000 · 3−0.1x and ﬁnd that they intersect at x = − 10 ln(2) ln(3) ≈ −6.3093. 0.1 ln(3) = − 10 ln(2) y = f (x) = 23x and y = g(x) = 161−x y = f (x) = 2000 and y = g(x) = 1000 · 3−0.1x 3. We ﬁrst note that we can rewrite the equation 9·3x = 72x as 32 ·3x = 72x to obtain 3x+2 = 72x. Since it is not convenient to express both sides as a power of 3 (or 7 for that matter) we use ln 3x+2 = ln 72x. The power rule gives (x + 2) ln(3) = 2x ln(7). Even the natural log: though this equation appears very complicated, keep in mind that ln(3) and ln(7) are just constants. The equation (x + 2) ln(3) = 2x ln(7) is actually a linear equation and as such we gather all of the terms with x on one side, and the constants on the other. We then divide both sides by the coeﬃcient of x, which we obtain by factoring. (x + 2) ln(3) = 2x ln(7) x ln(3) + 2 ln(3) = 2x ln(7) 2 ln(3) = 2x ln(7) − x ln(3) 2 ln(3) = x(2 ln(7) − ln(3)) Factor. x = 2 ln(3) 2 ln(7)−ln(3) Graphing f (x) = 9·3x and g(x) = 72x on the calculator, we see that these two graphs intersect at x = 2 ln(3) 2 ln(7)−ln(3) ≈ 0.7866. 4. Our objective in solving 75 = 100 is to ﬁrst isolate
the exponential. To that end, we clear denominators and get 75 1 + 3e−2t = 100. From this we get 75 + 225e−2t = 100, which leads to 225e−2t = 25, and ﬁnally, e−2t = 1 9. Taking the natural log of both sides 1+3e−2t 450 Exponential and Logarithmic Functions. Since natural log is log base e, ln e−2t = −2t. We can also use gives ln e−2t = ln 1 9 = − ln(9). Putting these two steps together, we simplify the Power Rule to write ln 1 9 to −2t = − ln(9). We arrive at our solution, t = ln(9) ln e−2t = ln 1 2 which simpliﬁes to 9 t = ln(3). (Can you explain why?) The calculator conﬁrms the graphs of f (x) = 75 and g(x) = 100 1+3e−2x intersect at x = ln(3) ≈ 1.099. y = f (x) = 9 · 3x and y = g(x) = 72x y = f (x) = 75 and y = g(x) = 100 1+3e−2x 5. We start solving 25x = 5x + 6 by rewriting 25 = 52 so that we have 52x = 5x + 6, or 52x = 5x + 6. Even though we have a common base, having two terms on the right hand side of the equation foils our plan of equating exponents or taking logs. If we stare at this long enough, we notice that we have three terms with the exponent on one term exactly twice that of another. To our surprise and delight, we have a ‘quadratic in disguise’. Letting u = 5x, we have u2 = (5x)2 = 52x so the equation 52x = 5x + 6 becomes u2 = u + 6. Solving this as u2 − u − 6 = 0 gives u = −2 or u = 3. Since u = 5x, we have 5x = −2 or 5x = 3. Since 5x = −2 has no real solution, (Why not?) we focus on 5x
= 3. Since it isn’t convenient to express 3 as a power of 5, we take natural logs and get ln (5x) = ln(3) so that x ln(5) = ln(3) or x = ln(3) ln(5). On the calculator, we see the graphs of f (x) = 25x and g(x) = 5x + 6 intersect at x = ln(3) ln(5) ≈ 0.6826. 2 6. At ﬁrst, it’s unclear how to proceed with ex−e−x = 5, besides clearing the denominator to obtain ex − e−x = 10. Of course, if we rewrite e−x = 1 ex, we see we have another denominator lurking in the problem: ex − 1 ex = 10. Clearing this denominator gives us e2x − 1 = 10ex, and once again, we have an equation with three terms where the exponent on one term is exactly twice that of another - a ‘quadratic in disguise.’ If we let u = ex, then u2 = e2x so the equation e2x − 1 = 10ex can be viewed as u2 − 1 = 10u. Solving u2 − 10u − 1 = 0, we obtain 26. Since 5 − by the quadratic formula u = 5 ± 26 < 0, we get no real solution to ex = 5 − 26, we take natural logs to obtain x = ln 5 + and g(x) = 5, we see that the graphs intersect at x = ln 5 + 26. From this, we have ex = 5 ± √ 26, but for ex = 5 + 26. If we graph f (x) = ex−e−x √ 26 ≈ 2.312 √ √ √ √ √ 2 6.3 Exponential Equations and Inequalities 451 y = f (x) = 25x and y = g(x) = 5x + 6 y = f (x) = ex−e−x y = g(x) = 5 2 and The authors would be remiss not to mention that Example 6.3.1 still holds great educational value. Much can be learned about logarithms and exponentials by verifying the solutions obtained in Example 6.3.1 analy
tically. For example, to verify our solution to 2000 = 1000 · 3−0.1t, we substitute t = − 10 ln(2) ln(3) and obtain 2000 2000 2000 2000 2000 − 10 ln(2) ln(3) −0.1 ln(2) ln(3)? = 1000 · 3? = 1000 · 3? = 1000 · 3log3(2)? = 1000 · 2 = 2000 Change of Base Inverse Property The other solutions can be veriﬁed by using a combination of log and inverse properties. Some fall out quite quickly, while others are more involved. We leave them to the reader. Since exponential functions are continuous on their domains, the Intermediate Value Theorem 3.1 applies. As with the algebraic functions in Section 5.3, this allows us to solve inequalities using sign diagrams as demonstrated below. Example 6.3.2. Solve the following inequalities. Check your answer graphically using a calculator. 1. 2x2−3x − 16 ≥ 0 Solution. 2. ex ex − 4 ≤ 3 3. xe2x < 4x 1. Since we already have 0 on one side of the inequality, we set r(x) = 2x2−3x − 16. The domain of r is all real numbers, so in order to construct our sign diagram, we need to ﬁnd the zeros of r. Setting r(x) = 0 gives 2x2−3x − 16 = 0 or 2x2−3x = 16. Since 16 = 24 we have 2x2−3x = 24, so by the one-to-one property of exponential functions, x2 − 3x = 4. Solving x2 − 3x − 4 = 0 gives x = 4 and x = −1. From the sign diagram, we see r(x) ≥ 0 on (−∞, −1] ∪ [4, ∞), which corresponds to where the graph of y = r(x) = 2x2−3x − 16, is on or above the x-axis. 452 Exponential and Logarithmic Functions (+) 0 (−) 0 (+) −1 4 y = r(x) = 2x2−3x − 16 2. The ﬁrst step we need to take to solve ex ex−4 ≤ 3 is to get 0 on one
side of the inequality. To that end, we subtract 3 from both sides and get a common denominator ex ex − 4 ≤ 3 ex ex − 4 − − 3 ≤ 0 ex ex − 4 3 (ex − 4) ex − 4 12 − 2ex ex − 4 ≤ 0 Common denomintors. ≤ 0 We set r(x) = 12−2ex ex−4 and we note that r is undeﬁned when its denominator ex − 4 = 0, or when ex = 4. Solving this gives x = ln(4), so the domain of r is (−∞, ln(4)) ∪ (ln(4), ∞). To ﬁnd the zeros of r, we solve r(x) = 0 and obtain 12 − 2ex = 0. Solving for ex, we ﬁnd ex = 6, or x = ln(6). When we build our sign diagram, ﬁnding test values may be a little tricky since we need to check values around ln(4) and ln(6). Recall that the function ln(x) is increasing4 which means ln(3) < ln(4) < ln(5) < ln(6) < ln(7). While the prospect of determining the sign of r (ln(3)) may be very unsettling, remember that eln(3) = 3, so r (ln(3)) = 12 − 2eln(3) eln(3) − 4 = 12 − 2(3) 3 − 4 = −6 We determine the signs of r (ln(5)) and r (ln(7)) similarly.5 From the sign diagram, we ﬁnd our answer to be (−∞, ln(4)) ∪ [ln(6), ∞). Using the calculator, we see the graph of f (x) = ex ex−4 is below the graph of g(x) = 3 on (−∞, ln(4)) ∪ (ln(6), ∞), and they intersect at x = ln(6) ≈ 1.792. 4This is because the base of ln(x) is e > 1. If the base b were in the interval 0 < b < 1, then logb(x) would decreasing. 5We could, of course, use the calculator, but what fun
would that be? 6.3 Exponential Equations and Inequalities 453 (−) (+) 0 (−) ln(4) ln(6) y = f (x) = ex ex−4 y = g(x) = 3 3. As before, we start solving xe2x < 4x by getting 0 on one side of the inequality, xe2x − 4x < 0. We set r(x) = xe2x − 4x and since there are no denominators, even-indexed radicals, or logs, the domain of r is all real numbers. Setting r(x) = 0 produces xe2x − 4x = 0. We factor to get x e2x − 4 = 0 which gives x = 0 or e2x − 4 = 0. To solve the latter, we isolate the exponential and take logs to get 2x = ln(4), or x = ln(4) 2 = ln(2). (Can you explain the last equality using properties of logs?) As in the previous example, we need to be careful about choosing test values. Since ln(1) = 0, we choose ln 1 and ln(3). Evaluating,6 we get 2, ln 3 2 r ln 1 2 = ln 1 2 = ln 1 2 = ln 1 2 4 ln 1 = 1 2 ) − 4 ln 1 e2 ln( 1 2 2 )2 − 4 ln 1 eln( 1 2 eln( 1 4 ) − 4 ln 1 2 = − 15 − 4 ln 1 2 2 4 ln 1 2 Power Rule 2 < 1, ln 1 ) is (+), so r(x) < 0 on (0, ln(2)). The calculator Since 1 conﬁrms that the graph of f (x) = xe2x is below the graph of g(x) = 4x on these intervals.7 < 0 and we get r(ln 1 2 2 (+) 0 (−) 0 (+) 0 ln(2) y = f (x) = xe2x and y = g(x) = 4x 6A calculator can be used at this point. As usual, we proceed without apologies, with the analytical method. 7Note: ln(2) ≈ 0.693. 454 Exponential and Logarithmic Functions
Example 6.3.3. Recall from Example 6.1.2 that the temperature of coﬀee T (in degrees Fahrenheit) t minutes after it is served can be modeled by T (t) = 70 + 90e−0.1t. When will the coﬀee be warmer than 100◦F? 3 so that t = −10 ln 1 Solution. We need to ﬁnd when T (t) > 100, or in other words, we need to solve the inequality 70 + 90e−0.1t > 100. Getting 0 on one side of the inequality, we have 90e−0.1t − 30 > 0, and we set r(t) = 90e−0.1t − 30. The domain of r is artiﬁcially restricted due to the context of the problem to [0, ∞), so we proceed to ﬁnd the zeros of r. Solving 90e−0.1t − 30 = 0 results in which, after a quick application of the Power Rule leaves us with e−0.1t = 1 t = 10 ln(3). If we wish to avoid using the calculator to choose test values, we note that since 1 < 3, 0 = ln(1) < ln(3) so that 10 ln(3) > 0. So we choose t = 0 as a test value in [0, 10 ln(3)). Since 3 < 4, 10 ln(3) < 10 ln(4), so the latter is our choice of a test value for the interval (10 ln(3), ∞). Our sign diagram is below, and next to it is our graph of y = T (t) from Example 6.1.2 with the horizontal line y = 100. 3 (+) 0 0 (−) 10 ln(3) y 180 160 140 120 80 60 40 20 y = 100 H.A. y = 70 2 4 6 8 10 12 14 16 18 20 t y = T (t) In order to interpret what this means in the context of the real world, we need a reasonable approximation of the number 10 ln(3) ≈ 10.986. This means it takes approximately 11 minutes for the coﬀee to cool to 100◦F. Until then, the coﬀee is warmer than that.8 We close this
section by ﬁnding the inverse of a function which is a composition of a rational function with an exponential function. Example 6.3.4. The function f (x) = 5ex ex + 1 your answer graphically using your calculator. Solution. We start by writing y = f (x), and interchange the roles of x and y. To solve for y, we ﬁrst clear denominators and then isolate the exponential function. is one-to-one. Find a formula for f −1(x) and check 8Critics may point out that since we needed to use the calculator to interpret our answer anyway, why not use it earlier to simplify the computations? It is a fair question which we answer unfairly: it’s our book. 6.3 Exponential Equations and Inequalities 455 Switch x and y y = x = 5ex ex + 1 5ey ey + 1 x (ey + 1) = 5ey xey + x = 5ey x = 5ey − xey x = ey(5 − x) ey = x 5 − x ln (ey) = ln y = ln −x We claim f −1(x) = ln. To verify this analytically, we would need to verify the compositions f −1 ◦ f (x) = x for all x in the domain of f and that f ◦ f −1 (x) = x for all x in the domain of f −1. We leave this to the reader. To verify our solution graphically, we graph y = f (x) = 5ex ex+1 and y = g(x) = ln on the same set of axes and observe the symmetry about the line y = x. Note the domain of f is the range of g and vice-versa. x 5−x y = f (x) = 5ex ex+1 and y = g(x) = ln x 5−x 456 Exponential and Logarithmic Functions 6.3.1 Exercises In Exercises 1 - 33, solve the equation analytically. 1. 24x = 8 4. 42x = 1 2 7. 37x = 814−2x 10. 5−x = 2 13. (1.005)12x = 3 16. 500 1 − e2x = 250 19. 100ex ex + 2 x 22. 25 4 5 = 50 =
10 25. 3(x−1) = 2x 2. 3(x−1) = 27 5. 8x = 1 128 8. 9 · 37x = 1 9 2x 11. 5x = −2 14. e−5730k = 1 2 3. 52x−1 = 125 6. 2(x3−x) = 1 9. 32x = 5 12. 3(x−1) = 29 15. 2000e0.1t = 4000 17. 70 + 90e−0.1t = 75 18. 30 − 6e−0.1x = 20 20. 5000 1 + 2e−3t = 2500 23. e2x = 2ex 26. 3(x−1) = 1 2 (x+5) 21. 150 1 + 29e−0.8t = 75 24. 7e2x = 28e−6x 27. 73+7x = 34−2x 28. e2x − 3ex − 10 = 0 29. e2x = ex + 6 30. 4x + 2x = 12 31. ex − 3e−x = 2 32. ex + 15e−x = 8 33. 3x + 25 · 3−x = 10 In Exercises 34 - 39, solve the inequality analytically. 34. ex > 53 36. 2(x3−x) < 1 38. 150 1 + 29e−0.8t ≤ 130 35. 1000 (1.005)12t ≥ 3000 37. 25 4 5 x ≥ 10 39. 70 + 90e−0.1t ≤ 75 In Exercises 40 - 45, use your calculator to help you solve the equation or inequality. 40. 2x = x2 41. ex = ln(x) + 5 43. e−x − xe−x ≥ 0 44. 3(x−1) < 2x √ 42. e x = x + 1 45. ex < x3 − x 46. Since f (x) = ln(x) is a strictly increasing function, if 0 < a < b then ln(a) < ln(b). Use this fact to solve the inequality e(3x−1) > 6 without a sign diagram. Use this technique to solve the inequalities in Exercises 34 - 39. (NOTE: Isolate the exponential function ﬁrst!) 47. Compute
the inverse of f (x) = ex − e−x 2. State the domain and range of both f and f −1. 6.3 Exponential Equations and Inequalities 457 48. In Example 6.3.4, we found that the inverse of f (x) = we left a few loose ends for you to tie up. 5ex ex + 1 was f −1(x) = ln x 5 − x but (a) Show that f −1 ◦ f (x) = x for all x in the domain of f and that f ◦ f −1 (x) = x for all x in the domain of f −1. (b) Find the range of f by ﬁnding the domain of f −1. 5x x + 1 (c) Let g(x) = and h(x) = ex. Show that f = g ◦ h and that (g ◦ h)−1 = h−1 ◦ g−1. (We know this is true in general by Exercise 31 in Section 5.2, but it’s nice to see a speciﬁc example of the property.) 49. With the help of your classmates, solve the inequality ex > xn for a variety of natural numbers n. What might you conjecture about the “speed” at which f (x) = ex grows versus any polynomial? 458 Exponential and Logarithmic Functions 6.3.2 Answers 1. x = 3 4 4. x = − 1 4 7. x = 16 15 10. x = − ln(2) ln(5) 13. x = ln(3) 12 ln(1.005) 2. x = 4 5. x = − 7 3 8. x = − 2 11 11. No solution. 14. k = ln( 1 2 ) −5730 = ln(2) 5730 3. x = 2 6. x = −1, 0, 1 9. x = ln(5) 2 ln(3) 12. x = ln(29)+ln(3) ln(3) 15. t = ln(2) 0.1 = 10 ln(2) 17. t = 18 ) ln( 1 −0.1 = 10 ln(18) 19. x = ln(2) 21. t
= ln( 1 29 ) −0.8 = 5 4 ln(29) 23. x = ln(2) 25. x = ln(3) ln(3)−ln(2) 27. x = 4 ln(3)−3 ln(7) 7 ln(7)+2 ln(3) 16. x = 1 2 ln 1 2 = − 1 2 ln(2) 18. x = −10 ln 5 3 = 10 ln 3 5 20. t = 1 3 ln(2) 22. x = = ln(2)−ln(5) ln(4)−ln(5) 24. x = − 1 = 1 4 ln(2) ln( 2 5 ) ln( 4 5 ) 8 ln 1 ln(3)+5 ln( 1 2 ) ln(3)−ln( 1 2 ) 4 26. x = = ln(3)−5 ln(2) ln(3)+ln(2) 28. x = ln(5) 31. x = ln(3) 34. (ln(53), ∞) 36. (−∞, −1) ∪ (0, 1) 29. x = ln(3) 32. x = ln(3), ln(5) 35. 30. x = ln(3) ln(2) 33. x = ln(5) ln(3) −∞, ln(2)−ln(5) ln(4)−ln(5) = ln(3) 12 ln(1.005), ∞ 37. −∞, ln( 2 5 ) ln( 4 5 ) 38. −∞, ln( 2 377 ) −0.8 = −∞, 5 4 ln 377 2 39. ln( 1 18 ) −0.1, ∞ = [10 ln(18), ∞) 40. x ≈ −0.76666, x = 2, x = 4 41. x ≈ 0.01866, x ≈ 1.7115 42. x = 0 44. ≈ (−∞, 2.7095) 43. (−∞, 1] 45. ≈ (2.3217, 4.3717
) 46. x > 1 3 (ln(6) + 1) √ 47. f −1 = ln x + x2 + 1. Both f and f −1 have domain (−∞, ∞) and range (−∞, ∞). 6.4 Logarithmic Equations and Inequalities 459 6.4 Logarithmic Equations and Inequalities In Section 6.3 we solved equations and inequalities involving exponential functions using one of two basic strategies. We now turn our attention to equations and inequalities involving logarithmic functions, and not surprisingly, there are two basic strategies to choose from. For example, suppose we wish to solve log2(x) = log2(5). Theorem 6.4 tells us that the only solution to this equation is x = 5. Now suppose we wish to solve log2(x) = 3. If we want to use Theorem 6.4, we need to 23 = log2(8). rewrite 3 as a logarithm base 2. We can use Theorem 6.3 to do just that: 3 = log2 Our equation then becomes log2(x) = log2(8) so that x = 8. However, we could have arrived at the same answer, in fewer steps, by using Theorem 6.3 to rewrite the equation log2(x) = 3 as 23 = x, or x = 8. We summarize the two common ways to solve log equations below. Steps for Solving an Equation involving Logarithmic Functions 1. Isolate the logarithmic function. 2. (a) If convenient, express both sides as logs with the same base and equate the arguments of the log functions. (b) Otherwise, rewrite the log equation as an exponential equation. Example 6.4.1. Solve the following equations. Check your solutions graphically using a calculator. 1. log117(1 − 3x) = log117 x2 − 3 2. 2 − ln(x − 3) = 1 3. log6(x + 4) + log6(3 − x) = 1 4. log7(1 − 2x) = 1 − log7(3 − x) 5. log2(x + 3) = log2(6 − x) + 3 6. 1 + 2 log4(x + 1) = 2 log2(x) Solution. 1. Since we have the
same base on both sides of the equation log117(1 − 3x) = log117 x2 − 3, we equate what’s inside the logs to get 1 − 3x = x2 − 3. Solving x2 + 3x − 4 = 0 gives x = −4 and x = 1. To check these answers using the calculator, we make use of the change ln(x2−3) of base formula and graph f (x) = ln(1−3x) ln(117) and we see they intersect only at x = −4. To see what happened to the solution x = 1, we substitute it into our original equation to obtain log117(−2) = log117(−2). While these expressions look identical, neither is a real number,1 which means x = 1 is not in the domain of the original equation, and is not a solution. ln(117) and g(x) = 2. Our ﬁrst objective in solving 2−ln(x−3) = 1 is to isolate the logarithm. We get ln(x−3) = 1, which, as an exponential equation, is e1 = x − 3. We get our solution x = e + 3. On the calculator, we see the graph of f (x) = 2 − ln(x − 3) intersects the graph of g(x) = 1 at x = e + 3 ≈ 5.718. 1They do, however, represent the same family of complex numbers. We stop ourselves at this point and refer the reader to a good course in Complex Variables. 460 Exponential and Logarithmic Functions y = f (x) = log117(1 − 3x) and x2 − 3 y = g(x) = log117 y = f (x) = 2 − ln(x − 3) and y = g(x) = 1 3. We can start solving log6(x + 4) + log6(3 − x) = 1 by using the Product Rule for logarithms to rewrite the equation as log6 [(x + 4)(3 − x)] = 1. Rewriting this as an exponential equation, we get 61 = (x + 4)(3 − x). This reduces to x2 + x − 6 = 0, which gives x = −3 and x = 2. Graphing y
= f (x) = ln(x+4) and y = g(x) = 1, we see they intersect twice, at x = −3 and x = 2. ln(6) + ln(3−x) ln(6) y = f (x) = log6(x + 4) + log6(3 − x) and y = g(x) = 1 4. Taking a cue from the previous problem, we begin solving log7(1 − 2x) = 1 − log7(3 − x) by ﬁrst collecting the logarithms on the same side, log7(1 − 2x) + log7(3 − x) = 1, and then using the Product Rule to get log7[(1 − 2x)(3 − x)] = 1. Rewriting this as an exponential equation gives 71 = (1−2x)(3−x) which gives the quadratic equation 2x2 −7x−4 = 0. Solving, we ﬁnd x = − 1 intersect ln(7) only at x = − 1 2. Checking x = 4 in the original equation produces log7(−7) = 1 − log7(−1), which is a clear domain violation. 2 and x = 4. Graphing, we ﬁnd y = f (x) = ln(1−2x) and y = g(x) = 1− ln(3−x) ln(7) 5. Starting with log2(x + 3) = log2(6 − x) + 3, we gather the logarithms to one side and get log2(x + 3) − log2(6 − x) = 3. We then use the Quotient Rule and convert to an exponential equation log2 x + 3 6 − x = 3 ⇐⇒ 23 = x + 3 6 − x This reduces to the linear equation 8(6 − x) = x + 3, which gives us x = 5. When we graph f (x) = ln(x+3) ln(2) + 3, we ﬁnd they intersect at x = 5. ln(2) and g(x) = ln(6−x) 6.4 Logarithmic Equations and Inequalities 461 y = f
(x) = log7(1 − 2x) and y = g(x) = 1 − log7(3 − x) y = f (x) = log2(x + 3) and y = g(x) = log2(6 − x) + 3 6. Starting with 1 + 2 log4(x + 1) = 2 log2(x), we gather the logs to one side to get the equation 1 = 2 log2(x) − 2 log4(x + 1). Before we can combine the logarithms, however, we need a common base. Since 4 is a power of 2, we use change of base to convert log4(x + 1) = log2(x + 1) log2(4) = 1 2 log2(x + 1) Hence, our original equation becomes 2 log2(x + 1) 1 = 2 log2(x) − 2 1 1 = 2 log2(x) − log2(x + 1) x2 − log2(x + 1) 1 = log2 x2 x + 1 1 = log2 Power Rule Quotient Rule Rewriting this in exponential form, we get x2 formula, we get x = 1 ± √ 3. Graphing f (x) = 1 + 2 ln(x+1) x+1 = 2 or x2 − 2x − 2 = 0. Using the quadratic ln(2), we see the 3 < 0, which means if and g(x) = 2 ln(x) ln(4) 3 ≈ 2.732. The solution x = 1 − √ graphs intersect only at x = 1 + substituted into the original equation, the term 2 log2 √ 3 is undeﬁned. 1 − √ y = f (x) = 1 + 2 log4(x + 1) and y = g(x) = 2 log2(x) 462 Exponential and Logarithmic Functions If nothing else, Example 6.4.1 demonstrates the importance of checking for extraneous solutions2 when solving equations involving logarithms. Even though we checked our answers graphically, extraneous solutions are easy to spot - any supposed solution which causes a negative number inside a logarithm needs to be discarded. As with the equations in Example 6.3.1, much can be learned from checking
all of the answers in Example 6.4.1 analytically. We leave this to the reader and turn our attention to inequalities involving logarithmic functions. Since logarithmic functions are continuous on their domains, we can use sign diagrams. Example 6.4.2. Solve the following inequalities. Check your answer graphically using a calculator. 1. 1 ln(x) + 1 ≤ 1 Solution. 2. (log2(x))2 < 2 log2(x) + 3 3. x log(x + 1) ≥ x 1. We start solving 1 1 1 ln(x) ln(x)+1 − ln(x)+1 ln(x)+1 ≤ 0 which reduces to − ln(x) ln(x)+1 ≤ 1 by getting 0 on one side of the inequality: Getting a common denominator yields ln(x)+1 ≥ 0. We deﬁne r(x) = ln(x) ln(x)+1 − 1 ≤ 0. ln(x)+1 ≤ 0, ln(x)+1 and set about ﬁnding the domain and the zeros or In order to keep the of r. Due to the appearance of the term ln(x), we require x > 0. denominator away from zero, we solve ln(x) + 1 = 0 so ln(x) = −1, so x = e−1 = 1 e. Hence, the domain of r is 0, 1 ln(x)+1 = 0 so that e ln(x) = 0, and we ﬁnd x = e0 = 1. In order to determine test values for r without resorting to the calculator, we need to ﬁnd numbers between 0, 1 e, and 1 which have a base of e. Since, we use the fact that e ≈ 2.718 > 1, 0 < 1 e < 1√ = −2 ln 1 −2+1 = 2, which is (+). The rest of the test values e2 ∪ [1, ∞). are determined similarly. From our sign diagram, we ﬁnd the solution to be 0, 1 e Graphing f (x) = 1 ln(x)+1 and g(x) = 1, we see the graph of f is below the graph of g on the solution intervals
, and that the graphs intersect at x = 1. e2 < 1 = ln e−2 = −2, and ﬁnd r 1 e2 e, ∞. To ﬁnd the zeros of r, we set r(x) = ln(x) e < 1 < e. To determine the sign of r 1 ∪ 1 e2 (+) (−) 0 (+) 0 1 e 1 2Recall that an extraneous solution is an answer obtained analytically which does not satisfy the original equation. y = f (x) = 1 ln(x)+1 and y = g(x) = 1 6.4 Logarithmic Equations and Inequalities 463 2. Moving all of the nonzero terms of (log2(x))2 < 2 log2(x) + 3 to one side of the inequality, we have (log2(x))2 − 2 log2(x) − 3 < 0. Deﬁning r(x) = (log2(x))2 − 2 log2(x) − 3, we get the domain of r is (0, ∞), due to the presence of the logarithm. To ﬁnd the zeros of r, we set r(x) = (log2(x))2 − 2 log2(x) − 3 = 0 which results in a ‘quadratic in disguise.’ We set u = log2(x) so our equation becomes u2 − 2u − 3 = 0 which gives us u = −1 and u = 3. Since u = log2(x), we get log2(x) = −1, which gives us x = 2−1 = 1 2, and log2(x) = 3, which yields x = 23 = 8. We use test values which are powers of 2: 0 < 1 2 < 1 < 8 < 16, and from our 2 sign diagram, we see r(x) < 0 on 1 is below the graph of y = g(x) = 2 ln(x) 2, 8. Geometrically, we see the graph of f (x) = ln(2) + 3 on the solution interval. ln(x) ln(2) 4 < 1 (+) 0 (−) 1 2 0 (+) 8 0 y = f (x) = (log2
(x))2 and y = g(x) = 2 log2(x) + 3 3. We begin to solve x log(x+1) ≥ x by subtracting x from both sides to get x log(x+1)−x ≥ 0. We deﬁne r(x) = x log(x+1)−x and due to the presence of the logarithm, we require x+1 > 0, or x > −1. To ﬁnd the zeros of r, we set r(x) = x log(x + 1) − x = 0. Factoring, we get x (log(x + 1) − 1) = 0, which gives x = 0 or log(x+1)−1 = 0. The latter gives log(x+1) = 1, or x + 1 = 101, which admits x = 9. We select test values x so that x + 1 is a power of 10, and we obtain −1 < −0.9 < 0 < 10 − 1 < 9 < 99. Our sign diagram gives the solution to be (−1, 0] ∪ [9, ∞). The calculator indicates the graph of y = f (x) = x log(x + 1) is above y = g(x) = x on the solution intervals, and the graphs intersect at x = 0 and x = 9. √ (+) 0 (−) 0 (+) −1 0 9 y = f (x) = x log(x + 1) and y = g(x) = x 464 Exponential and Logarithmic Functions Our next example revisits the concept of pH ﬁrst seen in Exercise 77 in Section 6.1. Example 6.4.3. In order to successfully breed Ippizuti ﬁsh the pH of a freshwater tank must be at least 7.8 but can be no more than 8.5. Determine the corresponding range of hydrogen ion concentration, and check your answer using a calculator. Solution. Recall from Exercise 77 in Section 6.1 that pH = − log[H+] where [H+] is the hydrogen ion concentration in moles per liter. We require 7.8 ≤ − log[H+] ≤ 8.5 or −7.8 ≥ log[H+] ≥ −8.5. To solve this compound inequality we solve −7.
8 ≥ log[H+] and log[H+] ≥ −8.5 and take the intersection of the solution sets.3 The former inequality yields 0 < [H+] ≤ 10−7.8 and the latter yields [H+] ≥ 10−8.5. Taking the intersection gives us our ﬁnal answer 10−8.5 ≤ [H+] ≤ 10−7.8. (Your Chemistry professor may want the answer written as 3.16 × 10−9 ≤ [H+] ≤ 1.58 × 10−8.) After carefully adjusting the viewing window on the graphing calculator we see that the graph of f (x) = − log(x) lies between the lines y = 7.8 and y = 8.5 on the interval [3.16 × 10−9, 1.58 × 10−8]. The graphs of y = f (x) = − log(x), y = 7.8 and y = 8.5 We close this section by ﬁnding an inverse of a one-to-one function which involves logarithms. Example 6.4.4. The function f (x) = log(x) 1 − log(x) is one-to-one. Find a formula for f −1(x) and check your answer graphically using your calculator. Solution. We ﬁrst write y = f (x) then interchange the x and y and solve for y. y = f (x) y = x = log(x) 1 − log(x) log(y) 1 − log(y) x (1 − log(y)) = log(y) x − x log(y) = log(y) Interchange x and y. x = x log(y) + log(y) x = (x + 1) log(y) x x + 1 = log(y) y = 10 x x+1 Rewrite as an exponential equation. 3Refer to page 4 for a discussion of what this means. 6.4 Logarithmic Equations and Inequalities 465 We have f −1(x) = 10 x x+1. Graphing f and f −1 on the same viewing window yields y = f (x) = log(x) 1 − log(x) and y = g(x) = 10 x x+1 466 Exponential and Logar
ithmic Functions 6.4.1 Exercises In Exercises 1 - 24, solve the equation analytically. 1. log(3x − 1) = log(4 − x) 3. ln 8 − x2 = ln(2 − x) 5. log3(7 − 2x) = 2 7. ln x2 − 99 = 0 9. log125 3x − 2 2x + 3 = 1 3 11. − log(x) = 5.4 13. 6 − 3 log5(2x) = 0 2. log2 x3 = log2(x) 4. log5 18 − x2 = log5(6 − x) 6. log 1 2 (2x − 1) = −3 8. log(x2 − 3x) = 1 10. log x 10−3 = 4.7 12. 10 log x 10−12 = 150 14. 3 ln(x) − 2 = 1 − ln(x) 15. log3(x − 4) + log3(x + 4) = 2 16. log5(2x + 1) + log5(x + 2) = 1 17. log169(3x + 7) − log169(5x − 9) = 1 2 18. ln(x + 1) − ln(x) = 3 19. 2 log7(x) = log7(2) + log7(x + 12) 20. log(x) − log(2) = log(x + 8) − log(x + 2) 21. log3(x) = log 1 3 (x) + 8 23. (log(x))2 = 2 log(x) + 15 22. ln(ln(x)) = 3 24. ln(x2) = (ln(x))2 In Exercises 25 - 30, solve the inequality analytically. 25. 1 − ln(x) x2 < 0 27. 10 log x 10−12 ≥ 90 29. 2.3 < − log(x) < 5.4 26. x ln(x) − x > 0 28. 5.6 ≤ log x 10−3 ≤ 7.1 30. ln(x2) ≤ (ln(x))2 In Exercises 31 - 34, use your calculator to help you solve the equation or inequality. 31
. ln(x) = e−x 33. ln(x2 + 1) ≥ 5 √ 32. ln(x) = 4 x 34. ln(−2x3 − x2 + 13x − 6) < 0 6.4 Logarithmic Equations and Inequalities 467 35. Since f (x) = ex is a strictly increasing function, if a < b then ea < eb. Use this fact to solve the inequality ln(2x + 1) < 3 without a sign diagram. Use this technique to solve the inequalities in Exercises 27 - 29. (Compare this to Exercise 46 in Section 6.3.) 36. Solve ln(3 − y) − ln(y) = 2x + ln(5) for y. 37. In Example 6.4.4 we found the inverse of f (x) = log(x) 1 − log(x) to be f −1(x) = 10 x x+1. (a) Show that f −1 ◦ f (x) = x for all x in the domain of f and that f ◦ f −1 (x) = x for all x in the domain of f −1. (b) Find the range of f by ﬁnding the domain of f −1. x 1 − x (c) Let g(x) = and h(x) = log(x). Show that f = g ◦ h and (g ◦ h)−1 = h−1 ◦ g−1. (We know this is true in general by Exercise 31 in Section 5.2, but it’s nice to see a speciﬁc example of the property.) 38. Let f (x) = 1 2 ln 1 + x 1 − x. Compute f −1(x) and ﬁnd its domain and range. 39. Explain the equation in Exercise 10 and the inequality in Exercise 28 above in terms of the Richter scale for earthquake magnitude. (See Exercise 75 in Section 6.1.) 40. Explain the equation in Exercise 12 and the inequality in Exercise 27 above in terms of sound intensity level as measured in decibels. (See Exercise 76 in Section 6.1.) 41. Explain the equation in Exercise 11 and the inequality in Exercise 29 above in terms of the pH of a solution. (See Exercise 77
in Section 6.1.) √ 42. With the help of your classmates, solve the inequality n x > ln(x) for a variety of natural numbers n. What might you conjecture about the “speed” at which f (x) = ln(x) grows versus any principal nth root function? 468 Exponential and Logarithmic Functions 6.4.2 Answers 1. x = 5 4 4. x = −3, 4 7. x = ±10 10. x = 101.7 13. x = 25 2 16. x = 1 2 19. x = 6 22. x = ee3 25. (e, ∞) 28. 102.6, 104.1 31. x ≈ 1.3098 2. x = 1 5. x = −1 8. x = −2, 5 11. x = 10−5.4 14. x = e3/4 17. x = 2 20. x = 4 3. x = −2 6. x = 9 2 9. x = − 17 7 12. x = 103 15. x = 5 18. x = 1 e3−1 21. x = 81 23. x = 10−3, 105 24. x = 1, x = e2 26. (e, ∞) 27. 10−3, ∞ 29. 10−5.4, 10−2.3 30. (0, 1] ∪ [e2, ∞) 32. x ≈ 4.177, x ≈ 5503.665 33. ≈ (−∞, −12.1414) ∪ (12.1414, ∞) 34. ≈ (−3.0281, −3)∪(0.5, 0.5991)∪(1.9299, 2) 35. − 1 2 < x < e3 − 1 2 e2x − 1 e2x + 1 38. f −1(x) = ex − e−x ex + e−x. (To see why we rewrite this in this form, see Exercise 51 in Section 11.10.) The domain of f −1 is (−∞, ∞) and its range is the same as the domain of f, namely (−1, 1). = 36. y = 3 5e2x + 1 6.5 Applications of Exponential and Logarithmic Functions 4
69 6.5 Applications of Exponential and Logarithmic Functions As we mentioned in Section 6.1, exponential and logarithmic functions are used to model a wide variety of behaviors in the real world. In the examples that follow, note that while the applications are drawn from many diﬀerent disciplines, the mathematics remains essentially the same. Due to the applied nature of the problems we will examine in this section, the calculator is often used to express our answers as decimal approximations. 6.5.1 Applications of Exponential Functions Perhaps the most well-known application of exponential functions comes from the ﬁnancial world. Suppose you have $100 to invest at your local bank and they are oﬀering a whopping 5 % annual percentage interest rate. This means that after one year, the bank will pay you 5% of that $100, or $100(0.05) = $5 in interest, so you now have $105.1 This is in accordance with the formula for simple interest which you have undoubtedly run across at some point before. Equation 6.1. Simple Interest The amount of interest I accrued at an annual rate r on an investmenta P after t years is The amount A in the account after t years is given by I = P rt aCalled the principal A = P + I = P + P rt = P (1 + rt) Suppose, however, that six months into the year, you hear of a better deal at a rival bank.2 Naturally, you withdraw your money and try to invest it at the higher rate there. Since six months is one half of a year, that initial $100 yields $100(0.05) 1 = $2.50 in interest. You take your 2 $102.50 oﬀ to the competitor and ﬁnd out that those restrictions which may apply actually do apply to you, and you return to your bank which happily accepts your $102.50 for the remaining six months of the year. To your surprise and delight, at the end of the year your statement reads $105.06, not $105 as you had expected.3 Where did those extra six cents come from? For the ﬁrst six months of the year, interest was earned on the original principal of $100, but for the second six months, interest was earned on $102.50, that is, you earned interest on your interest. This is the basic
concept behind compound interest. In the previous discussion, we would say that the interest was compounded twice, or semiannually.4 If more money can be earned by earning interest on interest already earned, a natural question to ask is what happens if the interest is compounded more often, say 4 times a year, which is every three months, or ‘quarterly.’ In this case, the money is in the account for three months, or 1 4 of a year, at a time. After the ﬁrst quarter, we = $101.25. We now invest the $101.25 for the next three have A = P (1 + rt) = $100 1 + 0.05 · 1 4 1How generous of them! 2Some restrictions may apply. 3Actually, the ﬁnal balance should be $105.0625. 4Using this convention, simple interest after one year is the same as compounding the interest only once. 470 Exponential and Logarithmic Functions months and ﬁnd that at the end of the second quarter, we have A = $101.25 1 + 0.05 · 1 ≈ $102.51. 4 Continuing in this manner, the balance at the end of the third quarter is $103.79, and, at last, we obtain $105.08. The extra two cents hardly seems worth it, but we see that we do in fact get more money the more often we compound. In order to develop a formula for this phenomenon, we need to do some abstract calculations. Suppose we wish to invest our principal P at an annual rate r and th of a year compound the interest n times per year. This means the money sits in the account 1 n between compoundings. Let Ak denote the amount in the account after the kth compounding. Then. After the second compounding, we use A1 A1 = P 1 + r 1. Continuing in as our new principal and get A2 = A1 this fashion, we get A3 = P 1 + r k. Since we compound the interest n times per year, after t years, we have nt compoundings. We have just derived the general formula for compound interest below. which simpliﬁes to A1 = P 1 + r, and so on, so that Ak =, A4 = Equation 6.2. Compounded Interest: If an initial principal P is invested at an annual rate r and
the interest is compounded n times per year, the amount A in the account after t years is A(t) = P 1 + nt r n If we take P = 100, r = 0.05, and n = 4, Equation 6.2 becomes A(t) = 100 1 + 0.05 which 4 reduces to A(t) = 100(1.0125)4t. To check this new formula against our previous calculations, we ﬁnd A 1 4 ≈ $103.79, and A(1) ≈ $105.08. ≈ $102.51, A 3 4 4 ) = 101.25, A 1 2 = 100(1.0125)4( 1 4t Example 6.5.1. Suppose $2000 is invested in an account which oﬀers 7.125% compounded monthly. 1. Express the amount A in the account as a function of the term of the investment t in years. 2. How much is in the account after 5 years? 3. How long will it take for the initial investment to double? 4. Find and interpret the average rate of change5 of the amount in the account from the end of the fourth year to the end of the ﬁfth year, and from the end of the thirty-fourth year to the end of the thirty-ﬁfth year. Solution. 1. Substituting P = 2000, r = 0.07125, and n = 12 (since interest is compounded monthly) into 12t Equation 6.2 yields A(t) = 2000 1 + 0.07125 = 2000(1.0059375)12t. 12 2. Since t represents the length of the investment in years, we substitute t = 5 into A(t) to ﬁnd A(5) = 2000(1.0059375)12(5) ≈ 2852.92. After 5 years, we have approximately $2852.92. 5See Deﬁnition 2.3 in Section 2.1. 6.5 Applications of Exponential and Logarithmic Functions 471 3. Our initial investment is $2000, so to ﬁnd the time it takes this to double, we need to ﬁnd t when A(t) = 4000. We get 2000(1.0059375)12t = 4000, or (1.00
59375)12t = 2. Taking natural 12 ln(1.0059375) ≈ 9.75. Hence, it takes approximately 9 years logs as in Section 6.3, we get t = 9 months for the investment to double. ln(2) 5−4 4. To ﬁnd the average rate of change of A from the end of the fourth year to the end of the ﬁfth year, we compute A(5)−A(4) ≈ 195.63. Similarly, the average rate of change of A from the end of the thirty-fourth year to the end of the thirty-ﬁfth year is A(35)−A(34) ≈ 1648.21. This means that the value of the investment is increasing at a rate of approximately $195.63 per year between the end of the fourth and ﬁfth years, while that rate jumps to $1648.21 per year between the end of the thirty-fourth and thirty-ﬁfth years. So, not only is it true that the longer you wait, the more money you have, but also the longer you wait, the faster the money increases.6 35−34 We have observed that the more times you compound the interest per year, the more money you will earn in a year. Let’s push this notion to the limit.7 Consider an investment of $1 invested at 100% interest for 1 year compounded n times a year. Equation 6.2 tells us that the amount of money in the account after 1 year is A = 1 + 1 n n. Below is a table of values relating n and A. A n 2 1 2 2.25 4 ≈ 2.4414 12 ≈ 2.6130 360 ≈ 2.7145 1000 ≈ 2.7169 10000 ≈ 2.7181 100000 ≈ 2.7182 As promised, the more compoundings per year, the more money there is in the account, but we also observe that the increase in money is greatly diminishing. We are witnessing a mathematical ‘tug of war’. While we are compounding more times per year, and hence getting interest on our interest more often, the amount of time between compoundings is getting smaller and smaller, so there is less time to build up additional interest. With Calculus, we can show8 that as n → ∞, A
= 1 + 1 n → e, where e is the natural base ﬁrst presented in Section 6.1. Taking the number n of compoundings per year to inﬁnity results in what is called continuously compounded interest. Theorem 6.8. If you invest $1 at 100% interest compounded continuously, then you will have $e at the end of one year. 6In fact, the rate of increase of the amount in the account is exponential as well. This is the quality that really deﬁnes exponential functions and we refer the reader to a course in Calculus. 7Once you’ve had a semester of Calculus, you’ll be able to fully appreciate this very lame pun. 8Or deﬁne, depending on your point of view. 472 Exponential and Logarithmic Functions Using this deﬁnition of e and a little Calculus, we can take Equation 6.2 and produce a formula for continuously compounded interest. Equation 6.3. Continuously Compounded Interest: If an initial principal P is invested at an annual rate r and the interest is compounded continuously, the amount A in the account after t years is A(t) = P ert If we take the scenario of Example 6.5.1 and compare monthly compounding to continuous compounding over 35 years, we ﬁnd that monthly compounding yields A(35) = 2000(1.0059375)12(35) which is about $24,035.28, whereas continuously compounding gives A(35) = 2000e0.07125(35) which is about $24,213.18 - a diﬀerence of less than 1%. Equations 6.2 and 6.3 both use exponential functions to describe the growth of an investment. Curiously enough, the same principles which govern compound interest are also used to model In Biology, The Law of Uninhibited Growth states as short term growth of populations. its premise that the instantaneous rate at which a population increases at any time is directly proportional to the population at that time.9 In other words, the more organisms there are at a given moment, the faster they reproduce. Formulating the law as stated results in a diﬀerential equation, which requires Calculus to solve. Its solution is stated below. Equation 6.4. Uninhibited Growth: If a population increases according to The Law of Un
inhibited Growth, the number of organisms N at time t is given by the formula N (t) = N0ekt, where N (0) = N0 (read ‘N nought’) is the initial number of organisms and k > 0 is the constant of proportionality which satisﬁes the equation (instantaneous rate of change of N (t) at time t) = k N (t) It is worth taking some time to compare Equations 6.3 and 6.4. In Equation 6.3, we use P to denote the initial investment; in Equation 6.4, we use N0 to denote the initial population. In Equation 6.3, r denotes the annual interest rate, and so it shouldn’t be too surprising that the k in Equation 6.4 corresponds to a growth rate as well. While Equations 6.3 and 6.4 look entirely diﬀerent, they both represent the same mathematical concept. Example 6.5.2. In order to perform arthrosclerosis research, epithelial cells are harvested from discarded umbilical tissue and grown in the laboratory. A technician observes that a culture of twelve thousand cells grows to ﬁve million cells in one week. Assuming that the cells follow The Law of Uninhibited Growth, ﬁnd a formula for the number of cells, N, in thousands, after t days. Solution. We begin with N (t) = N0ekt. Since N is to give the number of cells in thousands, we have N0 = 12, so N (t) = 12ekt. In order to complete the formula, we need to determine the 9The average rate of change of a function over an interval was ﬁrst introduced in Section 2.1. Instantaneous rates of change are the business of Calculus, as is mentioned on Page 161. 6.5 Applications of Exponential and Logarithmic Functions 473 growth rate k. We know that after one week, the number of cells has grown to ﬁve million. Since t measures days and the units of N are in thousands, this translates mathematically to N (7) = 5000. 3 ). Of We get the equation 12e7k = 5000 which gives k = 1 course, in practice, we would approximate k to some desired accuracy, say k ≈ 0.8618, which we can interpret as an
86.18% daily growth rate for the cells.. Hence, N (t) = 12e 7 ln 1250 7 ln( 1250 3 t Whereas Equations 6.3 and 6.4 model the growth of quantities, we can use equations like them to describe the decline of quantities. One example we’ve seen already is Example 6.1.1 in Section 6.1. There, the value of a car declined from its purchase price of $25,000 to nothing at all. Another real world phenomenon which follows suit is radioactive decay. There are elements which are unstable and emit energy spontaneously. In doing so, the amount of the element itself diminishes. The assumption behind this model is that the rate of decay of an element at a particular time is directly proportional to the amount of the element present at that time. In other words, the more of the element there is, the faster the element decays. This is precisely the same kind of hypothesis which drives The Law of Uninhibited Growth, and as such, the equation governing radioactive decay is hauntingly similar to Equation 6.4 with the exception that the rate constant k is negative. Equation 6.5. Radioactive Decay The amount of a radioactive element A at time t is given by the formula A(t) = A0ekt, where A(0) = A0 is the initial amount of the element and k < 0 is the constant of proportionality which satisﬁes the equation (instantaneous rate of change of A(t) at time t) = k A(t) Example 6.5.3. Iodine-131 is a commonly used radioactive isotope used to help detect how well the thyroid is functioning. Suppose the decay of Iodine-131 follows the model given in Equation 6.5, and that the half-life10 of Iodine-131 is approximately 8 days. If 5 grams of Iodine-131 is present initially, ﬁnd a function which gives the amount of Iodine-131, A, in grams, t days later. Solution. Since we start with 5 grams initially, Equation 6.5 gives A(t) = 5ekt. Since the half-life is 8 days, it takes 8 days for half of the Iodine-131 to decay, leaving half of it behind. Hence, A(8) = 2.5 8 ln 1 which means 5e
8k = 2.5. Solving, we get k = 1 8 ≈ −0.08664, which we can interpret as a loss of material at a rate of 8.664% daily. Hence, A(t) = 5e− t ln(2) 8 ≈ 5e−0.08664t. = − ln(2) 2 We now turn our attention to some more mathematically sophisticated models. One such model is Newton’s Law of Cooling, which we ﬁrst encountered in Example 6.1.2 of Section 6.1. In that example we had a cup of coﬀee cooling from 160◦F to room temperature 70◦F according to the formula T (t) = 70 + 90e−0.1t, where t was measured in minutes. In this situation, we know the physical limit of the temperature of the coﬀee is room temperature,11 and the diﬀerential equation 10The time it takes for half of the substance to decay. 11The Second Law of Thermodynamics states that heat can spontaneously ﬂow from a hotter object to a colder one, but not the other way around. Thus, the coﬀee could not continue to release heat into the air so as to cool below room temperature. 474 Exponential and Logarithmic Functions which gives rise to our formula for T (t) takes this into account. Whereas the radioactive decay model had a rate of decay at time t directly proportional to the amount of the element which remained at time t, Newton’s Law of Cooling states that the rate of cooling of the coﬀee at a given time t is directly proportional to how much of a temperature gap exists between the coﬀee at time t and room temperature, not the temperature of the coﬀee itself. In other words, the coﬀee cools faster when it is ﬁrst served, and as its temperature nears room temperature, the coﬀee cools ever more slowly. Of course, if we take an item from the refrigerator and let it sit out in the kitchen, the object’s temperature will rise to room temperature, and since the physics behind warming and cooling is the same, we combine both cases in the equation below. Equation 6.6. Newton’s Law
of Cooling (Warming): The temperature T of an object at time t is given by the formula T (t) = Ta + (T0 − Ta) e−kt, where T (0) = T0 is the initial temperature of the object, Ta is the ambient temperaturea and k > 0 is the constant of proportionality which satisﬁes the equation (instantaneous rate of change of T (t) at time t) = k (T (t) − Ta) aThat is, the temperature of the surroundings. If we re-examine the situation in Example 6.1.2 with T0 = 160, Ta = 70, and k = 0.1, we get, according to Equation 6.6, T (t) = 70 + (160 − 70)e−0.1t which reduces to the original formula given. The rate constant k = 0.1 indicates the coﬀee is cooling at a rate equal to 10% of the diﬀerence between the temperature of the coﬀee and its surroundings. Note in Equation 6.6 that the constant k is positive for both the cooling and warming scenarios. What determines if the function T (t) is increasing or decreasing is if T0 (the initial temperature of the object) is greater than Ta (the ambient temperature) or vice-versa, as we see in our next example. Example 6.5.4. A 40◦F roast is cooked in a 350◦F oven. After 2 hours, the temperature of the roast is 125◦F. 1. Assuming the temperature of the roast follows Newton’s Law of Warming, ﬁnd a formula for the temperature of the roast T as a function of its time in the oven, t, in hours. 2. The roast is done when the internal temperature reaches 165◦F. When will the roast be done? Solution. 1. The initial temperature of the roast is 40◦F, so T0 = 40. The environment in which we are placing the roast is the 350◦F oven, so Ta = 350. Newton’s Law of Warming tells us T (t) = 350 + (40 − 350)e−kt, or T (t) = 350 − 310e−kt. To determine k, we use the fact that after 2 hours, the roast is 125◦F, which means
T (2) = 125. This gives rise to the equation 350 − 310e−2k = 125 which yields k = − 1 ≈ 0.1602. The temperature function is T (t) = 350 − 310e 2 ln 45 62 2 ln( 45 62 ) ≈ 350 − 310e−0.1602t. t 6.5 Applications of Exponential and Logarithmic Functions 475 2. To determine when the roast is done, we set T (t) = 165. This gives 350 − 310e−0.1602t = 165 ≈ 3.22. It takes roughly 3 hours and 15 minutes to cook 0.1602 ln 37 62 whose solution is t = − 1 the roast completely. If we had taken the time to graph y = T (t) in Example 6.5.4, we would have found the horizontal asymptote to be y = 350, which corresponds to the temperature of the oven. We can also arrive at this conclusion by applying a bit of ‘number sense’. As t → ∞, −0.1602t ≈ very big (−) so that e−0.1602t ≈ very small (+). The larger the value of t, the smaller e−0.1602t becomes so that T (t) ≈ 350 − very small (+), which indicates the graph of y = T (t) is approaching its horizontal asymptote y = 350 from below. Physically, this means the roast will eventually warm up to 350◦F.12 The function T is sometimes called a limited growth model, since the function T remains bounded as t → ∞. If we apply the principles behind Newton’s Law of Cooling to a biological example, it says the growth rate of a population is directly proportional to how much room the population has to grow. In other words, the more room for expansion, the faster the growth rate. The logistic growth model combines The Law of Uninhibited Growth with limited growth and states that the rate of growth of a population varies jointly with the population itself as well as the room the population has to grow. Equation 6.7. Logistic Growth: If a population behaves according to the assumptions of logistic growth, the number of organisms N at time t is given by the equation N (t) = L 1 + Ce−kLt, where N (0) = N0 is the initial
population, L is the limiting population,a C is a measure of how much room there is to grow given by C = L N0 − 1. and k > 0 is the constant of proportionality which satisﬁes the equation (instantaneous rate of change of N (t) at time t) = k N (t) (L − N (t)) aThat is, as t → ∞, N (t) → L The logistic function is used not only to model the growth of organisms, but is also often used to model the spread of disease and rumors.13 Example 6.5.5. The number of people N, in hundreds, at a local community college who have heard the rumor ‘Carl is afraid of Virginia Woolf’ can be modeled using the logistic equation N (t) = 84 1 + 2799e−t, 12at which point it would be more toast than roast. 13Which can be just as damaging as diseases. 476 Exponential and Logarithmic Functions where t ≥ 0 is the number of days after April 1, 2009. 1. Find and interpret N (0). 2. Find and interpret the end behavior of N (t). 3. How long until 4200 people have heard the rumor? 4. Check your answers to 2 and 3 using your calculator. Solution. 1. We ﬁnd N (0) = 100. Since N (t) measures the number of people who have heard the rumor in hundreds, N (0) corresponds to 3 people. Since t = 0 corresponds to April 1, 2009, we may conclude that on that day, 3 people have heard the rumor.14 1+2799e0 = 84 2800 = 3 84 2. We could simply note that N (t) is written in the form of Equation 6.7, and identify L = 84. However, to see why the answer is 84, we proceed analytically. Since the domain of N is restricted to t ≥ 0, the only end behavior of signiﬁcance is t → ∞. As we’ve seen before,15 as t → ∞, we have 1997e−t → 0+ and so N (t) ≈ ≈ 84. Hence, as t → ∞, N (t) → 84. This means that as time goes by, the number of people who will have heard the rumor approaches 8400. 84
1+very small (+) 3. To ﬁnd how long it takes until 4200 people have heard the rumor, we set N (t) = 42. Solving 1+2799e−t = 42 gives t = ln(2799) ≈ 7.937. It takes around 8 days until 4200 people have heard the rumor. 84 4. We graph y = N (x) using the calculator and see that the line y = 84 is the horizontal asymptote of the graph, conﬁrming our answer to part 2, and the graph intersects the line y = 42 at x = ln(2799) ≈ 7.937, which conﬁrms our answer to part 3. y = f (x) = 84 1+2799e−x and y = 84 y = f (x) = 84 1+2799e−x and y = 42 14Or, more likely, three people started the rumor. I’d wager Jeﬀ, Jamie, and Jason started it. So much for telling your best friends something in conﬁdence! 15See, for example, Example 6.1.2. 6.5 Applications of Exponential and Logarithmic Functions 477 If we take the time to analyze the graph of y = N (x) above, we can see graphically how logistic growth combines features of uninhibited and limited growth. The curve seems to rise steeply, then at some point, begins to level oﬀ. The point at which this happens is called an inﬂection point or is sometimes called the ‘point of diminishing returns’. At this point, even though the function is still increasing, the rate at which it does so begins to decline. It turns out the point of diminishing returns always occurs at half the limiting population. (In our case, when y = 42.) While these concepts are more precisely quantiﬁed using Calculus, below are two views of the graph of y = N (x), one on the interval [0, 8], the other on [8, 15]. The former looks strikingly like uninhibited growth; the latter like limited growth. y = f (x) = 84 1+2799e−x for 0 ≤ x ≤ 8 y = f (x) = 84 1+2799e−x for 8 ≤ x ≤ 16
6.5.2 Applications of Logarithms Just as many physical phenomena can be modeled by exponential functions, the same is true of logarithmic functions. In Exercises 75, 76 and 77 of Section 6.1, we showed that logarithms are useful in measuring the intensities of earthquakes (the Richter scale), sound (decibels) and acids and bases (pH). We now present yet a diﬀerent use of the a basic logarithm function, password strength. Example 6.5.6. The information entropy H, in bits, of a randomly generated password consisting of L characters is given by H = L log2(N ), where N is the number of possible symbols for each character in the password. In general, the higher the entropy, the stronger the password. 1. If a 7 character case-sensitive16 password is comprised of letters and numbers only, ﬁnd the associated information entropy. 2. How many possible symbol options per character is required to produce a 7 character password with an information entropy of 50 bits? Solution. 1. There are 26 letters in the alphabet, 52 if upper and lower case letters are counted as diﬀerent. There are 10 digits (0 through 9) for a total of N = 62 symbols. Since the password is to be 7 characters long, L = 7. Thus, H = 7 log2(62) = 7 ln(62) ln(2) ≈ 41.68. 16That is, upper and lower case letters are treated as diﬀerent characters. 478 Exponential and Logarithmic Functions 2. We have L = 7 and H = 50 and we need to ﬁnd N. Solving the equation 50 = 7 log2(N ) gives N = 250/7 ≈ 141.323, so we would need 142 diﬀerent symbols to choose from.17 Chemical systems known as buﬀer solutions have the ability to adjust to small changes in acidity to maintain a range of pH values. Buﬀer solutions have a wide variety of applications from maintaining a healthy ﬁsh tank to regulating the pH levels in blood. Our next example shows how the pH in a buﬀer solution is a little more complicated than the pH we ﬁrst encountered in Exercise 77 in Section 6.1. Example 6.5.7. Blood is a
buﬀer solution. When carbon dioxide is absorbed into the bloodstream it produces carbonic acid and lowers the pH. The body compensates by producing bicarbonate, a weak base to partially neutralize the acid. The equation18 which models blood pH in this situation is pH = 6.1 + log 800, where x is the partial pressure of carbon dioxide in arterial blood, measured x in torr. Find the partial pressure of carbon dioxide in arterial blood if the pH is 7.4. Solution. We set pH = 7.4 and get 7.4 = 6.1 + log 800 x x = 800 101.3 ≈ 40.09. Hence, the partial pressure of carbon dioxide in the blood is about 40 torr. = 1.3. Solving, we ﬁnd, or log 800 x Another place logarithms are used is in data analysis. Suppose, for instance, we wish to model the spread of inﬂuenza A (H1N1), the so-called ‘Swine Flu’. Below is data taken from the World Health Organization (WHO) where t represents the number of days since April 28, 2009, and N represents the number of conﬁrmed cases of H1N1 virus worldwide. 1 t 7 N 148 257 367 658 898 1085 1490 1893 2371 2500 3440 4379 4694 13 10 12 11 4 5 3 2 6 9 8 14 t N 5251 15 16 17 18 19 20 5728 6497 7520 8451 8480 8829 Making a scatter plot of the data treating t as the independent variable and N as the dependent variable gives Which models are suggested by the shape of the data? Thinking back Section 2.5, we try a Quadratic Regression, with pretty good results. 17Since there are only 94 distinct ASCII keyboard characters, to achieve this strength, the number of characters in the password should be increased. 18Derived from the Henderson-Hasselbalch Equation. See Exercise 43 in Section 6.2. Hasselbalch himself was studying carbon dioxide dissolving in blood - a process called metabolic acidosis. 6.5 Applications of Exponential and Logarithmic Functions 479 However, is there any scientiﬁc reason for the data to be quadratic? Are there other models which ﬁt the data equally well, or better? Scientists often use logarith
ms in an attempt to ‘linearize’ data sets - in other words, transform the data sets to produce ones which result in straight lines. To see how this could work, suppose we guessed the relationship between N and t was some kind of power function, not necessarily quadratic, say N = BtA. To try to determine the A and B, we can take the natural log of both sides and get ln(N ) = ln BtA. Using properties of logs to expand the right hand side of this equation, we get ln(N ) = A ln(t) + ln(B). If we set X = ln(t) and Y = ln(N ), this equation becomes Y = AX + ln(B). In other words, we have a line with slope A and Y -intercept ln(B). So, instead of plotting N versus t, we plot ln(N ) versus ln(t). ln(t) ln(N ) 0 0.693 1.099 1.386 1.609 1.792 1.946 2.079 2.197 2.302 2.398 2.485 2.565 4.997 5.549 5.905 6.489 6.800 6.989 7.306 7.546 7.771 7.824 8.143 8.385 8.454 ln(t) ln(N ) 2.639 8.566 2.708 2.773 2.833 2.890 2.944 2.996 8.653 8.779 8.925 9.042 9.045 9.086 Running a linear regression on the data gives The slope of the regression line is a ≈ 1.512 which corresponds to our exponent A. The y-intercept b ≈ 4.513 corresponds to ln(B), so that B ≈ 91.201. Hence, we get the model N = 91.201t1.512, something from Section 5.3. Of course, the calculator has a built-in ‘Power Regression’ feature. If we apply this to our original data set, we get the same model we arrived at before.19 19Critics may question why the authors of the book have chosen to even discuss linearization of data when the calculator has a Power Regression built-in and ready to go. Our response: talk
to your science faculty. 480 Exponential and Logarithmic Functions This is all well and good, but the quadratic model appears to ﬁt the data better, and we’ve yet to mention any scientiﬁc principle which would lead us to believe the actual spread of the ﬂu follows any kind of power function at all. If we are to attack this data from a scientiﬁc perspective, it does seem to make sense that, at least in the early stages of the outbreak, the more people who have the ﬂu, the faster it will spread, which leads us to proposing an uninhibited growth model. If we assume N = BeAt then, taking logs as before, we get ln(N ) = At + ln(B). If we set X = t and Y = ln(N ), then, once again, we get Y = AX + ln(B), a line with slope A and Y -intercept ln(B). Plotting ln(N ) versus t gives the following linear regression. We see the slope is a ≈ 0.202 and which corresponds to A in our model, and the y-intercept is b ≈ 5.596 which corresponds to ln(B). We get B ≈ 269.414, so that our model is N = 269.414e0.202t. Of course, the calculator has a built-in ‘Exponential Regression’ feature which produces what appears to be a diﬀerent model N = 269.414(1.22333419)t. Using properties of exponents, we write e0.202t = e0.202t ≈ (1.223848)t, which, had we carried more decimal places, would have matched the base of the calculator model exactly. The exponential model didn’t ﬁt the data as well as the quadratic or power function model, but it stands to reason that, perhaps, the spread of the ﬂu is not unlike that of the spread of a rumor 6.5 Applications of Exponential and Logarithmic Functions 481 and that a logistic model can be used to model the data. The calculator does have a ‘Logistic Regression’ feature, and using it produces the model N = 10739.147 1+42.416e0.268t.
This appears to be an excellent ﬁt, but there is no friendly coeﬃcient of determination, R2, by which to judge this numerically. There are good reasons for this, but they are far beyond the scope of the text. Which of the models, quadratic, power, exponential, or logistic is the ‘best model’? If by ‘best’ we mean ‘ﬁts closest to the data,’ then the quadratic and logistic models are arguably the winners with the power function model a close second. However, if we think about the science behind the spread of the ﬂu, the logistic model gets an edge. For one thing, it takes into account that only a ﬁnite number of people will ever get the ﬂu (according to our model, 10,739), whereas the quadratic model predicts no limit to the number of cases. As we have stated several times before in the text, mathematical models, regardless of their sophistication, are just that: models, and they all have their limitations.20 20Speaking of limitations, as of June 3, 2009, there were 19,273 conﬁrmed cases of inﬂuenza A (H1N1). This is well above our prediction of 10,739. Each time a new report is issued, the data set increases and the model must be recalculated. We leave this recalculation to the reader. 482 Exponential and Logarithmic Functions 6.5.3 Exercises For each of the scenarios given in Exercises 1 - 6, Find the amount A in the account as a function of the term of the investment t in years. Determine how much is in the account after 5 years, 10 years, 30 years and 35 years. Round your answers to the nearest cent. Determine how long will it take for the initial investment to double. Round your answer to the nearest year. Find and interpret the average rate of change of the amount in the account from the end of the fourth year to the end of the ﬁfth year, and from the end of the thirty-fourth year to the end of the thirty-ﬁfth year. Round your answer to two decimal places. 1. $500 is invested in an account which oﬀers 0.75%, compounded monthly. 2. $500 is invested in
an account which oﬀers 0.75%, compounded continuously. 3. $1000 is invested in an account which oﬀers 1.25%, compounded monthly. 4. $1000 is invested in an account which oﬀers 1.25%, compounded continuously. 5. $5000 is invested in an account which oﬀers 2.125%, compounded monthly. 6. $5000 is invested in an account which oﬀers 2.125%, compounded continuously. 7. Look back at your answers to Exercises 1 - 6. What can be said about the diﬀerence between monthly compounding and continuously compounding the interest in those situations? With the help of your classmates, discuss scenarios where the diﬀerence between monthly and continuously compounded interest would be more dramatic. Try varying the interest rate, the term of the investment and the principal. Use computations to support your answer. 8. How much money needs to be invested now to obtain $2000 in 3 years if the interest rate in a savings account is 0.25%, compounded continuously? Round your answer to the nearest cent. 9. How much money needs to be invested now to obtain $5000 in 10 years if the interest rate in a CD is 2.25%, compounded monthly? Round your answer to the nearest cent. 10. On May, 31, 2009, the Annual Percentage Rate listed at Jeﬀ’s bank for regular savings accounts was 0.25% compounded monthly. Use Equation 6.2 to answer the following. (a) If P = 2000 what is A(8)? (b) Solve the equation A(t) = 4000 for t. (c) What principal P should be invested so that the account balance is $2000 is three years? 6.5 Applications of Exponential and Logarithmic Functions 483 11. Jeﬀ’s bank also oﬀers a 36-month Certiﬁcate of Deposit (CD) with an APR of 2.25%. (a) If P = 2000 what is A(8)? (b) Solve the equation A(t) = 4000 for t. (c) What principal P should be invested so that the account balance is $2000 in three years? (d) The Annual Percentage Yield is the simple interest rate that returns the same amount of interest after one year as the compound interest does. With the help of your classmates,
compute the APY for this investment. 12. A ﬁnance company oﬀers a promotion on $5000 loans. The borrower does not have to make any payments for the ﬁrst three years, however interest will continue to be charged to the loan at 29.9% compounded continuously. What amount will be due at the end of the three year period, assuming no payments are made? If the promotion is extended an additional three years, and no payments are made, what amount would be due? 13. Use Equation 6.2 to show that the time it takes for an investment to double in value does not depend on the principal P, but rather, depends only on the APR and the number of compoundings per year. Let n = 12 and with the help of your classmates compute the doubling time for a variety of rates r. Then look up the Rule of 72 and compare your answers to what that rule says. If you’re really interested21 in Financial Mathematics, you could also compare and contrast the Rule of 72 with the Rule of 70 and the Rule of 69. In Exercises 14 - 18, we list some radioactive isotopes and their associated half-lives. Assume that each decays according to the formula A(t) = A0ekt where A0 is the initial amount of the material and k is the decay constant. For each isotope: Find the decay constant k. Round your answer to four decimal places. Find a function which gives the amount of isotope A which remains after time t. (Keep the units of A and t the same as the given data.) Determine how long it takes for 90% of the material to decay. Round your answer to two decimal places. (HINT: If 90% of the material decays, how much is left?) 14. Cobalt 60, used in food irradiation, initial amount 50 grams, half-life of 5.27 years. 15. Phosphorus 32, used in agriculture, initial amount 2 milligrams, half-life 14 days. 16. Chromium 51, used to track red blood cells, initial amount 75 milligrams, half-life 27.7 days. 17. Americium 241, used in smoke detectors, initial amount 0.29 micrograms, half-life 432.7 years. 18. Uranium 235, used for nuclear power, initial amount 1 kg grams, half-life 704 million years. 21Awesome pun
! 484 Exponential and Logarithmic Functions 19. With the help of your classmates, show that the time it takes for 90% of each isotope listed in Exercises 14 - 18 to decay does not depend on the initial amount of the substance, but rather, on only the decay constant k. Find a formula, in terms of k only, to determine how long it takes for 90% of a radioactive isotope to decay. 20. In Example 6.1.1 in Section 6.1, the exponential function V (x) = 25 4 5 was used to model the value of a car over time. Use the properties of logs and/or exponents to rewrite the model in the form V (t) = 25ekt. x 21. The Gross Domestic Product (GDP) of the US (in billions of dollars) t years after the year 2000 can be modeled by: G(t) = 9743.77e0.0514t (a) Find and interpret G(0). (b) According to the model, what should have been the GDP in 2007? In 2010? (According to the US Department of Commerce, the 2007 GDP was $14, 369.1 billion and the 2010 GDP was $14, 657.8 billion.) 22. The diameter D of a tumor, in millimeters, t days after it is detected is given by: D(t) = 15e0.0277t (a) What was the diameter of the tumor when it was originally detected? (b) How long until the diameter of the tumor doubles? 23. Under optimal conditions, the growth of a certain strain of E. Coli is modeled by the Law of Uninhibited Growth N (t) = N0ekt where N0 is the initial number of bacteria and t is the elapsed time, measured in minutes. From numerous experiments, it has been determined that the doubling time of this organism is 20 minutes. Suppose 1000 bacteria are present initially. (a) Find the growth constant k. Round your answer to four decimal places. (b) Find a function which gives the number of bacteria N (t) after t minutes. (c) How long until there are 9000 bacteria? Round your answer to the nearest minute. 24. Yeast is often used in biological experiments. A research technician estimates that a sample of yeast suspension contains 2.5 million organisms per cubic centimeter (cc). Two hours later, she estimates the population density to
be 6 million organisms per cc. Let t be the time elapsed since the ﬁrst observation, measured in hours. Assume that the yeast growth follows the Law of Uninhibited Growth N (t) = N0ekt. (a) Find the growth constant k. Round your answer to four decimal places. (b) Find a function which gives the number of yeast (in millions) per cc N (t) after t hours. (c) What is the doubling time for this strain of yeast? 6.5 Applications of Exponential and Logarithmic Functions 485 25. The Law of Uninhibited Growth also applies to situations where an animal is re-introduced into a suitable environment. Such a case is the reintroduction of wolves to Yellowstone National Park. According to the National Park Service, the wolf population in Yellowstone National Park was 52 in 1996 and 118 in 1999. Using these data, ﬁnd a function of the form N (t) = N0ekt which models the number of wolves t years after 1996. (Use t = 0 to represent the year 1996. Also, round your value of k to four decimal places.) According to the model, how many wolves were in Yellowstone in 2002? (The recorded number is 272.) 26. During the early years of a community, it is not uncommon for the population to grow according to the Law of Uninhibited Growth. According to the Painesville Wikipedia entry, in 1860, the Village of Painesville had a population of 2649. In 1920, the population was 7272. Use these two data points to ﬁt a model of the form N (t) = N0ekt were N (t) is the number of Painesville Residents t years after 1860. (Use t = 0 to represent the year 1860. Also, round the value of k to four decimal places.) According to this model, what was the population of Painesville in 2010? (The 2010 census gave the population as 19,563) What could be some causes for such a vast discrepancy? For more on this, see Exercise 37. 27. The population of Sasquatch in Bigfoot county is modeled by P (t) = 120 1 + 3.167e−0.05t where P (t) is the population of Sasquatch t years after 2010. (a) Find and interpret P (0). (b) Find the population of Sasquatch in Bigfoot county in 2013. Round your answer
to the nearest Sasquatch. (c) When will the population of Sasquatch in Bigfoot county reach 60? Round your answer to the nearest year. (d) Find and interpret the end behavior of the graph of y = P (t). Check your answer using a graphing utility. 28. The half-life of the radioactive isotope Carbon-14 is about 5730 years. (a) Use Equation 6.5 to express the amount of Carbon-14 left from an initial N milligrams as a function of time t in years. (b) What percentage of the original amount of Carbon-14 is left after 20,000 years? (c) If an old wooden tool is found in a cave and the amount of Carbon-14 present in it is estimated to be only 42% of the original amount, approximately how old is the tool? (d) Radiocarbon dating is not as easy as these exercises might lead you to believe. With the help of your classmates, research radiocarbon dating and discuss why our model is somewhat over-simpliﬁed. 486 Exponential and Logarithmic Functions 29. Carbon-14 cannot be used to date inorganic material such as rocks, but there are many other methods of radiometric dating which estimate the age of rocks. One of them, RubidiumStrontium dating, uses Rubidium-87 which decays to Strontium-87 with a half-life of 50 billion years. Use Equation 6.5 to express the amount of Rubidium-87 left from an initial 2.3 micrograms as a function of time t in billions of years. Research this and other radiometric techniques and discuss the margins of error for various methods with your classmates. 30. Use Equation 6.5 to show that k = − ln(2) h where h is the half-life of the radioactive isotope. 31. A pork roast22 was taken out of a hardwood smoker when its internal temperature had reached 180◦F and it was allowed to rest in a 75◦F house for 20 minutes after which its internal temperature had dropped to 170◦F. Assuming that the temperature of the roast follows Newton’s Law of Cooling (Equation 6.6), (a) Express the temperature T (in ◦F) as a function of time t (in minutes). (b) Find the time at which the roast would have dropped to 140◦
F had it not been carved and eaten. 32. In reference to Exercise 44 in Section 5.3, if Fritzy the Fox’s speed is the same as Chewbacca the Bunny’s speed, Fritzy’s pursuit curve is given by y(x) = x2 − 1 4 1 4 ln(x) − 1 4 Use your calculator to graph this path for x > 0. Describe the behavior of y as x → 0+ and interpret this physically. 33. The current i measured in amps in a certain electronic circuit with a constant impressed voltage of 120 volts is given by i(t) = 2 − 2e−10t where t ≥ 0 is the number of seconds after the circuit is switched on. Determine the value of i as t → ∞. (This is called the steady state current.) 34. If the voltage in the circuit in Exercise 33 above is switched oﬀ after 30 seconds, the current is given by the piecewise-deﬁned function i(t) = 2 − 2e−10t 2 − 2e−300 e−10t+300 if 0 ≤ t < 30 if t ≥ 30 With the help of your calculator, graph y = i(t) and discuss with your classmates the physical signiﬁcance of the two parts of the graph 0 ≤ t < 30 and t ≥ 30. 22This roast was enjoyed by Jeﬀ and his family on June 10, 2009. This is real data, folks! 6.5 Applications of Exponential and Logarithmic Functions 487 35. In Exercise 26 in Section 2.3, we stated that the cable of a suspension bridge formed a parabola but that a free hanging cable did not. A free hanging cable forms a catenary and its basic shape is given by y = 1 2 (ex + e−x). Use your calculator to graph this function. What are its domain and range? What is its end behavior? Is it invertible? How do you think it is related to the function given in Exercise 47 in Section 6.3 and the one given in the answer to Exercise 38 in Section 6.4? When ﬂipped upside down, the catenary makes an arch. The Gateway Arch in St. Louis, Missouri has the shape y = 757.7 − 127.7 2 e x 127.7 + e− x 127.7 where x and y
27511 (a) Use your calculator to ﬁt a logistic model to these data, using x = 0 to represent the year 1860. (b) Graph these data and your logistic function on your calculator to judge the reasonable- ness of the ﬁt. (c) Use this model to estimate the population of Lake County in 2010. (The 2010 census gave the population to be 230,041.) (d) According to your model, what is the population limit of Lake County, Ohio? 39. According to facebook, the number of active users of facebook has grown signiﬁcantly since its initial launch from a Harvard dorm room in February 2004. The chart below has the approximate number U (x) of active users, in millions, x months after February 2004. For example, the ﬁrst entry (10, 1) means that there were 1 million active users in December 2004 and the last entry (77, 500) means that there were 500 million active users in July 2010. Month x Active Users in Millions U (x) 10 22 34 38 44 54 59 60 62 65 67 70 72 77 1 5.5 12 20 50 100 150 175 200 250 300 350 400 500 With the help of your classmates, ﬁnd a model for this data. 40. Each Monday during the registration period before the Fall Semester at LCCC, the Enrollment Planning Council gets a report prepared by the data analysts in Institutional Eﬀectiveness and Planning.23 While the ongoing enrollment data is analyzed in many diﬀerent ways, we shall focus only on the overall headcount. Below is a chart of the enrollment data for Fall Semester 2008. It starts 21 weeks before “Opening Day” and ends on “Day 15” of the semester, but we have relabeled the top row to be x = 1 through x = 24 so that the math is easier. (Thus, x = 22 is Opening Day.) Week x Total Headcount Week x Total Headcount 1 2 3 4 5 6 7 8 1194 1564 2001 2475 2802 3141 3527 3790 9 10 11 12 13 14 15 16 4065 4371 4611 4945 5300 5657 6056 6478 23The authors thank Dr. Wendy Marley and her staﬀ for this data and Dr. Marcia Ballinger for the permission to use it in this problem. 6.5 Applications of
Exponential and Logarithmic Functions 489 Week x Total Headcount 17 18 19 20 21 22 23 24 7161 7772 8505 9256 10201 10743 11102 11181 With the help of your classmates, ﬁnd a model for this data. Unlike most of the phenomena we have studied in this section, there is no single diﬀerential equation which governs the enrollment growth. Thus there is no scientiﬁc reason to rely on a logistic function even though the data plot may lead us to that model. What are some factors which inﬂuence enrollment at a community college and how can you take those into account mathematically? 41. When we wrote this exercise, the Enrollment Planning Report for Fall Semester 2009 had only 10 data points for the ﬁrst 10 weeks of the registration period. Those numbers are given below. Week x Total Headcount 1 2 3 4 5 6 7 8 9 10 1380 2000 2639 3153 3499 3831 4283 4742 5123 5398 With the help of your classmates, ﬁnd a model for this data and make a prediction for the Opening Day enrollment as well as the Day 15 enrollment. (WARNING: The registration period for 2009 was one week shorter than it was in 2008 so Opening Day would be x = 21 and Day 15 is x = 23.) 490 Exponential and Logarithmic Functions 6.5.4 Answers 1. 2. 3. 4. 5. 12t A(t) = 500 1 + 0.0075 12 A(5) ≈ $519.10, A(10) ≈ $538.93, A(30) ≈ $626.12, A(35) ≈ $650.03 It will take approximately 92 years for the investment to double. The average rate of change from the end of the fourth year to the end of the ﬁfth year is approximately 3.88. This means that the investment is growing at an average rate of $3.88 per year at this point. The average rate of change from the end of the thirtyfourth year to the end of the thirty-ﬁfth year is approximately 4.85. This means that the investment is growing at an average rate of $4.85 per year at this point. A(t) = 500e0.0075t A(5) ≈ $519.11, A(10)
≈ $538.94, A(30) ≈ $626.16, A(35) ≈ $650.09 It will take approximately 92 years for the investment to double. The average rate of change from the end of the fourth year to the end of the ﬁfth year is approximately 3.88. This means that the investment is growing at an average rate of $3.88 per year at this point. The average rate of change from the end of the thirtyfourth year to the end of the thirty-ﬁfth year is approximately 4.86. This means that the investment is growing at an average rate of $4.86 per year at this point. 12t A(t) = 1000 1 + 0.0125 12 A(5) ≈ $1064.46, A(10) ≈ $1133.07, A(30) ≈ $1454.71, A(35) ≈ $1548.48 It will take approximately 55 years for the investment to double. The average rate of change from the end of the fourth year to the end of the ﬁfth year is approximately 13.22. This means that the investment is growing at an average rate of $13.22 per year at this point. The average rate of change from the end of the thirtyfourth year to the end of the thirty-ﬁfth year is approximately 19.23. This means that the investment is growing at an average rate of $19.23 per year at this point. A(t) = 1000e0.0125t A(5) ≈ $1064.49, A(10) ≈ $1133.15, A(30) ≈ $1454.99, A(35) ≈ $1548.83 It will take approximately 55 years for the investment to double. The average rate of change from the end of the fourth year to the end of the ﬁfth year is approximately 13.22. This means that the investment is growing at an average rate of $13.22 per year at this point. The average rate of change from the end of the thirtyfourth year to the end of the thirty-ﬁfth year is approximately 19.24. This means that the investment is growing at an average rate of $19.24 per year at this point. A(t) = 5000 1 + 0.02
125 A(5) ≈ $5559.98, A(10) ≈ $6182.67, A(30) ≈ $9453.40, A(35) ≈ $10512.13 It will take approximately 33 years for the investment to double. 12t 12 6.5 Applications of Exponential and Logarithmic Functions 491 The average rate of change from the end of the fourth year to the end of the ﬁfth year is approximately 116.80. This means that the investment is growing at an average rate of $116.80 per year at this point. The average rate of change from the end of the thirtyfourth year to the end of the thirty-ﬁfth year is approximately 220.83. This means that the investment is growing at an average rate of $220.83 per year at this point. 6. A(t) = 5000e0.02125t A(5) ≈ $5560.50, A(10) ≈ $6183.83, A(30) ≈ $9458.73, A(35) ≈ $10519.05 It will take approximately 33 years for the investment to double. The average rate of change from the end of the fourth year to the end of the ﬁfth year is approximately 116.91. This means that the investment is growing at an average rate of $116.91 per year at this point. The average rate of change from the end of the thirtyfourth year to the end of the thirty-ﬁfth year is approximately 221.17. This means that the investment is growing at an average rate of $221.17 per year at this point. 8. P = 2000 e0.0025·3 ≈ $1985.06 9. P = 5000 (1+ 0.0225 12 )12·10 ≈ $3993.42 (a) A(8) = 2000 1 + 0.0025 12 10. 12·8 ≈ $2040.40 (b) t = (c) P = ln(2) 12 ln 1 + 0.0025 12 2000 1 + 0.0025 12 36 ≈ $1985.06 ≈ 277.29 years 11. (a) A(8) = 2000 1 + 0.0225 12 12·8 ≈ $2394.03 (b) t = (
c) P = ≈ 30.83 years ln(2) 12 ln 1 + 0.0225 12 2000 1 + 0.0225 12 12 ≈ 1.0227 so the APY is 2.27% 36 ≈ $1869.57 (d) 1 + 0.0225 12 12. A(3) = 5000e0.299·3 ≈ $12, 226.18, A(6) = 5000e0.299·6 ≈ $30, 067.29 14. 5.27 ≈ −0.1315 k = ln(1/2) A(t) = 50e−0.1315t t = ln(0.1) −0.1315 ≈ 17.51 years. 15. 14 ≈ −0.0495 k = ln(1/2) A(t) = 2e−0.0495t t = ln(0.1) −0.0495 ≈ 46.52 days. 492 16. 18. 27.7 ≈ −0.0250 k = ln(1/2) A(t) = 75e−0.0250t t = ln(0.1) −0.025 ≈ 92.10 days. Exponential and Logarithmic Functions 17. 432.7 ≈ −0.0016 k = ln(1/2) A(t) = 0.29e−0.0016t t = ln(0.1) −0.0016 ≈ 1439.11 years. 704 ≈ −0.0010 k = ln(1/2) A(t) = e−0.0010t t = ln(0.1) −0.0010 ≈ 2302.58 million years, or 2.30 billion years. 19. t = ln(0.1) k = − ln(10) k 20. V (t) = 25eln( 4 5 )t ≈ 25e−0.22314355t 21. (a) G(0) = 9743.77 This means that the GDP of the US in 2000 was $9743.77 billion dollars. (b) G(7) = 13963.24 and G(10) = 16291.25, so the model predicted a GDP of $13
, 963.24 billion in 2007 and $16, 291.25 billion in 2010. (a) D(0) = 15, so the tumor was 15 millimeters in diameter when it was ﬁrst detected. (b) t = ln(2) 0.0277 ≈ 25 days. 22. 23. (a) k = ln(2) 20 ≈ 0.0346 (b) N (t) = 1000e0.0346t (c) t = ln(9) 0.0346 ≈ 63 minutes 3 ln 118 52 24. ln(6) 2.5 ≈ 0.4377 (a) k = 1 2 (b) N (t) = 2.5e0.4377t (c) t = ln(2) 0.4377 ≈ 1.58 hours 25. N0 = 52, k = 1 ≈ 0.2731, N (t) = 52e0.2731t. N (6) ≈ 268. 26. N0 = 2649, k = 1 60 ln 7272 2649 ≈ 0.0168, N (t) = 2649e0.0168t. N (150) ≈ 32923, so the population of Painesville in 2010 based on this model would have been 32,923. 27. (a) P (0) = 120 (b) P (3) = 4.167 ≈ 29. There are 29 Sasquatch in Bigfoot County in 2010. 1+3.167e−0.05(3) ≈ 32 Sasquatch. 120 (c) t = 20 ln(3.167) ≈ 23 years. (d) As t → ∞, P (t) → 120. As time goes by, the Sasquatch Population in Bigfoot County will approach 120. Graphically, y = P (x) has a horizontal asymptote y = 120. 28. (a) A(t) = N e ln(2) 5730 t − ≈ N e−0.00012097t (b) A(20000) ≈ 0.088978 · N so about 8.9% remains (c) t ≈ ln(.42) −0.00012097 ≈ 7171 years old 29. A(t) = 2.
3e−0.0138629t 6.5 Applications of Exponential and Logarithmic Functions 493 31. (a) T (t) = 75 + 105e−0.005005t (b) The roast would have cooled to 140◦F in about 95 minutes. 32. From the graph, it appears that as x → 0+, y → ∞. This is due to the presence of the ln(x) term in the function. This means that Fritzy will never catch Chewbacca, which makes sense since Chewbacca has a head start and Fritzy only runs as fast as he does. y(x) = 1 4 x2 − 1 4 ln(x) − 1 4 33. The steady state current is 2 amps. 36. The linear regression on the data below is y = 1.74899x + 0.70739 with r2 ≈ 0.999995. This is an excellent ﬁt. 1 2.4849 2 4.1897 x ln(N (x)) N (x) = 2.02869(5.74879)x = 2.02869e1.74899x with r2 ≈ 0.999995. This is also an excellent ﬁt and corresponds to our linearized model because ln(2.02869) ≈ 0.70739. 8 14.7006 7 12.9497 10 18.2025 9 16.4523 6 11.1988 3 5.9454 4 7.6967 5 9.4478 37. (a) The calculator gives: y = 2895.06(1.0147)x. Graphing this along with our answer from Exercise 26 over the interval [0, 60] shows that they are pretty close. From this model, y(150) ≈ 25840 which once again overshoots the actual data value. (b) P (150) ≈ 18717, so this model predicts 17,914 people in Painesville in 2010, a more conservative number than was recorded in the 2010 census. As t → ∞, P (t) → 18691. So the limiting population of Painesville based on this model is 18,691 people. 38. (a) y = 242526 1 + 874.62e−0.07113x, where x is the number of years since 1860. (b
) The plot of the data and the curve is below. (c) y(140) ≈ 232889, so this model predicts 232,889 people in Lake County in 2010. (d) As x → ∞, y → 242526, so the limiting population of Lake County based on this model is 242,526 people. 494 Exponential and Logarithmic Functions Chapter 7 Hooked on Conics 7.1 Introduction to Conics In this chapter, we study the Conic Sections - literally ‘sections of a cone’. Imagine a doublenapped cone as seen below being ‘sliced’ by a plane. If we slice the cone with a horizontal plane the resulting curve is a circle. 496 Hooked on Conics Tilting the plane ever so slightly produces an ellipse. If the plane cuts parallel to the cone, we get a parabola. If we slice the cone with a vertical plane, we get a hyperbola. For a wonderful animation describing the conics as intersections of planes and cones, see Dr. Louis Talman’s Mathematics Animated Website. 7.1 Introduction to Conics 497 If the slicing plane contains the vertex of the cone, we get the so-called ‘degenerate’ conics: a point, a line, or two intersecting lines. We will focus the discussion on the non-degenerate cases: circles, parabolas, ellipses, and hyperbolas, in that order. To determine equations which describe these curves, we will make use of their deﬁnitions in terms of distances. 498 7.2 Circles Hooked on Conics Recall from Geometry that a circle can be determined by ﬁxing a point (called the center) and a positive number (called the radius) as follows. Deﬁnition 7.1. A circle with center (h, k) and radius r > 0 is the set of all points (x, y) in the plane whose distance to (h, k) is r. (x, y) r (h, k) From the picture, we see that a point (x, y) is on the circle if and only if its distance to (h, k) is r. We express this relationship algebraically using the Distance Formula, Equation 1.1, as r = (x − h)2 + (y −
k)2 By squaring both sides of this equation, we get an equivalent equation (since r > 0) which gives us the standard equation of a circle. Equation 7.1. The Standard Equation of a Circle: The equation of a circle with center (h, k) and radius r > 0 is (x − h)2 + (y − k)2 = r2. Example 7.2.1. Write the standard equation of the circle with center (−2, 3) and radius 5. Solution. Here, (h, k) = (−2, 3) and r = 5, so we get (x − (−2))2 + (y − 3)2 = (5)2 (x + 2)2 + (y − 3)2 = 25 Example 7.2.2. Graph (x + 2)2 + (y − 1)2 = 4. Find the center and radius. Solution. From the standard form of a circle, Equation 7.1, we have that x + 2 is x − h, so h = −2 and y − 1 is y − k so k = 1. This tells us that our center is (−2, 1). Furthermore, r2 = 4, so r = 2. Thus we have a circle centered at (−2, 1) with a radius of 2. Graphing gives us 7.2 Circles 499 y 4 3 2 1 −4 −3 −2 −1 1 x −1 If we were to expand the equation in the previous example and gather up like terms, instead of the easily recognizable (x + 2)2 + (y − 1)2 = 4, we’d be contending with x2 + 4x + y2 − 2y + 1 = 0. If we’re given such an equation, we can complete the square in each of the variables to see if it ﬁts the form given in Equation 7.1 by following the steps given below. To Write the Equation of a Circle in Standard Form 1. Group the same variables together on one side of the equation and position the constant on the other side. 2. Complete the square on both variables as needed. 3. Divide both sides by the coeﬃcient of the squares. (For circles, they will be the same.) Example 7.2.3. Complete the square to ﬁnd the center and radius of 3x2 − 6
x + 3y2 + 4y − 4 = 0. Solution. 3x2 − 6x + 3y2 + 4y − 4 = 0 3x2 − 6x + 3y2 + 4y = 4 3 x2 − 2x + 3 y2 + 4 3 y = 4 add 4 to both sides factor out leading coeﬃcients 3 x2 − 2x + 1 + 3 y2 + (x − 1)2 + 3 y + (x − 1)(1) + 3 4 9 complete the square in x, y = = 25 3 25 9 factor divide both sides by 3 From Equation 7.1, we identify x − 1 as x − h, so h = 1, and y + 2 center is (h, k) = 1, − 2 3. Furthermore, we see that r2 = 25 9 so the radius is r = 5 3. 3 as y − k, so k = − 2 3. Hence, the 500 Hooked on Conics It is possible to obtain equations like (x − 3)2 + (y + 1)2 = 0 or (x − 3)2 + (y + 1)2 = −1, neither of which describes a circle. (Do you see why not?) The reader is encouraged to think about what, if any, points lie on the graphs of these two equations. The next example uses the Midpoint Formula, Equation 1.2, in conjunction with the ideas presented so far in this section. Example 7.2.4. Write the standard equation of the circle which has (−1, 3) and (2, 4) as the endpoints of a diameter. Solution. We recall that a diameter of a circle is a line segment containing the center and two points on the circle. Plotting the given data yields y r (h, k) 4 3 2 1 −2 −1 1 2 3 x Since the given points are endpoints of a diameter, we know their midpoint (h, k) is the center of the circle. Equation 1.2 gives us (h, k) = = =,, y1 + y2 2 3 + 4 2 x1 + x2 2 −1 + 2 2 7 2 1 2, The diameter of the circle is the distance between the given points, so we know that half of the distance is the radius. Thus, (x2 − x1)2 + (y2 − y1
)2 (2 − (−1))2 + (4 − 3)2 32 + 12 √ 10 2 Finally, since 2 √ 10 2 = 10 4, our answer becomes = 10 4 7.2 Circles 501 We close this section with the most important1 circle in all of mathematics: the Unit Circle. Deﬁnition 7.2. The Unit Circle is the circle centered at (0, 0) with a radius of 1. The standard equation of the Unit Circle is x2 + y2 = 1. Example 7.2.5. Find the points on the unit circle with y-coordinate Solution. We replace y with √ 3 2 in the equation x2 + y2 = 1 to get √ 3 2. x2 + y2 = 1 √ 2 = 1 x2 + 3 2 3 4 + x2 = 1 1 4 x2 = x = ± 1 4 1 2 Our ﬁnal answers are √ 3 2 1 2, and −. √ 3 2 1 2, x = ± 1While this may seem like an opinion, it is indeed a fact. See Chapters 10 and 11 for details. 502 7.2.1 Exercises Hooked on Conics In Exercises 1 - 6, ﬁnd the standard equation of the circle and then graph it. 1. Center (−1, −5), radius 10 2. Center (4, −2), radius 3 3. Center −3, 7, radius 1 13 2 √ 2, radius π 5. Center −e, 4. Center (5, −9), radius ln(8) √ 6. Center (π, e2), radius 3 91 In Exercises 7 - 12, complete the square in order to put the equation into standard form. Identify the center and the radius or explain why the equation does not represent a circle. 7. x2 − 4x + y2 + 10y = −25 8. −2x2 − 36x − 2y2 − 112 = 0 9. x2 + y2 + 8x − 10y − 1 = 0 10. x2 + y2 + 5x − y − 1 = 0 11. 4x2 + 4y2 − 24y + 36 = 0 12. x2 + x + y2 − 6 5 y = 1 In Exercises 13 - 16, ﬁnd the standard equation of the circle
which satisﬁes the given criteria. 13. center (3, 5), passes through (−1, −2) 14. center (3, 6), passes through (−1, 4) 15. endpoints of a diameter: (3, 6) and (−1, 4) 16. endpoints of a diameter: 1 2, 4, 3 2, −1 17. The Giant Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height is 136 feet.2 Find an equation for the wheel assuming that its center lies on the y-axis and that the ground is the x-axis. 18. Verify that the following points lie on the Unit Circle: (±1, 0), (0, ±1), ± and √ 19. Discuss with your classmates how to obtain the standard equation of a circle, Equation 7.1, from the equation of the Unit Circle, x2 + y2 = 1 using the transformations discussed in Section 1.7. (Thus every circle is just a few transformations away from the Unit Circle.) 20. Find an equation for the function represented graphically by the top half of the Unit Circle. Explain how the transformations is Section 1.7 can be used to produce a function whose graph is either the top or bottom of an arbitrary circle. 21. Find a one-to-one function whose graph is half of a circle. (Hint: Think piecewise.) 2Source: Cedar Point’s webpage. 7.2 Circles 7.2.2 Answers 503 1. (x + 1)2 + (y + 5)2 = 100 2. (x − 4)2 + (y + 2)2 = 9 y 5 −11 −1 x 9 −5 −15 3. (x + 3)2 + y − 7 13 2 −5 4. (x − 5)2 + (y + 9)2 = (ln(8))2 y x 5 − ln(8) 5 5 + ln(8) y 27 26 7 13 1 26 x −9 + ln(8) −9 −9 − ln(8) = π2 6. (x − π)2 + y − e22 y = 91 2 3 22 y √ 2 + π − 7 2 −3 − 5 2 5. (x + e)2 + y − √ √ 2 x −e −
π −e −e + π √ 2 − π e2 + 3√ 91 e2 e2 − 3√ 91 π − 3√ 91 π x π + 3√ 91 504 Hooked on Conics 7. (x − 2)2 + (y + 5)2 = 4 Center (2, −5), radius r = 2 9. (x + 4)2 + (y − 5)2 = 42 Center (−4, 5), radius r = √ 42 11. x2 + (y − 3)2 = 0 This is not a circle. 8. (x + 9)2 + y2 = 25 Center (−9, 0), radius r = 5 2 10. x + 5 2 Center − 5 2 = 30 4, radius r = 2 12. x + 1 2 Center − 1 2 = 161 100, radius √ 30 2 √ 161 10 13. (x − 3)2 + (y − 5)2 = 65 14. (x − 3)2 + (y − 6)2 = 20 15. (x − 1)2 + (y − 5)2 = 5 17. x2 + (y − 72)2 = 4096 16. (x − 1)2 + y − 3 2 2 = 13 2 7.3 Parabolas 7.3 Parabolas 505 We have already learned that the graph of a quadratic function f (x) = ax2 + bx + c (a = 0) is called a parabola. To our surprise and delight, we may also deﬁne parabolas in terms of distance. Deﬁnition 7.3. Let F be a point in the plane and D be a line not containing F. A parabola is the set of all points equidistant from F and D. The point F is called the focus of the parabola and the line D is called the directrix of the parabola. Schematically, we have the following. F V D Each dashed line from the point F to a point on the curve has the same length as the dashed line from the point on the curve to the line D. The point suggestively labeled V is, as you should expect, the vertex. The vertex is the point on the parabola closest to the focus. We want to use only the distance deﬁnition of parabola
to derive the equation of a parabola and, if all is right with the universe, we should get an expression much like those studied in Section 2.3. Let p denote the directed1 distance from the vertex to the focus, which by deﬁnition is the same as the distance from the vertex to the directrix. For simplicity, assume that the vertex is (0, 0) and that the parabola opens upwards. Hence, the focus is (0, p) and the directrix is the line y = −p. Our picture becomes y (0, p) (0, 0) (x, y) x y = −p (x, −p) From the deﬁnition of parabola, we know the distance from (0, p) to (x, y) is the same as the distance from (x, −p) to (x, y). Using the Distance Formula, Equation 1.1, we get 1We’ll talk more about what ‘directed’ means later. 506 Hooked on Conics (x − 0)2 + (y − p)2 = x2 + (y − p)2 = x2 + (y − p)2 = (y + p)2 x2 + y2 − 2py + p2 = y2 + 2py + p2 (x − x)2 + (y − (−p))2 (y + p)2 x2 = 4py square both sides expand quantities gather like terms Solving for y yields y = x2 a = 1 4p and vertex (0, 0). 4p, which is a quadratic function of the form found in Equation 2.4 with We know from previous experience that if the coeﬃcient of x2 is negative, the parabola opens downwards. In the equation y = x2 4p this happens when p < 0. In our formulation, we say that p is a ‘directed distance’ from the vertex to the focus: if p > 0, the focus is above the vertex; if p < 0, the focus is below the vertex. The focal length of a parabola is \|p\|. If we choose to place the vertex at an arbitrary point (h, k), we arrive at the following formula using either transformations from Section 1.7 or re-deriving the formula from Deﬁnition
7.3. Equation 7.2. The Standard Equation of a Verticala Parabola: The equation of a (vertical) parabola with vertex (h, k) and focal length \|p\| is (x − h)2 = 4p(y − k) If p > 0, the parabola opens upwards; if p < 0, it opens downwards. aThat is, a parabola which opens either upwards or downwards. Notice that in the standard equation of the parabola above, only one of the variables, x, is squared. This is a quick way to distinguish an equation of a parabola from that of a circle because in the equation of a circle, both variables are squared. Example 7.3.1. Graph (x + 1)2 = −8(y − 3). Find the vertex, focus, and directrix. Solution. We recognize this as the form given in Equation 7.2. Here, x − h is x + 1 so h = −1, and y − k is y − 3 so k = 3. Hence, the vertex is (−1, 3). We also see that 4p = −8 so p = −2. Since p < 0, the focus will be below the vertex and the parabola will open downwards. y 5 4 3 2 1 −6 −5 −4 −3 −2 −1 1 2 3 4 x 7.3 Parabolas 507 The distance from the vertex to the focus is \|p\| = 2, which means the focus is 2 units below the vertex. From (−1, 3), we move down 2 units and ﬁnd the focus at (−1, 1). The directrix, then, is 2 units above the vertex, so it is the line y = 5. Of all of the information requested in the previous example, only the vertex is part of the graph of the parabola. So in order to get a sense of the actual shape of the graph, we need some more information. While we could plot a few points randomly, a more useful measure of how wide a parabola opens is the length of the parabola’s latus rectum.2 The latus rectum of a parabola is the line segment parallel to the directrix which contains the focus. The endpoints of the latus rectum are, then, two points on ‘opposite’ sides of the
parabola. Graphically, we have the following. the latus rectum F V D It turns out3 that the length of the latus rectum, called the focal diameter of the parabola is \|4p\|, which, in light of Equation 7.2, is easy to ﬁnd. In our last example, for instance, when graphing (x + 1)2 = −8(y − 3), we can use the fact that the focal diameter is \| − 8\| = 8, which means the parabola is 8 units wide at the focus, to help generate a more accurate graph by plotting points 4 units to the left and right of the focus. Example 7.3.2. Find the standard form of the parabola with focus (2, 1) and directrix y = −4. Solution. Sketching the data yields, y 1 −1 1 2 3 x −1 −2 −3 The vertex lies on this vertical line midway between the focus and the directrix 2No, I’m not making this up. 3Consider this an exercise to show what follows. 508 Hooked on Conics From the diagram, we see the parabola opens upwards. (Take a moment to think about it if you don’t see that immediately.) Hence, the vertex lies below the focus and has an x-coordinate of 2. To ﬁnd the y-coordinate, we note that the distance from the focus to the directrix is 1 − (−4) = 5, which means the vertex lies 5 2 units (halfway) below the focus. Starting at (2, 1) and moving down 5/2 units leaves us at (2, −3/2), which is our vertex. Since the parabola opens upwards, we know p is positive. Thus p = 5/2. Plugging all of this data into Equation 7.2 give us (x − 2)2 = 4 (x − 2)2 = 10 If we interchange the roles of x and y, we can produce ‘horizontal’ parabolas: parabolas which open to the left or to the right. The directrices4 of such animals would be vertical lines and the focus would either lie to the left or to the right of the vertex, as seen below. D V F Equation 7.3. The Standard Equation of a Horizontal Parabola: The equation
of a (horizontal) parabola with vertex (h, k) and focal length \|p\| is If p > 0, the parabola opens to the right; if p < 0, it opens to the left. (y − k)2 = 4p(x − h) 4plural of ‘directrix’ 7.3 Parabolas 509 Example 7.3.3. Graph (y − 2)2 = 12(x + 1). Find the vertex, focus, and directrix. Solution. We recognize this as the form given in Equation 7.3. Here, x − h is x + 1 so h = −1, and y − k is y − 2 so k = 2. Hence, the vertex is (−1, 2). We also see that 4p = 12 so p = 3. Since p > 0, the focus will be the right of the vertex and the parabola will open to the right. The distance from the vertex to the focus is \|p\| = 3, which means the focus is 3 units to the right. If we start at (−1, 2) and move right 3 units, we arrive at the focus (2, 2). The directrix, then, is 3 units to the left of the vertex and if we move left 3 units from (−1, 2), we’d be on the vertical line x = −4. Since the focal diameter is \|4p\| = 12, the parabola is 12 units wide at the focus, and thus there are points 6 units above and below the focus on the parabola5 −4 −3 −2 −1 −1 1 2 3 x −2 −3 −4 If we As with circles, not all parabolas will come to us in the forms in Equations 7.2 or 7.3. encounter an equation with two variables in which exactly one variable is squared, we can attempt to put the equation into a standard form using the following steps. To Write the Equation of a Parabola in Standard Form 1. Group the variable which is squared on one side of the equation and position the non- squared variable and the constant on the other side. 2. Complete the square if necessary and divide by the coeﬃcient of the perfect square. 3. Factor out the coeﬃcient of the non-squared variable from it and the constant. Example 7.3.4
. Consider the equation y2 + 4y + 8x = 4. Put this equation into standard form and graph the parabola. Find the vertex, focus, and directrix. Solution. We need a perfect square (in this case, using y) on the left-hand side of the equation and factor out the coeﬃcient of the non-squared variable (in this case, the x) on the other. 510 Hooked on Conics y2 + 4y + 8x = 4 y2 + 4y = −8x + 4 y2 + 4y + 4 = −8x + 4 + 4 complete the square in y only factor (y + 2)2 = −8x + 8 (y + 2)2 = −8(x − 1) Now that the equation is in the form given in Equation 7.3, we see that x − h is x − 1 so h = 1, and y − k is y + 2 so k = −2. Hence, the vertex is (1, −2). We also see that 4p = −8 so that p = −2. Since p < 0, the focus will be the left of the vertex and the parabola will open to the left. The distance from the vertex to the focus is \|p\| = 2, which means the focus is 2 units to the left of 1, so if we start at (1, −2) and move left 2 units, we arrive at the focus (−1, −2). The directrix, then, is 2 units to the right of the vertex, so if we move right 2 units from (1, −2), we’d be on the vertical line x = 3. Since the focal diameter is \|4p\| is 8, the parabola is 8 units wide at the focus, so there are points 4 units above and below the focus on the parabola. y 2 1 −2 −1 1 2 x −1 −2 −3 −4 −5 −6 In studying quadratic functions, we have seen parabolas used to model physical phenomena such as the trajectories of projectiles. Other applications of the parabola concern its ‘reﬂective property’ which necessitates knowing about the focus of a parabola. For example, many satellite dishes are formed in the shape of a paraboloid of revolution as depicted below. 7.3 Par
abolas 511 Every cross section through the vertex of the paraboloid is a parabola with the same focus. To see why this is important, imagine the dashed lines below as electromagnetic waves heading towards a parabolic dish. It turns out that the waves reﬂect oﬀ the parabola and concentrate at the focus which then becomes the optimal place for the receiver. If, on the other hand, we imagine the dashed lines as emanating from the focus, we see that the waves are reﬂected oﬀ the parabola in a coherent fashion as in the case in a ﬂashlight. Here, the bulb is placed at the focus and the light rays are reﬂected oﬀ a parabolic mirror to give directional light. F Example 7.3.5. A satellite dish is to be constructed in the shape of a paraboloid of revolution. If the receiver placed at the focus is located 2 ft above the vertex of the dish, and the dish is to be 12 feet wide, how deep will the dish be? Solution. One way to approach this problem is to determine the equation of the parabola suggested to us by this data. For simplicity, we’ll assume the vertex is (0, 0) and the parabola opens upwards. Our standard form for such a parabola is x2 = 4py. Since the focus is 2 units above the vertex, we know p = 2, so we have x2 = 8y. Visually, y 12 units wide 2 (6, y)? 6 x −6 Since the parabola is 12 feet wide, we know the edge is 6 feet from the vertex. To ﬁnd the depth, we are looking for the y value when x = 6. Substituting x = 6 into the equation of the parabola yields 62 = 8y or y = 36 2 = 4.5. Hence, the dish will be 4.5 feet deep. 8 = 9 512 7.3.1 Exercises Hooked on Conics In Exercises 1 - 8, sketch the graph of the given parabola. Find the vertex, focus and directrix. Include the endpoints of the latus rectum in your sketch. 1. (x − 3)2 = −16y 3. (y − 2)2 = −12(x + 3) 5. (x − 1)2
= 4(y + 3) 7. (y − 4)2 = 18(x − 2) 2. (y + 4)2 = 4x 6. (x + 2)2 = −20(y − 5) 8. y + 3 2 2 = −7 x + 9 2 In Exercises 9 - 14, put the equation into standard form and identify the vertex, focus and directrix. 9. y2 − 10y − 27x + 133 = 0 10. 25x2 + 20x + 5y − 1 = 0 11. x2 + 2x − 8y + 49 = 0 12. 2y2 + 4y + x − 8 = 0 13. x2 − 10x + 12y + 1 = 0 14. 3y2 − 27y + 4x + 211 4 = 0 In Exercises 15 - 18, ﬁnd an equation for the parabola which ﬁts the given criteria. 15. Vertex (7, 0), focus (0, 0) 16. Focus (10, 1), directrix x = 5 17. Vertex (−8, −9); (0, 0) and (−16, 0) are 18. The endpoints of latus rectum are (−2, −7) points on the curve and (4, −7) 19. The mirror in Carl’s ﬂashlight is a paraboloid of revolution. If the mirror is 5 centimeters in diameter and 2.5 centimeters deep, where should the light bulb be placed so it is at the focus of the mirror? 20. A parabolic Wi-Fi antenna is constructed by taking a ﬂat sheet of metal and bending it into a parabolic shape.5 If the cross section of the antenna is a parabola which is 45 centimeters wide and 25 centimeters deep, where should the receiver be placed to maximize reception? 21. A parabolic arch is constructed which is 6 feet wide at the base and 9 feet tall in the middle. Find the height of the arch exactly 1 foot in from the base of the arch. 22. A popular novelty item is the ‘mirage bowl.’ Follow this link to see another startling appli- cation of the reﬂective property of the parabola. 23. With the help of your classmates, research spinning liquid mirrors. To get you started, check out this website. 5This shape is called a �
�parabolic cylinder.’ 7.3 Parabolas 7.3.2 Answers 513 1. (x − 3)2 = −16y Vertex (3, 0) Focus (3, −4) Directrix y = 4 Endpoints of latus rectum (−5, −4), (11, −4) y 4 3 2 1 −5−4−3−2−1 − 10 11 x −2 −3 −4 2 Vertex − 7 3, − 5 Focus − 7 3, −2 Directrix y = −3 Endpoints of latus rectum − 10 2 3, −2, − 4 3, −2 3. (y − 2)2 = −12(x + 3) Vertex (−3, 2) Focus (−6, 2) Directrix x = 0 Endpoints of latus rectum (−6, 8), (−6, −4) y 2 1 −5 −4 −3 −2 −1 x −1 −2 −7−6−5−4−3−2−1 −1 x −2 −3 −4 514 Hooked on Conics 4. (y + 4)2 = 4x Vertex (0, −4) Focus (1, −4) Directrix x = −1 Endpoints of latus rectum (1, −2), (1, −6) 5. (x − 1)2 = 4(y + 3) Vertex (1, −3) Focus (1, −2) Directrix y = −4 Endpoints of latus rectum (3, −2), (−1, −2) 6. (x + 2)2 = −20(y − 5) Vertex (−2, 5) Focus (−2, 0) Directrix y = 10 Endpoints of latus rectum (−12, 0), (8, 0) 7. (y − 4)2 = 18(x − 2) Vertex (2, 4) Focus 13 2, 4 Directrix x = − 5 2 Endpoints of latus rectum 13 2, −5, 13 2, 13 y −1 1 2 3 4 x −1 −2 −3 −4 −5 −6 −7 −8 y −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 y 10 12 −10 −8 −6 −4 −2 2 4 6 8 x y 13
11 9 7 5 3 1 −1 −3 −5 7.3 Parabolas 515 8. y + 3 2 2 2 = − Vertex − 9 Focus − 25 Directrix x = − 11 4 Endpoints of latus rectum − 25 2 2 4, 2, − 25 4, −5 9. (y − 5)2 = 27(x − 4) Vertex (4, 5) 4, 5 Focus 43 Directrix x = − 11 4 11. (x + 1)2 = 8(y − 6) Vertex (−1, 6) Focus (−1, 8) Directrix y = 4 13. (x − 5)2 = −12(y − 2) Vertex (5, 2) Focus (5, −1) Directrix y = 5 15. y2 = −28(x − 7) 17. (x + 8)2 = 64 9 (y + 9) −5 −4 −3 −2 −1 y x 2 1 −1 −2 −3 −4 −5 5 (y − 1) 2 10 Vertex − 2 Focus − 2 5, 19 Directrix y = 21 20 20 12. (y + 1)2 = − 1 2 (x − 10) Vertex (10, −1) Focus 79 8, −1 Directrix x = 81 8 3 (x − 2) 14. y − 9 2 = − 4 2 Vertex 2, 9 2 Focus 5 3, 9 Directrix x = 7 3 2 16. (y − 1)2 = 10 x − 15 2 or 18. (x − 1)2 = 6 y + 17 2 (x − 1)2 = −6 y + 11 2 19. The bulb should be placed 0.625 centimeters above the vertex of the mirror. (As veriﬁed by Carl himself!) 20. The receiver should be placed 5.0625 centimeters from the vertex of the cross section of the antenna. 21. The arch can be modeled by x2 = −(y − 9) or y = 9 − x2. One foot in from the base of the arch corresponds to either x = ±2, so the height is y = 9 − (±2)2 = 5 feet. 516 7.4 Ellipses Hooked on Conics In the deﬁnition of a circle, Deﬁnition 7.1, we ﬁxed a
point called the center and considered all of the points which were a ﬁxed distance r from that one point. For our next conic section, the ellipse, we ﬁx two distinct points and a distance d to use in our deﬁnition. Deﬁnition 7.4. Given two distinct points F1 and F2 in the plane and a ﬁxed distance d, an ellipse is the set of all points (x, y) in the plane such that the sum of each of the distances from F1 and F2 to (x, y) is d. The points F1 and F2 are called the focia of the ellipse. athe plural of ‘focus’ (x, y) d1 F1 d2 F2 d1 + d2 = d for all (x, y) on the ellipse We may imagine taking a length of string and anchoring it to two points on a piece of paper. The curve traced out by taking a pencil and moving it so the string is always taut is an ellipse. The center of the ellipse is the midpoint of the line segment connecting the two foci. The major axis of the ellipse is the line segment connecting two opposite ends of the ellipse which also contains the center and foci. The minor axis of the ellipse is the line segment connecting two opposite ends of the ellipse which contains the center but is perpendicular to the major axis. The vertices of an ellipse are the points of the ellipse which lie on the major axis. Notice that the center is also the midpoint of the major axis, hence it is the midpoint of the vertices. In pictures we have, 7.4 Ellipses 517 s i x A r o n i M V1 Major Axis F1 C F2 V2 An ellipse with center C; foci F1, F2; and vertices V1, V2 Note that the major axis is the longer of the two axes through the center, and likewise, the minor axis is the shorter of the two. In order to derive the standard equation of an ellipse, we assume that the ellipse has its center at (0, 0), its major axis along the x-axis, and has foci (c, 0) and (−c, 0) and vertices
(−a, 0) and (a, 0). We will label the y-intercepts of the ellipse as (0, b) and (0, −b) (We assume a, b, and c are all positive numbers.) Schematically, y (0, b) (−a, 0) (−c, 0) (c, 0) (x, y) x (a, 0) (0, −b) Note that since (a, 0) is on the ellipse, it must satisfy the conditions of Deﬁnition 7.4. That is, the distance from (−c, 0) to (a, 0) plus the distance from (c, 0) to (a, 0) must equal the ﬁxed distance d. Since all of these points lie on the x-axis, we get distance from (−c, 0) to (a, 0) + distance from (c, 0) to (a, 0) = d (a + c) + (a − c) = d 2a = d 518 Hooked on Conics In other words, the ﬁxed distance d mentioned in the deﬁnition of the ellipse is none other than the length of the major axis. We now use that fact (0, b) is on the ellipse, along with the fact that d = 2a to get (0 − (−c))2 + (b − 0)2 + distance from (−c, 0) to (0, b) + distance from (c, 0) to (0, b) = 2a (0 − c)2 + (b − 0)2 = 2a b2 + c2 = 2a b2 + c2 + b2 + c2 = 2a 2 b2 + c2 = a √ √ √ √ From this, we get a2 = b2 + c2, or b2 = a2 − c2, which will prove useful later. Now consider a point (x, y) on the ellipse. Applying Deﬁnition 7.4, we get distance from (−c, 0) to (x, y) + distance from (c, 0) to (x, y) = 2a (x − c)2 + (y − 0)2 = 2a (x − c)2 + y
2 = 2a (x − (−c))2 + (y − 0)2 + (x + c)2 + y2 + In order to make sense of this situation, we need to make good use of Intermediate Algebra. (x + c)2 + y2 + (x + c)2 + y22 (x − c)2 + y2 = 2a (x + c)2 + y2 = 2a − 2a − (x + c)2 + y2 = 4a2 − 4a (x − c)2 + y2 = 4a2 + (x − c)2 − (x + c)2 (x − c)2 + y2 = 4a2 − 4cx (x − c)2 + y2 = a2 − cx = 4a 4a a (x − c)2 + y22 = a2 − cx2 (x − c)2 + y2 (x − c)2 + y22 (x − c)2 + y2 + (x − c)2 + y2 a a2 (x − c)2 + y2 = a4 − 2a2cx + c2x2 a2x2 − 2a2cx + a2c2 + a2y2 = a4 − 2a2cx + c2x2 a2x2 − c2x2 + a2y2 = a4 − a2c2 a2 − c2 x2 + a2y2 = a2 a2 − c2 We are nearly ﬁnished. Recall that b2 = a2 − c2 so that a2 − c2 x2 + a2y2 = a2 a2 − c2 b2x2 + a2y2 = a2b2 y2 b2 = 1 x2 a2 + 7.4 Ellipses 519 This equation is for an ellipse centered at the origin. To get the formula for the ellipse centered at (h, k), we could use the transformations from Section 1.7 or re-derive the equation using Deﬁnition 7.4 and the distance formula to obtain the formula below. Equation 7.4. The Standard Equation of an Ellipse: For positive unequal numbers a and b, the equation of an ellipse with center (h, k) is (x
− h)2 a2 + (y − k)2 b2 = 1 Some remarks about Equation 7.4 are in order. First note that the values a and b determine how far in the x and y directions, respectively, one counts from the center to arrive at points on the ellipse. Also take note that if a > b, then we have an ellipse whose major axis is horizontal, In this case, as we’ve seen in the and hence, the foci lie to the left and right of the center. √ a2 − b2. If b > a, derivation, the distance from the center to the focus, c, can be found by c = the roles of the major and minor axes are reversed, and the foci lie above and below the center. b2 − a2. In either case, c is the distance from the center to each focus, and In this case, c = bigger denominator − smaller denominator. Finally, it is worth mentioning that if we take c = the standard equation of a circle, Equation 7.1, and divide both sides by r2, we get √ √ Equation 7.5. The Alternate Standard Equation of a Circle: The equation of a circle with center (h, k) and radius r > 0 is (x − h)2 r2 + (y − k)2 r2 = 1 Notice the similarity between Equation 7.4 and Equation 7.5. Both equations involve a sum of squares equal to 1; the diﬀerence is that with a circle, the denominators are the same, and with an ellipse, they are diﬀerent. If we take a transformational approach, we can consider both Equations 7.4 and 7.5 as shifts and stretches of the Unit Circle x2 + y2 = 1 in Deﬁnition 7.2. Replacing x with (x − h) and y with (y − k) causes the usual horizontal and vertical shifts. Replacing x with x a and y with y b causes the usual vertical and horizontal stretches. In other words, it is perfectly ﬁne to think of an ellipse as the deformation of a circle in which the circle is stretched farther in one direction than the other.1 Example 7.4.1. Graph (x+1)2 and minor axes, the vertices
, the endpoints of the minor axis, and the foci. 9 + (y−2)2 25 = 1. Find the center, the lines which contain the major Solution. We see that this equation is in the standard form of Equation 7.4. Here x − h is x + 1 so h = −1, and y − k is y − 2 so k = 2. Hence, our ellipse is centered at (−1, 2). We see that a2 = 9 so a = 3, and b2 = 25 so b = 5. This means that we move 3 units left and right from the center and 5 units up and down from the center to arrive at points on the ellipse. As an aid to sketching, we draw a rectangle matching this description, called a guide rectangle, and sketch the ellipse inside this rectangle as seen below on the left. 1This was foreshadowed in Exercise 19 in Section 7.2. 520 Hooked on Conics 4 −3 −2 −1 1 2 x −4 −3 −2 −1 1 2 x −1 −2 −3 −1 −2 −3 Since we moved farther in the y direction than in the x direction, the major axis will lie along the vertical line x = −1, which means the minor axis lies along the horizontal line, y = 2. The vertices are the points on the ellipse which lie along the major axis so in this case, they are the points (−1, 7) and (−1, −3), and the endpoints of the minor axis are (−4, 2) and (2, 2). (Notice these points are the four points we used to draw the guide rectangle.) To ﬁnd the foci, we ﬁnd c = 16 = 4, which means the foci lie 4 units from the center. Since the major axis is vertical, the foci lie 4 units above and below the center, at (−1, −2) and (−1, 6). Plotting all this information gives the graph seen above on the right. 25 − 9 = √ √ Example 7.4.2. Find the equation of the ellipse with foci (2, 1) and (4, 1) and vertex (0, 1). Solution. Plotting the data given to us, we have y 1 1 2 3 4 5 x From this sketch, we know that the major axis is horizontal
, meaning a > b. Since the center is the midpoint of the foci, we know it is (3, 1). Since one vertex is (0, 1) we have that a = 3, so a2 = 9. All that remains is to ﬁnd b2. Since the foci are 1 unit away from the center, we know c = 1. Since 9 − b2, so b2 = 8. Substituting all of our ﬁndings into the a > b, we have c = equation (x−h)2 9 + (y−1)2 √ a2 − b2, or 1 = b2 = 1, we get our ﬁnal answer to be (x−3)2 a2 + (y−k)2 8 = 1. √ 7.4 Ellipses 521 As with circles and parabolas, an equation may be given which is an ellipse, but isn’t in the standard form of Equation 7.4. In those cases, as with circles and parabolas before, we will need to massage the given equation into the standard form. To Write the Equation of an Ellipse in Standard Form 1. Group the same variables together on one side of the equation and position the constant on the other side. 2. Complete the square in both variables as needed. 3. Divide both sides by the constant term so that the constant on the other side of the equation becomes 1. Example 7.4.3. Graph x2 + 4y2 − 2x + 24y + 33 = 0. Find the center, the lines which contain the major and minor axes, the vertices, the endpoints of the minor axis, and the foci. Solution. Since we have a sum of squares and the squared terms have unequal coeﬃcients, it’s a good bet we have an ellipse on our hands.2 We need to complete both squares, and then divide, if necessary, to get the right-hand side equal to 1. x2 + 4y2 − 2x + 24y + 33 = 0 x2 − 2x + 4y2 + 24y = −33 x2 − 2x + 4 y2 + 6y = −33 x2 − 2x + 1 + 4 y2 + 6y + 9 = −33 + 1 + 4(9) (x − 1
)2 + 4(y + 3)2 = 4 (x − 1)2 + 4(y + 3)2 4 4 4 = (x − 1)2 4 (x − 1)2 4 + (y + 3)2 = 1 + (y + 3)2 1 = 1 Now that this equation is in the standard form of Equation 7.4, we see that x − h is x − 1 so h = 1, and y − k is y + 3 so k = −3. Hence, our ellipse is centered at (1, −3). We see that a2 = 4 so a = 2, and b2 = 1 so b = 1. This means we move 2 units left and right from the center and 1 unit up and down from the center to arrive at points on the ellipse. Since we moved farther in the x direction than in the y direction, the major axis will lie along the horizontal line y = −3, which means the minor axis lies along the vertical line x = 1. The vertices are the points on the ellipse which lie along the major axis so in this case, they are the points (−1, −3) and (3, −3), and the endpoints of the minor axis are (1, −2) and (1, −4). To ﬁnd the foci, we ﬁnd c = 3, which means 4 − 1 = √ √ 2The equation of a parabola has only one squared variable and the equation of a circle has two squared variables with identical coeﬃcients. 522 Hooked on Conics √ √ the foci lie left and right of the center, at (1 − √ 3 units from the center. Since the major axis is horizontal, the foci lie √ 3 units to the 3, −3). Plotting all of this information gives 3, −3) and (1 + y −1 1 2 3 4 x −1 −2 −3 −4 As you come across ellipses in the homework exercises and in the wild, you’ll notice they come in all shapes in sizes. Compare the two ellipses below. Certainly, one ellipse is more round than the other. This notion of ‘roundness’ is quantiﬁed below. Deﬁnition 7.5. The eccentricity of an ell
ipse, denoted e, is the following ratio: e = distance from the center to a focus distance from the center to a vertex In an ellipse, the foci are closer to the center than the vertices, so 0 < e < 1. The ellipse above on the left has eccentricity e ≈ 0.98; for the ellipse above on the right, e ≈ 0.66. In general, the closer the eccentricity is to 0, the more ‘circular’ the ellipse; the closer the eccentricity is to 1, the more ‘eccentric’ the ellipse. Example 7.4.4. Find the equation of the ellipse whose vertices are (±5, 0) with eccentricity e = 1 4. Solution. As before, we plot the data given to us y x 7.4 Ellipses 523 From this sketch, we know that the major axis is horizontal, meaning a > b. With the vertices located at (±5, 0), we get a = 5 so a2 = 25. We also know that the center is (0, 0) because the center is the midpoint of the vertices. All that remains is to ﬁnd b2. To that end, we use the fact that the eccentricity e = 1 4 which means e = distance from the center to a focus distance from the center to a vertex = c a = c 5 = 1 4 from which we get c = 5 we get b2 = 375 ﬁnal answer x2 4. To get b2, we use the fact that c = a2 − b2, so 5 16. Substituting all of our ﬁndings into the equation (x−h)2 25 + 16y2 375 = 1. 4 = a2 + (y−k)2 25 − b2 from which b2 = 1, yields our √ √ As with parabolas, ellipses have a reﬂective property. If we imagine the dashed lines below representing sound waves, then the waves emanating from one focus reﬂect oﬀ the top of the ellipse and head towards the other focus. F1 F2 Such geometry is exploited in the construction of so-called ‘Whispering Galleries’. If a person whispers at one focus, a person standing at the
other focus will hear the ﬁrst person as if they were standing right next to them. We explore the Whispering Galleries in our last example. Example 7.4.5. Jamie and Jason want to exchange secrets (terrible secrets) from across a crowded whispering gallery. Recall that a whispering gallery is a room which, in cross section, is half of an ellipse. If the room is 40 feet high at the center and 100 feet wide at the ﬂoor, how far from the outer wall should each of them stand so that they will be positioned at the foci of the ellipse? Solution. Graphing the data yields 524 Hooked on Conics y 40 units tall 100 units wide x It’s most convenient to imagine this ellipse centered at (0, 0). Since the ellipse is 100 units wide and 40 units tall, we get a = 50 and b = 40. Hence, our ellipse has the equation x2 402 = 1. √ 900 = 30, so that the foci are 30 units We’re looking for the foci, and we get c = from the center. That means they are 50 − 30 = 20 units from the vertices. Hence, Jason and Jamie should stand 20 feet from opposite ends of the gallery. 502 + y2 502 − 402 = √ 7.4 Ellipses 7.4.1 Exercises 525 In Exercises 1 - 8, graph the ellipse. Find the center, the lines which contain the major and minor axes, the vertices, the endpoints of the minor axis, the foci and the eccentricity. = 1 1. 3. 5. 7. + x2 y2 169 25 (x − 2)2 4 (x − 1)2 10 (x + 2)2 16 + + + (y + 3)2 9 (y − 3)2 11 (y − 5)2 20 = 1 = 1 = 1 2. 4. 6. 8. + x2 y2 9 25 (x + 5)2 16 (x − 1)2 9 (x − 4)2 8 = 1 + + + (y − 4)2 1 (y + 3)2 4 (y − 2)2 18 = 1 = 1 = 1 In Exercises 9 - 14, put the equation in standard form. Find the center, the lines which contain the major and minor axes
, the vertices, the endpoints of the minor axis, the foci and the eccentricity. 9. 9x2 + 25y2 − 54x − 50y − 119 = 0 10. 12x2 + 3y2 − 30y + 39 = 0 11. 5x2 + 18y2 − 30x + 72y + 27 = 0 12. x2 − 2x + 2y2 − 12y + 3 = 0 13. 9x2 + 4y2 − 4y − 8 = 0 14. 6x2 + 5y2 − 24x + 20y + 14 = 0 In Exercises 15 - 20, ﬁnd the standard form of the equation of the ellipse which has the given properties. 15. Center (3, 7), Vertex (3, 2), Focus (3, 3) 16. Foci (0, ±5), Vertices (0, ±8). 17. Foci (±3, 0), length of the Minor Axis 10 18. Vertices (3, 2), (13, 2); Endpoints of the Minor Axis (8, 4), (8, 0) 19. Center (5, 2), Vertex (0, 2), eccentricity 1 2 20. All points on the ellipse are in Quadrant IV except (0, −9) and (8, 0). (One might also say that the ellipse is “tangent to the axes” at those two points.) 21. Repeat Example 7.4.5 for a whispering gallery 200 feet wide and 75 feet tall. 22. An elliptical arch is constructed which is 6 feet wide at the base and 9 feet tall in the middle. Find the height of the arch exactly 1 foot in from the base of the arch. Compare your result with your answer to Exercise 21 in Section 7.3. 526 Hooked on Conics 23. The Earth’s orbit around the sun is an ellipse with the sun at one focus and eccentricity e ≈ 0.0167. The length of the semimajor axis (that is, half of the major axis) is deﬁned to be 1 astronomical unit (AU). The vertices of the elliptical orbit are given special names: ‘aphelion’ is the vertex farthest from the sun, and ‘perihelion’ is the vertex closest to the sun. Find the
distance in AU between the sun and aphelion and the distance in AU between the sun and perihelion. 24. The graph of an ellipse clearly fails the Vertical Line Test, Theorem 1.1, so the equation of an ellipse does not deﬁne y as a function of x. However, much like with circles and horizontal parabolas, we can split an ellipse into a top half and a bottom half, each of which would indeed represent y as a function of x. With the help of your classmates, use your calculator to graph the ellipses given in Exercises 1 - 8 above. What diﬃculties arise when you plot them on the calculator? 25. Some famous examples of whispering galleries include St. Paul’s Cathedral in London, England, National Statuary Hall in Washington, D.C., and The Cincinnati Museum Center. With the help of your classmates, research these whispering galleries. How does the whispering effect compare and contrast with the scenario in Example 7.4.5? 26. With the help of your classmates, research “extracorporeal shock-wave lithotripsy”. It uses the reﬂective property of the ellipsoid to dissolve kidney stones. 7.4 Ellipses 7.4.2 Answers 1. 2. + = 1 y2 25 x2 169 Center (0, 0) Major axis along y = 0 Minor axis along x = 0 Vertices (13, 0), (−13, 0) Endpoints of Minor Axis (0, −5), (0, 5) Foci (12, 0), (−12, 0) e = 12 13 + = 1 x2 y2 25 9 Center (0, 0) Major axis along x = 0 Minor axis along y = 0 Vertices (0, 5), (0, −5) Endpoints of Minor Axis (−3, 0), (3, 0) Foci (0, −4), (0, 4) e = 4 5 3. (x − 2)2 4 + (y + 3)2 9 = 1 Center (2, −3) Major axis along x = 2 Minor axis along y = −3 Vertices (2, 0), (2, −6) Endpoints of Minor Axis (0, −3), (4, −3) Foci (2, −3 + 5), (2, −3 − 5
e = 3 5) √ √ √ 4. (x + 5)2 16 + (y − 4)2 1 = 1 Center (−5, 4) Major axis along y = 4 Minor axis along x = −5 Vertices (−9, 4), (−1, 4) Endpoints of Minor Axis (−5, 3), (−5, 5) 15, 4), (−5 − Foci (−5 + √ e = 15, 4) √ √ 15 4 527 x 13 −13 y 1 5 4 3 2 1 −1 −1 −2 −3 −4 −5 y 5 4 3 2 1 −3−2−1 −1 1 2 3 x −2 −3 −4 −5 y 1 2 3 4 x −1 −2 −3 −4 −5 −6 y 5 4 3 2 1 −9 −8 −7 −6 −5 −4 −3 −2 −1 x 528 Hooked on Conics 5. (x − 1)2 10 + (y − 3)2 11 = 1 √ Center (1, 3) Major axis along x = 1 Minor axis along y = 3 11), (1, 3 − Vertices (1, 3 + Endpoints of the Minor Axis (1 − Foci (1, 2), (1, 4) e = 10, 3), (1 + 10, 3) √ √ √ 11 11 √ 11) y 6 5 4 3 2 1 6. (x − 1)2 9 + (y + 3)2 4 = 1 Center (1, −3) Major axis along y = −3 Minor axis along x = 1 Vertices (4, −3), (−2, −3) Endpoints of the Minor Axis (1, −1), (1, −5) Foci (1 + √ e = 5, −3), (1 − 5, −3) √ √ 5 3 7. (x + 2)2 16 + (y − 5)2 20 = 1 Center (−2, 5) Major axis along x = −2 Minor axis along y = 5 Vertices (−2, 5 + 2 Endpoints of the Minor Axis (−6, 5), (2, 5) Foci (−2, 7), (−2, 3) e = 5), (−2, 5 − 2 5) √ √ √ 5 5 −2 −1 1 2 3 4 y
−2 −1 1 2 3 4 x x −1 −2 −3 −4 −5 y 10 6 −5 −4 −3 −2 −1 1 2 x 7.4 Ellipses 529 8. (x − 4)2 8 + (y − 2)2 18 = 1 √ Center (4, 2) Major axis along x = 4 Minor axis along y = 2 Vertices (4, 2 + 3 Endpoints of the Minor Axis (4 − 2 2, 2), (4 + 2 √ Foci (4, 2 + √ 5 e = 3 2, 2) 10), (4, 2 − √ √ √ 2), (4, 2 − 3 10) √ 21 −2 −. 11. = 1 10. (x − 3)2 25 + (y − 1)2 9 Center (3, 1) Major Axis along y = 1 Minor Axis along x = 3 Vertices (8, 1), (−2, 1) Endpoints of Minor Axis (3, 4), (3, −2) Foci (7, 1), (−1, 1) e = 4 5 = 1 12. (x − 3)2 18 + (y + 2)2 5 √ Center (3, −2) Major axis along y = −2 Minor axis along x = 3 Vertices (3 − 3 2, −2) Endpoints of Minor Axis (3, −2 + (3, −2 − Foci (3 − √ e = √ 5) √ 13, −2), (3 + 2, −2), (3 + 3 13, −2) √ √ 26 6 √ 5), + = 1 (y − 5)2 12 x2 3 Center (0, 5) Major axis along x = 0 Minor axis along y = 5 Vertices (0, 5 − 2 Endpoints of Minor Axis (− Foci (0, 2), (0, 8) e = √ √ 3 2 3), (0, 5 + 2 √ √ 3) 3, 5), ( √ 3, 5) (x − 1)2 16 + (y − 3)2 8 = 1 Center (1, 3) Major Axis along y = 3 Minor Axis along x = 1 Vertices (5, 3), (−3, 3) Endpoints of Minor Axis (1, 3 + 2 (1, 3 − 2 F

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