CHAPTER 5 LESSON 5.1 SUPPORT EXAMPLE EXERCISES REFRESHING YOUR SKILLS WORKING WITH POSITIVE SQUARE ROOTS

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1 5. Sample answer: Using the graphs of f () and g () 1, ou can find f (g ()) 1. Start at on the -ais, move to the corresponding point on the graph of g (), then move to, then to the graph of f (), and lastl to 1 on the -ais = f() = g() = 6 Compositions are essentiall a series of input-output functions. Drawing a vertical line up to the graph of g () gives the value of g (). Drawing a horizontal line to the graph of makes that -value into an -value. Drawing a vertical line to the graph of f () evaluates f () for that output value, and the horizontal line to the -ais reveals the answer. CHAPTER 5 REFRESHING YOUR SKILLS WORKING WITH POSITIVE SQUARE ROOTS 1. a , 3 _ 11, _ b. _ 36 _ 7 _ 9, 6 _ 7 7, _ c. _ 9 55 _ 6, 7 _ 55 8, _ d. _ 6 _ 67 _ 81, 8 _ 67 9, _ a b c d a. 6 6 b c d e a. iii. B.; b. i., C.; c. iv., D.; d. ii., A.. a , , so b. 8 _ 5 _ 6 5 _ 30, 5 8 _ 5 _ 8 00, so 8 _ c. 30 _ , , so d _ 157, or approimatel 1.5, , so LESSON 5.1 SUPPORT EXAMPLE 5 EXERCISES 1. a. f (5).753(0.91) b. g (1) 38(1.37) c. h () 7.3(0.835).3.9 d. j (37) 5(1.085) (1.085) a. 16, 1, 9; 16(0.75). Starting with the initial value of 16, repeatedl multipl the previous term, u n 1, b 0.75 to find the net term, u n. Thus u 0 16, u 1 16(0.75) 1, and u u 1 (0.75) 1(0.75) 9. The eplicit formula for the sequence is u n 16(0.75) n. The eponent, n, represents the repeated multiplication of the common ratio, The equation of the continuous function through the points of this sequence is 16(0.75). b., 36, 5; (1.5). Starting with the initial value of, repeatedl multipl the previous term, u n 1, b 1.5 to find the net term, u n. Thus u 0, u 1 u 0 (1.5) 36, and u u 1 (1.5) 5. The eplicit formula for the sequence is given b u n (1.5) n. The equation of the continuous function through the points of this sequence is (1.5). 3. a. 15, 75, 5; u 0 15 and u n 0.6u n 1, where n 1. Starting with the initial value of 15, f (0), repeatedl multipl the previous term b 0.6 to find the net term. Thus f (0) 15(0.6) 0 15, f (1) 15(0.6) 1 75, and f () 15(0.6) 5. The recursive formula for the pattern is u 0 15 and u n 0.6u n 1, where n 1. b. 3, 6, 1; u 0 3 and u n u n 1, where n 1. Starting with the initial value of 3, f (0), repeatedl multipl the previous term b to find the net term. Thus f (0) 3() 0 3, f (1) 3() 1 6, and f () 3() 1. The recursive formula for the pattern is given b u 0 3 and u n u n 1, where n 1.. a. The ratio of 36 to 8 is , so this is a 5% decrease. b. The ratio of 7 to 5 is _ _ _, so this is a 33.3% increase. c. The ratio of 7 to 50 is , so this is a 6% decrease. d. The ratio of 50 to 7 is , so this is approimatel a 6.38% increase. Discovering Advanced Algebra Solutions Manual CHAPTER 5 85

2 5. a. u and u n ( )u n 1 where n 1. The starting population in 1995, u 0, is 1.11 billion. Each ear the population is 1.5% greater than the population of the previous ear. Multipl the previous ear s population, u n 1, b ( ) to calculate the current ear s population, u n. Thus the recursive formula is u and u n ( )u n 1, where n 1. b. Year Population (in billions) Using the recursive formula in 5a, the population in 1995 is u The population in 1996 is u 1 ( )u (1.11) 1.9. Continue to use this formula to complete the table. c. 1.11( ). Let represent the number of ears since 1995, and let represent the population in billions. The starting population in 1995 is 1.11, and each ear the population is multiplied b ( ). Epress the population growth with the eplicit formula 1.11( ). The eponent,, represents the repeated multiplication b ( ). To see that this formula works for our table of values, choose two data points, such as 1998 and 000. In 1998 it had been 3 ears since Check that the population (in billions) in 1998 equals 1.11( ) 3. Calculate this to get 1.11( ) , which is the same population calculated using the recursive formula. In 000 it had been 5 ears since Check that the population (in billions) in 000 equals 1.11( ) 5. Calculate this to get 1.11( ) , which also matches the table in 5b. d. Using the recursive or eponential formula, the equations predict that the population of China in 006 was 1.6 billion. This is larger than the actual value. This means that the population is growing at a slower rate that it was in a..56(.5) ; 50 cm; 65 cm. Calculate the ratio of the plant s height each da to its height the previous da: , 0.5,.5, _ _ 16 and Let represent the number of the da, and let represent the height in centimeters. The plant has height.56 cm on da 0 and its height increases b a factor of.5 ever da, so the eponential equation is.56(.5). On the fifth da,.56(.5) 5 50, so the plant s height is 50 cm. On the sith da,.56(.5) 6 65, so the plant s height is 65 cm. b..56(.5) cm. Jack measured the height of the plant at 8:00 A.M. ever da, so when Jack s brother measured the plant at 8:00 P.M. on the third da, it had been 3 das 1 hours, or 3.5 das. Use the equation.56(.5) to find the height of the plant..56(.5) , so its height is approimatel 63.5 cm at 8:00 P.M. on the third da. c. Approimatel 78 cm. At noon on the sith da, it has been 6 das hours, or approimatel 6.16 das. The height is.56(.5) _ , or approimatel 78 cm. d da, or 18 hours. Using the equation.56(.5), find the value of when Using a table on a graphing calculator, ou can see that 5.15 when 0.757, so the doubling time is approimatel 0.76 da, or 18 hours. e. 11 das 13 hours, or 9 P.M. on da 11. To find the da and time when the stalk reached its final height of 1 km, or 100,000 cm, ou can use guess-and-check, ou can graph the equation and zoom in to find the time, or ou can graph a second function, 100,000, and find the intersection point of the two graphs. Multipl the decimal part of our answer b to get the number of hours past 8 A.M. The plant will reach a height of 1 km at 11 das 13 hours, or at 9 P.M. on da CHAPTER 5 Discovering Advanced Algebra Solutions Manual

3 7. a d. e. As the base increases, the graph becomes steeper. The curves all intersect the -ais at (0, 1). f. The graph of 6 should be the steepest of all of these. It will contain the points (0, 1) and (1, 6). c. 0.5 ( 1), or 0. 5 ( 1). This is the graph of 0.5 translated verticall units and horizontall 1 unit. d. _ ( ) or ( ). This is the graph of 0.5 dilated verticall b a factor of 3 and dilated horizontall b a factor of. 11. a. The common ratio is f (1) f (0) b. f () 30(0.9). The general form of an eponential function is f () ab, where a is f (0) and b is the common ratio. c. 30 g() f() a. 3. This is the graph of translated horizontall 3 units. b., or. This is the graph of translated verticall units. c. 3, or _ 3. This is the graph of dilated verticall b a factor of 3. d. 3. This is the graph of dilated horizontall b a factor of a d. e. As the base increases, the graph flattens out. The curves all intersect the -ais at (0, 1). f. The graph of 0.1 should be the steepest of all of these. It will contain the points (0, 1) and ( 1, 10). 10. a , or This is the graph of 0.5 translated verticall 5 units. b , or ( 5) 0.5. This is the graph of 0.5 reflected across the -ais and then translated verticall 5 units. d. g () f ( ) f (0) 30 e. Because g () f ( ), g () 30(0.9). f. You can use the - and -values of an point on the curve, 1, 1, and the common ratio to write the equation. 1. Answers will var. All will have b 1.8. Use (0, ) and (1, 7.): 7. b 1 0, 1.8 b, so Use (1, 7.) and (3, 3.38): b 3 1, 3. b, 1.8 b, so a. Let represent time in seconds, and let represent 10 (0, 10) (7, 10) distance in meters. Janell starts 10 m from the motion sensor at time 5 0 s, so the graph starts at (3.5, 3) (0, 10). Janell then moves at a constant rate of m/s 5 toward the motion sensor, so using the equation d rt, or t d_ r, relating distance, rate, and time, it takes Janell 7_ 3.5 s to get 3 m from the sensor. Connect (0, 10) to the new point, (3.5, 3). When she turns around, it takes another 3.5 s for her to get back to where she started from, 10 m from the motion sensor. Connect (3.5, 3) to the new point, (7, 10). b. The domain is 0 7 because Janell starts at time 0 s and it takes 7 s for her round-trip walk. The range is 3 10 because her distance from the motion sensor ranges between 3 m and 10 m. c This is the graph of stretched verticall b a factor of and translated verticall 3 units and horizontall 3.5 units. Discovering Advanced Algebra Solutions Manual CHAPTER 5 87

4 1. a. The point (3, 8.5) is on the graph of f, so f (3) 8.5. b ( 3), ( 6), or The slope of the line is 6 3, or 0.5. Using the point-slope form and the point (3, 8.5), the equation of the line is ( 3). Or using the point (6, 10), the equation is ( 6). Both equations are equivalent to a. A 5000 ( ) 5 b. S 500 ( ) 5 c. The will have the same amount at about 13.5 ears, when 5000 ( ) 500 (1 0.03). So, Austin will have more mone after 1 ears. 16. a. The rectangle diagram shows that the area of a rectangle with side lengths ( 6) and ( ) is 6, or. 6 6 b. Distribute ( ) across the sum ( 6). ( )( 6) ( 6) ( 6) c. Yes 6 d. A rectangle diagram also uses the distributive propert. Each term in the first binomial is multiplied b each term in the second binomial. EXTENSIONS A. Here are some eamples of growth and deca in the real world: Deca: the intensit of a star s light as it travels to Earth; the applause of a crowd after a show has ended Growth: the intensit of the sound of an airplane as it flies toward ou; the interest on an unpaid credit card balance B. Research results will var. LESSON 5. SUPPORT EXAMPLES 1. a. Power; b. Eponential; () (8) EXERCISES 1. a b. 6 (6 6) 36 c d. ( 1) 1 ( 1) 1 ( 1)( 1) 1 1 e f a. B the product propert of eponents, a 8 a 3 a 8 ( 3) a 5 b. B the quotient propert of eponents, b 6 b b 6 b c. B the power of a power propert, c 5 c 5 c 0. d. B the definition of zero eponent and the definition of negative eponents, d 0 1 e 3 e e a. False. Valid reasons: You must have the same base for the product propert of eponents; ,831,808. b. False. You must raise to the power before multipling. 88 CHAPTER 5 Discovering Advanced Algebra Solutions Manual

5 c. False. Valid reasons: You must raise to the power before dividing; onl one factor can be divided out; it should be ( ) d. True.. a , so b c , so d a b c / d / (0.8) e f () a (1) 1 b. 6 6 ( ) c. 5 3 ( 5)( ) d e f Sample answer: (a b) n is not necessaril equivalent to a n b n. For eample, ( + 3) 5 but However, the are equivalent when n 1, or when a b 0 and n is odd. 8. a. 9, , , , 33 b ; ; ; The sequence is not arithmetic because there is not a common difference. c. 1.67; 1.67; 1.67; The ratio of consecutive terms is alwas the same, so the difference is growing eponentiall. d. Possible answers: Non-integer powers ma produce non-integer values. If the eponents form an arithmetic sequence, the decimal powers form a geometric sequence. 9. a d. Discovering Advanced Algebra Solutions Manual CHAPTER 5 89

6 e. Sample answer: As the eponents increase, the graphs get narrower horizontall. The even-power functions are U-shaped and alwas in the first and second quadrants, whereas the odd-power functions have onl the right half of the U, with the left half pointed down in the third quadrant. The all pass through (0, 0) and (1, 1). f. Sample answer: The graph of 6 will be U-shaped, will be narrower than, and will pass through (0, 0), (1, 1), ( 1, 1), (, 6), and (, 6). Sample answer: The graph of 7 will be in the first and third quadrants, will be narrower than 3, 5, and will pass through (0, 0), (1, 1), ( 1, 1), (, 18), and (, 18). g. Power functions go through the origin and have long run values of infinit. Eponential functions have -intercepts at 1 (or a) and go to zero either as increases or decreases. 10. a. 3, or 3. This is the graph of 3 translated up units. b. ( ) 3. This is the graph of 3 translated left units. c. 1_ 3, or 3. This is the graph of 3 dilated verticall b a factor of. d. 1_ 8 3, 8( ) 3, or 1_ 3. This is the graph of 3 dilated verticall b a factor of 8 and then translated down units. 11. a. 7(0.9) 7(0.9) 1 (0.9) 1 7(0.9)(0.9) 1 b the product propert of eponents; 7(0.9)(0.9) 1 7(0.9) (0.9) 1.3(0.9) 1. b. 7(0.9)(0.9)(0.9) 7(0.9)(0.9) (0.9) 38.07(0.9) c. The coefficients are equal to the values of f 1 () corresponding to the number subtracted from in the eponent. If 1, 1 is on the curve, then an equation 1 b 1 1 is an eponential equation for the curve. 1. a. 30.0r 3 ; 5.r 6. The ball rebounds to 30 cm on the third bounce, so 1 30 and 1 3. The ball rebounds to 5. cm on the sith bounce, so 1 5. and 1 6. b. 30.0r 3 5.r 6 ; r r 3 5.r 6 Set the two equations as equal r 6 5. r 3 Divide both sides b 5. and r r 6 3 r 3 Divide r 6 b r r Raise both sides to the power of 1_ 3. r Evaluate. c. The ball was dropped from approimatel 173 cm. The equation for the rebound height is 30.0(0.5576) 3. Before the ball is dropped, it is at the zero bounce, so set equal to (0.5576) a. 7 ( 3) b c ( 5) 3 ( 5) 5 ( 5) a b. (0.976) 00 c. 39.8(0.976) CHAPTER 5 Discovering Advanced Algebra Solutions Manual

7 d. Using the first equation, (0.976) Using the second equation, 39.8(0.976) Both equations give approimatel 137. rads. e. Using the first equation, (0.976) Using the second equation, 39.8(0.976) Both equations give approimatel 7.3 rads. f. The first equation is (0.976) 00. B the product propert, (0.976)(0.976) B multipling (0.976), 39.8(0.976) 003. This is the second equation. 15. a. 7. Move the decimal in 3.7 seven places to the right to get 37,000,000. b.. Move the decimal in 8.01 four places to the left to get c.. Move the decimal in.75 four places to the right to get 7,500. d..61. Move the decimal in.61 two places to the left to get ( ) 3 ( ) ( ) a. Let represent time in seconds, and let represent distance in meters b. All ou need is the slope of the median-median line, which is determined b M 1 (8, 1.6) and M 3 (31, 6.). The slope is 0.. The speed is approimatel 0. m/s. IMPROVING YOUR REASONING SKILLS (Eas ) 1 ( ter ) 0 Egg (ter)0 Egg (Eas) 1Egg Eas 8 6, or one egg over eas. EXTENSIONS A. 1 The even-power functions are in the first and second quadrants and go through ( 1, 1) and (1, 1). The odd-power functions are in the first and third quadrants and go through ( 1, 1) and (1, 1). The two parts of each graph get closer and closer to both aes. The smaller the absolute value of the eponent, the faster the graph approaches the aes. B. See the solution to Take Another Look Activit 1 on page 31. LESSON 5.3 SUPPORT EXAMPLE Possible answers: 3, 3 1,, 1 3, EXERCISES a e j; b d g; 5 c i; f h; a. Power; the base is a variable. b. Power; the base is a variable. c. Eponential; the eponent is a variable. d. Power; is equivalent to 1, so the base is a variable. e. Power; a square root is equivalent to the power of 1_, so the base is a variable. f. Power; f (t) t t 3 t t 1 (t ) 1, so it is a translation of g (t) t. g. Eponential; 1 is equivalent to 1(3) 3 t. t 8 h. Power; w 5 is equivalent to 8(w 5) 1. i. Power; 8 is equivalent to 8. Discovering Advanced Algebra Solutions Manual CHAPTER 5 91

8 j. Neither; the independent variable appears twice, raised to two different powers. k. Power; 5 w 3 is equivalent to 1 5 w 3 5. l. Eponential; the eponent contains a variable. 3. a. a 1 6 b. b 8 10, b 5, or b 0.8 c. c 1, or c 0.5 d. d 7 5, or d 1.. a. a a 1 6. a b. b a a 1 6 a Evaluate. b b c. c b c c d. d c d d d Raise both sides to the power of b 8 b 5 Raise both sides to the power of 5_. Evaluate. 1 _ c 1 c Raise both sides to the power of. Evaluate. 5 d 7 d 7 5 Raise both sides to the power of 5_ 7. Evaluate W/cm. Let represent the number of gels, and let f () represent the intensit in W/cm. Epress f as an eponential function. Start b substituting the two points into the point-ratio form, 1 b 1 : 900 b 3 and 600 b 5 Equating the two epressions and solving, 900 b b 5 b 3 b a d. e. Each curve is less steep than the prior one. The graphs of 1 and 1 are onl in the first quadrant, whereas the graphs of 1 3 and 1 5 are in the first and third quadrants. The all pass through the points (0, 0) and (1, 1), and the graphs of 1 3 and 1 5 both pass through the point ( 1, 1). f. 1 7 will be less steep than the others graphed and will be in the first and third quadrants. It will pass through the points (0, 0), (1, 1), and ( 1, 1). g. The domains of 1 and 1 are 0 because ou can t take a square root or fourth root of a negative number. The domains of 1 3 and 1 5 are all real numbers. 7. a d. e. Each graph is steeper and less curved than the previous one. All of the functions go through (0, 0) and (1, 1). f. 5 should be steeper than the others and curve upward. b ( 3) ( 5) 3 b 3 b 3 So W cm At 6, CHAPTER 5 Discovering Advanced Algebra Solutions Manual

9 8. Sample answer: Power functions with rational eponents can have limited domain. When the eponent is between 0 and 1, the curve increases slowl with a shape similar to the graph of. Eponential curves alwas have a steadil increasing or decreasing slope, unlike power functions. 9. a. Eponential. The graph has a steadil increasing slope. b. Neither. The graph looks like a parabola with the middle section reflected across a horizontal line. This is like neither an eponential function nor a power function. c. Eponential. The graph has a steadil decreasing slope. d. Power. The graph appears in the first and third quadrants. The shape in the first quadrant is similar to the graph of. The shape in the third quadrant is similar to the graph of reflected across both aes. 10. a. 3 ( ) 3. The black graph is translated horizontall units and verticall 3 units to get the red graph. b. 1 ( 5) 3. The black graph is reflected across the -ais, then translated horizontall 5 units and verticall 1 unit to get the red graph. c. _ 3. The black graph is dilated horizontall b a factor of, and then translated verticall units to get the red graph. d. _ 3 3, or 3 3. The black graph is dilated horizontall b a factor of and verticall b a factor of. Then it is translated horizontall 3 units to get the red graph. 11. a b Original equation. Subtract from both sides. Divide both sides b Raise both sides to the power of 5. Original equation Square both sides Divide both sides b Raise both sides b the power of 1_. c Original equation Divide both sides b Raise both sides to the power of 3_. 1. a. Approimatel 0.73 AU. Let t represent the orbital time in ears, and let r represent the radius in AU. r t 3. Substitute for t and solve for r. r (0.615) b. Approimatel 9.75 r. Substitute 9.5 for r and solve for t. 9.5 t 3 t c. See 1a and b for eamples of how to fill in the blanks. Planet Mercur Venus Earth Mars Orbital radius (AU) Orbital time (r) Planet Jupiter Saturn Uranus Neptune Orbital radius (AU) Orbital time (r) a. P kv 1 Original equation. P k V V 1 V 1 PV k Multipl both sides b V. b. P 0 and V 1.3, so using PV k from 13a, k (0)(1.3) 9. c. 8. L. Substitute 60.0 for P and 9 for k in the equation PV K. 60.0V 9 V d. 3.8 mm Hg. Substitute 15 for V and 9 for k in the equation P kv 1. P 9(15) a b c d e Discovering Advanced Algebra Solutions Manual CHAPTER 5 93

10 15. a. ( ). This is translated left units. b. 1. This is translated up 1 unit. c. ( 5). This is reflected across the -ais and translated left 5 units and up units. d. ( 3). This is translated right 3 units and down units. e. 3. This is translated left 3 units. f. 1. This is translated down 1 unit. g. 1. This is translated left units and up 1 unit. h This is reflected across the -ais and translated right 1 unit and down 1 unit. 16. About 80. The dart plaers in at least the 98th percentile are the best % of all registered dart plaers. % of,000 is a. u 1 0 and u n (1 0.)u n 1 1.u n 1, where n b. About 86 rats. You can use Home screen recursion or Sequence mode to find that u c. Follow the pattern in 17b to see that u n 1. n 1 0 where n 1. Let represent the ear number, and let represent the number of rats. 0(1.) 1. EXTENSIONS A. See the Sketchpad demonstration Rational Eponents. B. See the solution to Take Another Look Activit on page 31. LESSON 5. SUPPORT EXAMPLE a. 3 1, so b. 8 17, so c. 1 5, so 5 5 d. 3 1, so EXERCISES 1. a. 5 50, so b , so c. No real solution. 11 and the square of a real number is never negative.. a. 1 3, so 1 5, and b c d e a b c a. 100 r 6 b. 100 r 6 50; r 6 0.5, so r c. The reduction rate is 100% minus the percent passed: , or 10.9%. 5. a. She must replace with ( 7) and replace 1 with 1 7 ; b 1. b. 7 (105 7) b 1 7 b ( 1) b c. Possible answers: 0, 00. b 0.508;, 57, b 0.510; 3, 31, b CHAPTER 5 Discovering Advanced Algebra Solutions Manual

11 6. a. d. Possible answer: The mean of the b-values is This gives the equation ŷ 7 98(0.511) 1. b. Sample answer: ŷ You might start b graphing the data with the line. The data curve upward slightl, so ou know the eponent must be greater then 1, but the points lie below the line, so ou also need to appl a scale factor less than 1. The equation ŷ appears to be a good fit. c. Approimatel 1,9,00 km. (Answers will var depending on the equation found in 6b.) Substitute for in the equation found in 6b This is in units of 100,000 km, so the orbit radius is approimatel 1,9,00 km. d. Approimatel das. (Answers will var depending on the equation found in 6b.) Substitute 19.5 for in the equation found in 6b (19.5) The orbital time is approimatel das. 7. a. Approimatel 39.3 tons. Substitute 5 for L in the given equation. W b. Approimatel 5.9 ft. Substitute 5 for W in the given equation L , L 3.18 L 38, a. Approimatel cm. Substitute 18 for h in the given equation d d 0.8 d b. Approimatel 3.75 m. The circumference and diameter are related b C d, so d 87. Substitute 87 for d in the given equation. h a. Approimatel 1.9 g. Substitute 15 for M in the given equation. F b. Approimatel 1.8%. Divide b 15. Use this unrounded number for better accurac % Approimatel 9% per ear. Use the point-ratio equation, 3.7(1 r) 007, where r is the rate of inflation. A gallon of milk is predicted to be $5.8 in 017, so (1 r) (1 r) 10 1 r r To find the monthl inflation, change the eponent to reflect the number of months instead of the number of ears and solve for r. This rate of inflation is approimatel 0.319%. 11. a , or.66% per ear. If r is the deca rate, then use the two data points to write the equation (1 r) 3. Solve for r. (1 r) r Discovering Advanced Algebra Solutions Manual CHAPTER 5 95

12 b. Approimatel 6.6 g. If 0 is the initial amount, then ( ) 0 (0.953). Solve for c. 6.6(0.953). The equation is in the form 0 (1 r). d. Approimatel 0.6 g. Substitute 50 for in the equation from 11c. 6.6(0.953) e. About 1.5 r. Look for an -value for which the corresponding -value is 3.3. Trace the graph on a graphing calculator to estimate the time. Substitute.5 for and.75 for z in the second original equation and solve for So.5,, and z.75. IMPROVING YOUR REASONING SKILLS To take the half root is the same as squaring. 1 cin nati (cin) nati cin cin nati He should look in Cincinnati, Ohio. EXTENSIONS A. Answers will var. B. Answers will var. LESSON a Temperature ( F) b. 18.9, 9.15, 0.1, 50.35, 57. c. Range ; IQR d. The data do not support Juan s conjecture. Ten cities fall below 3 F and onl seven fall above 50 F ,, z.75. Eliminate one variable and then solve a sstem of equations in two variables. z First equation. z 1 Second equation. 3z 3.75 Subtract z 1.75 Multipl the second equation b z 5 Third equation. 9z 6.75 Add. Now ou have two equations without. 1z 15 Multipl the first equation without b. 9z 6.75 Second equation without. 3z 8.5 Subtract. z.75 Divide b 3. Substitute.75 for z in the first equation without and solve for. 3(.75) SUPPORT EXAMPLES ( 3 ) 1 ( 3 ) 1 1 ( 3 ) f () f 3 1 ( ) So, f (6) 3 6 1, and f 1 ( ) EXERCISES 1. ( 3, ), ( 1, 0), (, ), (6, ). Switch the - and -coordinates of the known points.. a. g () 5 () 5 9 b. g 1 (9). The inverse of g (t) 5 t is g 1 (t) t 5 g 1 (9) 9 5. Or, from a, g () 9, so g 1 (9) must equal. c. g 1 (0) CHAPTER 5 Discovering Advanced Algebra Solutions Manual

13 3. Graph c is the inverse because the - and -coordinates have been switched from the original graph so that the graphs are smmetric across line.. a and e are inverses; b and d are inverses; c and g are inverses; f and h are inverses. To find the inverse of a, switch and and then solve for ( 6). This is e. To find the inverse of b, switch and and then solve for. 6 6 ( ) 6 6. This is d. To find the inverse of c, switch and and then solve for. 6( ) This is g. 6 To find the inverse of f, switch and and then solve for. 6 ( 6) This is h. 5. a. f (7) 0.5(7 3) g () 3 ( ) b. The might be inverse functions. c. f (1) 0.5(1 3) 0.5 g ( ) 3 ( ) d. The are not inverse functions, at least not over their entire domains and ranges. e. f () for 3 and g () for are inverse functions. Sketch a graph of the two functions and the line. (, 3) (3, ) 8 g() = f() f () is a whole verticall oriented parabola, but g () is onl the top half of a horizontall oriented parabola. Onl the right half of the graph of f () is the inverse of g (), so the restricted domain of f () is 3. The domain of g () is. 6. a. 3 ( ) Set f () equal to 1. ( ) Subtract from both sides Raise both sides to the 5_ 3 power and evaluate. 3 b. f 1 () ( ) 5 3 Add to both sides. ( ) 3 5 Switch and. ( ) 3 5 Subtract from both sides. ( ) 5 3 ( ) 5 3 Raise both sides to the 5_ 3 power. Add to both sides. c. Sample answer: The steps are the same, but ou don t have to do the numerical calculations when ou find an inverse. 7. a. f ( 3) 1, f ( 1) 0, f (0) 1, f () b. ( 1, 3), (0, 1), (1, 0), (, ) c. Yes, it is a function. It passes the vertical line test. 8. a. Original function: f () 3. Inverse function: f 1 () 3, or f 1 () 1_ 3_. To find the inverse, switch the and in the original function and solve for Discovering Advanced Algebra Solutions Manual CHAPTER 5 97

14 The inverse is a function, so f 1 () 3, or f 1 () 1_ 3_. 3 b. Original function: f (). Inverse function: f 1 () 3, or f 1 () _ 3 _ 3. To find the inverse, switch the and in the original function and solve for The inverse is a function, so f 1 () 3, or f 1 () _ 3 _ 3. c. Original function: f () 3. Inverse: 3 ; not a function. To find the inverse, switch the and in the original function and solve for The inverse isn t a function, so f 1 () notation doesn t appl. 9. a. i. f 1 10 () 6.3. To find f 1 (), switch the dependent and independent variables , or f 1 () ii. f f 1 (15.75) iii. f 1 ( f (15.75)) iv. f f 1 () f 1 f () f f 1 () f ( 10) 10 f 1 ( f ()) f 1 (6.3 10) (6.3 10) b. i. f 1 3 () 1.8. To find f 1 (), switch the dependent and independent variables , or f 1 () ii. f f 1 (15.75) iii. f 1 ( f (15.75)) iv. f f 1 () f 1 ( f ()) f f 1 () f ( 3) 3 f 1 ( f ()) f 1 (1.8 3) (1.8 3) a. Enter the data for altitude in meters and temperature in C into lists in our calculator, and calculate the equation of the median-median line. The equation is f () b. f () , or f 1 () Echange and to find the inverse and solve for , or f 1 () c. Enter the data for altitude in feet and temperature in F into lists in our calculator, and calculate the equation of the median-median line. The equation is g () d. g () , or g 1 () ,591. Echange and to find the inverse and solve for , or g 1 () , CHAPTER 5 Discovering Advanced Algebra Solutions Manual

15 e. 1. F. Because the altitude is given in meters, use the function in 10a to find the temperature in C first. f () (619) C. Then use the function from 9b to change C to F: 1.8 3, so 1.8( 5.80) 3 1. F. 11. a. 100 C b. 100 F F 3 F 1.8C 3, F 3 1.8C, and C 1.8. So substitute for C in the equation for part a to F 3 get Your friend s score is 1. Sample answers are given for eplanations of incorrect answers. Problem 1 is correct. Problem is incorrect. The notation f 1 () indicates the inverse function related to f (), not the eponent 1. Problem 3 is incorrect. The epression can be rewritten as Problem is incorrect. The epression 0 0 is undefined. 13. a. i, ii, iii b. ii, v c. i, iv d. i, ii, iii 1. a. c () , where c () is the cost in dollars and is the number of thousand gallons. b. c (8) (8) $ c. g () 3.98, where g () is the number of thousands of gallons and is the cost in dollars. This is the inverse function of c, so switch and in and solve for , so c () d. 1,000 gallons. g (5.9) e. g (c ()) g ( ) c (g ()) c f. Approimatel $6. (50 gallons/da)(30 das) 1500 gallons; 3.98(1.5) g. Answers will var, but volume should equal , or 36,500 in 3 (approimatel 00 ft 3 ). For eample, the container could be 30 in. b 50 in. b 31 in. 15. Possible answers: 3 15, 3 15, 5, , f () 1.6(1.5), or f ().55(1.5) 5. Use the point-ratio equation with the point (, 1.6): 1.6r. Now use the second point, (5,.55), to solve for r r r a. 9.5 r Replace with and 8 with 3. 9 Power of a power propert. 9 Common base propert of equalit. 9.5 Divide both sides b. b Rewrite 9 as Power of a power propert. 1 1 Common base propert of equalit. Subtract and 1 from both sides Divide both sides b. c. 1 3 Rewrite 1_ as. 3 Power of a power propert Common base propert of equalit. Add and 3 to both sides. 1 Divide both sides b ( 3) + and 1_ 9 ( ) 3 The vertical parabola has equation a( 3). Use the point (, 5) to find a. 5 a( 3) 3 a The equation is 3( 3). The horizontal parabola has equation b ( ) 3. Use the point (, 5) to find b. b (5 ) 3 b 1 9 The equation is 1_ 9 ( ) 3. Discovering Advanced Algebra Solutions Manual CHAPTER 5 99

16 19. 1, 1, z 0. Rewrite the second equation as z 0, and then add it to the first equation to eliminate. Also, divide the third equation b.. z Second equation. z 0 3 z 3z Subtract z from both sides. First equation. Add the two equations. Now use the third equation to solve for and z...z. Third equation. z 1 Divide both sides b.. 1 z Subtract z from both sides. (1 z) 3z Substitute (1 z) for in 3z. z 3z Distribute. 7z 0 z 0 Subtract from both sides. Divide both sides b 7. Substitute 0 for z in the third equation...(0). 1 Substitute 0 for z and 1 for in the first or second equation. (0) 1 Second equation. 1 Subtract 1 from both sides. The solution is 1, 1, z 0. EXTENSIONS A. Answers will var. B. Answers will var. c b 7 and a d b and a 8 e. 5. b = 5 and a f b 6 and a 1. a , so 3 b , so c. 7 1, so 7.65 d , so e f , so 0 3. a. log ; 3 log 100 b. log 5 100; log log 5 c. log 35 8; log 35 8 log log 35 log 5 d. log 0. 5; log log 0. log 0.03 e. log ; log log log 0.5 f. log ; log log 17. a. This is a translation of log left units. b. This is a vertical dilation of log b a factor of 3. LESSON 5.6 SUPPORT EXAMPLES log 3 log 16 log 3 log 16 log log 3 c. This is a reflection of log across the -ais, followed b a translation verticall units. EXERCISES 1. a Recall that log 1000 log , so b 10 and a b b 5 and a CHAPTER 5 Discovering Advanced Algebra Solutions Manual

17 d. This is a translation of 10 left units. e. This is a vertical stretch of 10 b a factor of 3. f. This is a reflection of 10 across the -ais, followed b a translation down units. 5. a. False; log 6 1 b. False; 5 log 5.5 c. False; log = log 3 because d. False; log Approimatel 5 min. Set g () equal to 5 and solve for. 5 3(0.9) log log log a Substitute 1000 for in the given equation and solve for ( ) log log log Because represents ears after 1900, according to the model the debt passed $1 trillion in b. Approimatel 8.3%. (1 r) , so r c. Approimatel 5.6 r. The doubling time depends onl on the ratio, so ou can ignore the and assume the function doubles from 1 to. Set equal to, and solve for. ( ) log log log The doubling time is about 8.7 r. 8. a. 100( ). Start with the equation ab, and use the data points (0, 100) and (5730, 50) to find a and b. Using the first point, 100 ab 0, so a 100. Now use the second point to find b b b 5730 b The equation is 100( ). b. About 600 ears ago. The technique is approimate and assumes that the carbon-1 concentration in the atmosphere has not changed over the last 600 ears ( ) log log log a. 88.7(1.0077). Start with the equation ab, and use the data points (0, 88.7) and (6, 9.9) to find a and b. Using the first point, 88.7 ab 0, so a Now use the second point to find b b b 6 b The equation is 88.7(1.0077). b. 3 or clicks (1.0077) log log log Discovering Advanced Algebra Solutions Manual CHAPTER 5 101

18 10. a , so 35 b a Passengers Years since 1990 ŷ b. 1.85, 0.15, 6.75, 17.5, 1.5, 6.5, 5.5, 19.5,.75, 3.95, 6.5, 51.75, 36.65, 6.75, 1.75, 15.5 l w a. l w 7 b. l 5, w 3.5; the length is 5 in. and the width is 3.5 in. Substitute the second equation into the first to get (w 7) w 155 w 1 w 155 6w 11 w 3.5 Substitute 3.5 for w into the second equation to find l. l (3.5) 7 l a. The are parallel c d. 86. million riders; (116) a. C 1 3.7, C 65., C , C , C , C To get the net frequenc, multipl the previous term b, or to get a previous frequenc divide b. b (), where represents C-note number and represents frequenc in ccles per second. To find this equation, start with the point-ratio equation 61.6(). This is equivalent to (). 13. a. 1 3, or. This is the graph of translated down 1 unit and right 3 units. b. ( 5), or ( 5). This is the graph of translated down units and left 5 units. c. 6, or 6. This is the graph of translated up units and left 6 units. d. 7, or 7. This is the graph of translated up 7 units and right units. b. Possible answer: A(0, 3); P (1, 1); Q (, 3) c. Possible answer: Translate right 1 unit and up units; ( 1) 3( ) 9. d. Possible answer: Translate right units and up 6 units; ( ) 3( 6) 9. e. Possible answer: ( 1) 3( ) , which is the equation of. ( ) 3( 6) , which is the equation of. EXTENSIONS A. Answers will var. B. Answers will var. LESSON 5.7 SUPPORT EXAMPLES 1. 3 log log 5 log 3 log 5 log 8 log 5 log 8_ 5. Possible answer: log 5 log (5 9) log 5 log 9 3. log log EXERCISES 1. a. log 5 log 11 log (5 11) log 55 b. 3 log log 3 log 8 c. log 8 log 7 log 8 7 log d. log 6 log 6 log 1 6 log CHAPTER 5 Discovering Advanced Algebra Solutions Manual

19 e. log 7 log 3 log 7 log 3 log(7 9) log 63. a. log log ( 11) log log 11 b. Answers will var. Possible answer: log 13 log 6 log 6 log c. log 39 log (3 13) log 3 log 13 d. Answers will var. Possible answer: log 7 log 1 log 1 log 3. a. log 5 log 5 b. log log c. log 3 log 3 1 1_ log 3 d. log 7 log 7 log 7. a. True; product propert of logarithms b. False; possible answer: log 5 log 3 log 15 c. True; power propert of logarithms d. True; quotient propert of logarithms e. False; possible answer: log 9 log 3 log 9_ 3 log 3 f. False; possible answer: log 7 log 7 1 1_ log 7 g. False; possible answer: log 35 log 5 log 7 h. True; log 1_ log 1 log log. (log 1 0) i. False; possible answer: log 3 log log 3_ j. True; log 6 log 6 log 6 log(16) log 16 b the power propert of logarithms 5. a. g h g k ; product propert of eponents b. log st; product propert of logarithms c. f w v ; quotient propert of eponents d. log h log k; quotient propert of logarithms e. j st ; power propert of eponents f. g log b; power propert of logarithms g. k m/n ; definition of rational eponents h. log u t; change-of-base propert i. w t s ; product propert of eponents 1 j. h ; definition of negative eponents p 6. a. 100( ). Start with the equation ab, and use the data points (0, 100) and (5750, 50) to find a and b. Use the first point to find a. 100 ab 0, so a 100. Now use the second point to find b b b 5750 b The equation is 100( ). b. Approimatel 11,60 ears old. Substitute 5 for and solve for ( ) log log c B.C.E. Substitute 6.5 for and solve for ( ) log log Subtract this value from 1981 to get the ear the ark was constructed: , or about 1911 B.C.E. d. 100( ) 100,000, There is virtuall nothing left to measure, so ou could not use carbon-1 for dating coal unless ou had ver sensitive instruments to detect the radioactivit. 7. a. Let represent the note s number of steps above middle C, and let represent the note s frequenc in hertz because the starting value is 61.6 and there are 1 intermediate frequencies to get to the last C note, which has double that frequenc. b. Note Frequenc (Hz) Do C 61.6 C# 77. Re D 93.6 D# Mi E 39.6 Fa F 39. F# Sol G 39.0 G# 15.3 La A 0.0 A# 66.1 Ti B 93.8 Do C a Take the logarithm of both sides to get log 5.1 log 7. Use the power propert of logarithms to get log 5.1 log 7. Then log log 5.1 b Subtract 17 from both sides to get Take the logarithm of both sides to get log 1.5 log 13. Use the power propert of logarithms to get log 1.5 log 13. Then log log 1.5 Discovering Advanced Algebra Solutions Manual CHAPTER 5 103

20 c Divide both sides b 7 to get _ 9. Take the logarithm of both sides to get log 0.93 log _ 9. Use the power propert of logarithms to get log 0.93 log _ 9. Divide both sides b log 0.93 to get d Subtract 3 from both sides and then divide both sides b 5 to get Take the logarithm of both sides to get log 1.0 log 1 5. Use the power propert of logarithms to get log 1.0 log 1 5. Divide both sides b log 1.0 to get a. 1.7( ). The point-ratio equation is 1.7 r. Use the data point (, 9.6) to find r. b r r Air pressure lb/in r (0, 1.7) (, 9.67) Altitude (mi) Altitude (mi) (9.67, ) (1.7, 0) Air pressure lb/in. c. Approimatel 8.91 lb/in. Convert the altitude to miles and substitute that value in the equation 1,000 ft from 9a..7 mi; 5,80 ft/mi 1.7( ) d. Approimatel 6.3 mi. Substitute 3.65 for and solve for log log a. 100% 3.5% 96.5% remains after 1 min. b. 100(0.965), where is time in minutes and is percentage of carbon-11 remaining. To find the equation, start with the point-ratio equation, ab, and use the data points (0, 100) and (1, 96.5) to find a and b. 100 ab 0, so a 100, and 100b b 1, so b = 0.965, and 100(0.965). c. Approimatel min (0.965) log log d. In one da the carbon-11 is virtuall gone, so ou could never date an archaeological find. 11. Graphs will var. If a horizontal line intersects the graph in more than one point, its inverse is not a function. 1. a The common difference, or slope, is 3, so the -intercept is Therefore the equation is 5 3. b. 3. The -intercept is and the common ratio is 3, so the equation is a. The graph has been verticall dilated b a factor of 3, then translated right 1 unit and down units b. The graph has been horizontall dilated b a factor of 3, reflected across the -ais, 5 and translated up units. 1. a. False. If everone got a grade 5 of 86% or better, one would have to have gotten a much higher grade to be in the 86th percentile. b. False. Consider the data set {5, 6, 9, 10, 11}. The mean is 8.; the median is 9. c. False. Consider the data set {0,, 8}. The range is 8; the difference between the mean, 10, and the maimum, 8, is 18. d. True 15. a. Let h represent the length of time in hours, and let c represent the driver s cost in dollars. c 1h 0. The domain is the set of possible values of the number of hours, h 0. The range is the set of possible values of the cost paid to the driver, c 0. b. Let c represent the driver s cost in dollars, and let a represent the agenc s charge in dollars. a 1.15c 5. The domain is the mone paid to the driver if she had been booked directl, c 0. The range is the amount charged b the agenc, a 1.15(0) 5 3 5; a 8. c. a 1.15(1h 0) 5, or a 16.1h 8 EXTENSIONS A. Research results will var. B. Results will var. 10 CHAPTER 5 Discovering Advanced Algebra Solutions Manual

21 LESSON 5.8 SUPPORT EXAMPLE a (1.1) log 1.1 log log log 1.1 b log 5 log 1.75 EXERCISES log 1.75 log a. log 800 log 10, so Check: b. log 08 log, so Check: c. log 16 log 0.5, so Check: d , so log log 18.5 log 10 log log 10 log log log 16 log 0.5 log 10, and 78 log Check: 18.05(10) e , so log log 1.89, and 155 log.0 log Check:.0(1.89) f , so log log 0.1, and log 19.1 log Check: 19.1(0.1) a. 10 n p 10 n 10 p Original equation. log 10 n p log 10 n 10 p Take the logarithm of both sides. (n p)log 10 log 10 n log 10 p Power propert and product propert of logarithms. (n p)log 10 n log 10 p log 10 Power propert of logarithms. (n p)log 10 (n p)log 10 Combine terms. b. 10d 10 e 10 d e Original equation. log 10d 10 e log 10 d e Take the logarithm of both sides. log 10 d log 10 e log 10 d e Quotient propert of logarithms. d log 10 e log 10 (d e)log 10 Power propert of logarithms. (d e)log 10 (d e)log 10 Combine terms. 3. About mo, or about 16 r mo. The equation for mone invested at an annual rate r, compounded n times per ear, with initial amount P, is A P 1 r n t, where t is the number of compoundings. In this case it is A t, where t is the number of months. Substitute 9000 for A and solve for t t t log 3 t log mo 1 mo/r months is r, and 0.31 r is (0.35 r)(1 mo/r) 3.9 mo. So, tripling our mone will take about months, or about 16 ears months.. a. h 16( ) T. Two points on the curve are (, 16) and (, ). Use the pointratio equation h 1 b T 1, where h represents the number of hours and T represents the temperature. Using the first point, h 16b T. Substitute (, ) into the equation to find b; 16b 18, so 16 b 18, and b The function is h 16( ) T. b. h 16( ) 30.1 hr at 30 C; h 16( ) hr at 16 C. c. Approimatel 3.9 C 17 16( ) T T T log log T Discovering Advanced Algebra Solutions Manual CHAPTER 5 105

22 d. e. A realistic domain is 0 to 100 C; these are the freezing and boiling points of water. 5. a. f (0) After 0 das, 133 games have been sold. b. f (80) After 80 das, 7969 games have been sold. c. 7. After 7 das, 6000 games have been sold. Make a table to approimate a value for when 7. d (1.09 ) (1.09) (1.09) 99(1.09) e. log 1.09 log 1 99 log log 1.09 c. Approimatel W/cm log I log I I I d. About 3.16 times as loud. Let I 1 be the power of the sound at 7 db, and let I be the power of the sound at db. I 7 10 log 1 I log I.7 log 1 I log I I I 1 I 7. a I I b. Points in the form (log, ) give a linear graph. c. ŷ 6 0 log. Use our calculator to approimate a line of fit. Sample answer: The number of games sold starts out increasing slowl, then speeds up, then slows down as everone who wants the game has purchased one. 6. a. D 10 log log db b. D 10 log log 3,160, db 106 CHAPTER 5 Discovering Advanced Algebra Solutions Manual

23 Two points on the line are (0.5, 16) and (, 6), so the slope is 0. The -intercept is 6, so the equation of the line is 6 0. The -values used to find this line were actuall the logarithms of the -values of the original data. Substitute log for. The final equation is ŷ 6 0 log. 9. Let represent height in meters, and let represent viewing distance in kilometers. a. The data are the most linear when plotted in the form (log, log ). d. Sample answer: Yes; the graph shows that the equation is a good model for the data. 8. a. Let represent time in minutes, and let represent temperature in degrees Fahrenheit. b. ŷ Use our calculator to find that the median-median line through the points in the form (log, log ) is approimatel log log. Solving for gives log log log ŷ Check that this equation fits the original data. b. ŷ (0.9318). Because the curve is both reflected and translated, first graph points in the form (, ) to account for the reflection. Then translate the points up so that the data approach a long-term value of zero. Then graph points in the form (, log( 7)), which appear to be linear. 10. a. After 1 da: 16(1 0.15) qt. After das: 1.6(1 0.15) qt. The recursive formula is u 0 16 and u n u n 1 (1 0.15) 1, where n 1. b. Use our calculator to find the medianmedian line through the points in the form (, log( 7)), or approimatel log( 7) Solving for gives the eponential equation ŷ , or ŷ (0.9337). Check that this equation fits the original data Discovering Advanced Algebra Solutions Manual CHAPTER 5 107

24 Chlorine (qt) Das after first treatment The sequence is a shifted geometric sequence, which corresponds to a translated eponential equation as an eplicit model. To determine the amount of translation, find the long-run value of the sequence. Recall that ou do this b assigning the same variable to u n and u n 1. c 0.85c 1 c 0.85c 1 Substitute c for both u n and u n 1, and rewrite (1 0.15) as Subtract 0.85c from both sides. 0.15c 1 Subtract. c 6. 6 Divide both sides b On our calculator, plot points in the form, These points form an eponential curve with long-run value 0. The eplicit eponential equation through these points will be in the form ab. The points still deca at the same rate, so b is still (1 0.15), or The starting value has been translated, so a is , or The equation through the points in the form, 6. 6 is (0.85). Solving for gives 9. 3 (0.85) Check that this equation fits the original data. 1 18b 10 8 b 6 b The equation is 18. b. log log 18 log log log.5 log,. log log 1 log 10, or Notice that the points (18, ) and (1, 10) are on the graph of the inverse of the function from 11a. Start with 18, switch and, and solve for log 18 ( ) log log 18 log log 18 log log log 18 log 1. a. Let represent the amount of fish sticks in pounds, and let represent cost or income in dollars. Cost: ,000; income: 1.9. b. 13. Cost/income ($) 1,000, , ,000 50, ,000,000 Fish sticks (lb) c 111,765 pounds. Find the point where the two lines intersect ,000 and 1.9, so , Therefore, 19, , and 111,765. d. $66,000. The profit is the income minus the cost: profit 1.9 ( ,000) ,000. At 500,000 fish, profit 0.17(500,000) 19,000 66,000. ( 5, 10) 1_ 3_ ( 5, 8) (, 8) 10 ( 5, 6) ( 5, 8) 11. a. 18, 1 10, or.5. Start b using the point-ratio equation with the first point, 18b, and use the second point to find b CHAPTER 5 Discovering Advanced Algebra Solutions Manual