111 Chapter 5 Logarithmic Functions Homework 5.1 1.f -1(7) = 4 3.f (4) = 6 5.f -1(4) = 5, since f (5) = 4 7. 1( )2 64 56 48 310 2x f x 9.g(2) = 6 11.g -1(2) = 1, since f(1) = 2 13. 1( )2 16 218 354 4162 5486 6x g x 15. 1( )4 610 516 422 328 234 1x f x 17.Begin by creating a table of values for f(x), then build a table for 1( ) f x from that information. ( )0 31 62 123 244 48x f x 1( )3 06 112 224 348 4x f x 19.f(3) = 24 21.f -1(3) = 0, since f (0) = 3 23. 25. Homework 5.1SSM: Intermediate Algebra 112 27. 29. 31. 33. 35. 37. 39.a.f(t) = 0.81t 45.86 11Replace( ) with:0.81 45.86Solve for:45.86 0.8145.860.8145.860.81 0.81Replace with( )( ) 1.23 56.62f t pp ttp tptptt f pf p p= + =+== += + b. f(100) = 0.81(100) 45.86 =35.14 According to the model, in 2000, 35.1% of births were out of wedlock. c.1(100) 1.23(100) 56.62 179.62 f = + = According to the inverse model, all births will be out of wedlock in 2080. This is probably a model breakdown. SSM: Intermediate AlgebraHomework 5.1 113 41.a. Linear Function b.Use linear regression ( ) 0.78 52.56 f t t = c.( ) 0.78 52.56 f t t = 11Replace( ) with:0.78 52.56Solve for:52.56 0.7852.560.7852.560.78 0.78Replace with( )( ) 1.28 67.38f t pp ttp tptptt f pf p p= + =+== += + d.Half the people is 50 percent, substitute 50 for p and solve. ( ) 0.78 52.5650 0.78 52.56102.56 0.78131.49p f t tttt= = = = The year will be 2031. e. 1(50) 1.28(50) 67.38 131.49 f = + The year will be 2031 f.The results are the same in parts d and e. 43.a. Linear Function c.Use linear regression ( ) 0.38 20.13 f t t = c.( ) 0.38 20.13 f t t = 11Replace( ) with:0.38 20.13Solve for:20.13 0.3820.130.3820.130.38 0.38Replace with( )( ) 2.63 52.97f t pp ttp tptptt f pf p p= + =+== += + d.(109) 0.38(109) 20.13 21.29 f = =In 2009, the price of CDs will be $21.29. e. 1(25) 2.63(25) 52.97 118.72 f = + =The price of CDs will be $25 in 2019. 45.f(x) = x 3 111Replace( ) with:3Solve for:3Replace with( ) :( ) 3Write equation in terms of:( ) 3f x yy xxy xx f yf y yxf x x= + == += + Homework 5.1SSM: Intermediate Algebra 114 47.f(x) = x + 8 111Replace( ) with:8Solve for:8Replace with( ) :( ) 8Write equation in terms of:( ) 8f x yy xxy xx f yf y yxf x x= + == = 49.f(x) = 4x 111Replace( ) with:4Solve for:4Replace with( ) :( )4Write equation in terms of:( )4f x yy xxyxx f yyf yxxf x==== 51.( )7xf x = 111Replace( ) with:7Solve for:7Replace with( ) :( ) 7Write equation in terms of:( ) 7f x yxyxy xx f yf y yxf x x==== 53.f(x) = 4x + 5 Replace( ) with:4 5Solve for:5 4f x yy xxy x= + = 1115454 4Replace with( ) :1 5( )4 4Write equation in terms of:1 5( )4 4yxyxx f yf y yxf x x== = = 55.f(x) = -6x 2 111Replace( ) with:6 2Solve for:2 66 2261 26 6Replace with( ) :1 1( )6 3Write equation in terms of:1 1( )6 3f x yy xxy xx yyxx yx f yf y yxf x x= + = = ++== = = 57.f(x) = 0.4x 7.9 111Replace( ) with:0.4 7.9Solve for:7.9 0.47.9 0.47.90.47.90.4 0.4Replace with( ) :( ) 2.5 19.75Write equation in terms of:( ) 2.5 19.75f x yy xxy xy xyxyxx f yf y yxf x x= + =+ =+== += += + SSM: Intermediate AlgebraHomework 5.1 115 59.f(x) = -72.83x + 99.23 111Replace( ) with:-72.8399.23Solve for:99.23 72.8399.23 72.8399.2372.8399.2372.83 72.83Replace with( ) :( ) 0.014 1.36Write equation in terms of:( ) 0.014 1f x yy xxy xy xyxyxx f yf y yxf x x= + = = == += += + .36 61. 7( ) 13f x x = + ( )111Replace( ) with:713Solve for:7133173 37Replace with( ) :3 3( )7 7Write equation in terms of:3 3( )7 7f x yy xxy xy xyxx f yf y yxf x x= + = === = 63. 5( ) 36f x x = Replace( ) with:536Solve for:536f x yy xxy x= + = ( )1116356 185Replace with( ) :6 18( )5 5Write equation in terms of:6 18( )5 5y xyxx f yf y yxf x x + = == = 65. 6 2( )5xf x= 111Replace( ) with:6 25Solve for:5 6 25 2 65 265 26 6Replace with( ) :5 1( )6 3Write equation in terms of:5 1( )6 3f x yxyxy xy xyxyxx f yf y yxf x x== + =+== += += + 67.f(x) = -3(2x 5) 1Replace( ) with:3(2 5)Solve for:12 5315 231 152 31 56 2Replace with( ) :f x yy xxy xy xy xx yx f y= = + = + = = +Homework 5.2SSM: Intermediate Algebra 116 111 5( )6 2Write equation in terms of:1 5( )6 2f y yxf x x= += + 69.f(x) = 7 8(x + 1) 111Replace( ) with:7 8( 1)Solve for:7 8 88 11 8181 18 8Replace with( ) :1 1( )8 8Write equation in terms of:1 1( )8 8f x yy xxy xy xy xyxx yx f yf y yxf x x= += = + = +== = = 71.f(x) = x 111Replace( ) with:Solve for:Replace with( ) :( )Write equation in terms of:( )f x yy xxx yx f yf y yxf x x==== 73.Answers may vary. Example: It makes sense that ( ) 5 g x x = is the inverse function of ( ) 5 f x x = + because addition and subtraction are inverse operations of each other (or, 5 and -5 are additive inverses). 75.f(x) = 3 is not one-to-one. For example,(1) 3 f =and(2) 3 f = . So, the inverse of f would send 3 to both 1 and 2, making the inverse of fnot a function.Since f -1 is not a function, f was not invertible. 77.a. f(x) = mx + b 111Replace( ) with:Solve for:11Replace with( ) :1( )Write equation in terms of:1( )f x yy mx bxy b mxy bxmy bxmbx ym mx f ybf y ym mxbf x xm m= + ==== = = b.The function f(x) given in part a is a linear function with nonzero slope (since0 m ). The inverse function is also a linear function as its slope is nonzero (1mis nonzero since 0 m ). Homework 5.2 1. 29log (81) 2, since 9 81 = =3. 33log (27) 3 since 3 27 = =5.52log (32) 5 since 2 32 = =7. 36log (216) 3 since 6 216 = =9. 26log (36) 2 since 6 36 = =11. 210log (100) 2 since 10 100 = =13. 37log (343) 3 since 7 343 = =15. 65log (15, 625) 6 since 5 15, 625 = =17. -141 1log 1 since 44 4 = = SSM: Intermediate AlgebraHomework 5.2 117 19. -321 1log 3 since 28 8 = = 21. -1101 1log 1 since 1010 10 = = 23. 05log (1) 0 since 5 1 = =25. 1291log (3)since 9 32= =27. 1381log (2)since 8 23= =29. ( )1271log 7since 7 72= =31. ( )14 4 451log 5since 5 54= =33. 42 2 22log (log (16)) log (4) since 2 16= 2since 2 4= == 35. 110 10 100log (log (10)) log (1) since 10 10= 0since 10 1= == 37. 1log ( ) 1 since bb b b = =39. ( )2 2 2log 2 since bb b b = =41. ( )121logsince 2bb b b = =43. 10log (log ( )) log (1) since b= 0since b 1b b bb b = == 45.( ) 3xf x = 13( ) log ( ) f x x=47.( ) 10xh x = 1( ) log( ) h x x=49. 5( ) log ( ) f x x = 1( ) 5xf x=51.( ) log( ) h x x = 1( ) 10xh x=53. 2(2) 2 4 f = =55. 1 12(2) log (2) 1, since 2 2 f= = =57.3(3) log (3) 1 g = =59. 1 3(3) 3 27 g= =61.(1) 3 f =63. 13(1) log (1) 0 f= =65. 68. 69. Homework 5.3SSM: Intermediate Algebra 118 71. 15( ) log ( ) f x x=First make a table of values for f(x), then build a table for 1( ) f x. ( )0 11 52 253 1254 625x f x so 1( )1 05 125 2125 3625 4x f x 73.a. x = 2, since 52 = 25 b.5log (25) 2, =since 52 = 25 c. 5x and 5log ( ) xare inverses of each other. 75.a. ( )80084.0 10loglog 4.0 108.6ARA = = = The Richter number for the Gansu earthquake was 8.6. b. ( )70071.6 10loglog 1.6 107.2ARA = = = The Richter number for the Naplesearthquake was 7.2. c. 8.61.197.2 =The ratio of the Richter numbers is 1.2. d. 874.0 10251.6 10= The ratio of the Gansu earthquakes amplitude to the Naples earthquakes amplitude is 25. e.Answers may vary. Example: The Gansu earthquake released 25 times the energy of the Naples earthquake. 77. 0110log LI = 212108642Intesity of Sound Decibel ReadingSound(L) (w/m )Faintest sound 10 0heard by humansWhisper10 20Inside a running10 40car Conversation10 60Noisy street corner 10 80Soft rock concert 10 100Threshold of pain 1 120 79.Answers may vary. Example:log ( )bais the exponent of b that gives a. A logarithm can undo the output of an exponential function if they have the same base. Homework 5.3 1. 3log (243) 5 = 53 243 =3.log(100) 2 = 210 100 =5. log ( )ba c = cb a =SSM: Intermediate AlgebraHomework 5.3 119 7.log( ) m n =10nm =9. 35 125 = 5log (125) 3 =11. 310 1000 =log(1000) 3 =13. wy x =log ( )yx w =15.10pq =log( ) q p =17.( )4log 2 is equivalent to: x = 2416xx== 19.( )2log 4 is equivalent to: x = 42116xx== 21.( ) log 2 is equivalent to: x = 21010.01100xx== = 23.( )4log 0 is equivalent to: x = 041xx== 25.( )8log 4 1 is equivalent to: x = 18 44 82xxx=== 27. ( )36log 2 is equivalent to: x = 2 3313636363.3019xxxx=== 29.( ) ( )2 3log log 3 is equivalent to: x = 33382 log ( )log ( ) 8 is equivalent to:36561xxxx==== 31.( ) log 49 2b= 21249497bbb=== 33.( ) log 8 3b= 313882bbb=== 35.( ) log 16 4b= 41416162bbb=== 37.( ) log 3.6 6b= 6163.63.61.2380bbb== Homework 5.3SSM: Intermediate Algebra 120 39. ( ) ( )( ) ( )( )( )4 9log 4 log 9log 4 log 9log 9log 41.5850xxxxx==== 41. ( )( ) ( )( ) ( )( )( )5 4 804 16log 4 log 16log 4 log 16log 16log 42xxxxxx====== 43. ( )( )( )( )11 7 540054007115400log 7 log115400log 7 log1171log5log 73.1842xxxxxx = = = = = 45. ( )( )( )( )( )8 5 2 795 2 71712571log 2 log571log 2 log571log5log 23.8278xxxxxxx+ === = = = 47. ( )( )( )3.83(2.18 ) 170.91170.912.183.83170.91log 2.18 log3.83170.91log 2.18 log3.83170.91log3.83log 2.184.8738xxxxxx== = = = 49. ( ) ( )( ) ( ) ( )( )( )( )( )( )( )4 54 52 17log 2 log 174 5 log 2 log 17log 174 5log 2log 174 5log 2log 175log 240.2281xxxxxxx++==+ =+ == = SSM: Intermediate AlgebraHomework 5.3 121 51. ( ) ( )( ) ( )( )( )( )3 2 13 2 15 15 14 4 1004 1004 100log 4 log 1005 1 log 4 225 1log 425 1log 421log 450.8644x xx xxxxxxxx+ ==== = == + + = 53.3 8x= Empty set solution. 3 raised to any power will always be positive. 55.log (81) 4b= 41481813bbb=== 57.3log ( ) 4 x = 4381xx== 59. 4log ( ) 3 x = 3464xx== 61. 7log ( ) 1 x = 177xx== 63.7 0x=Empty set solution. 7 raised to any power will always be positive, never zero. 65. 27131log ( 1)327 11 34xxxx == == 69. 21log 122212 21xxxx = === 69. ( )515log 2 5221.1487bbbb=== 71.Line 3:log[3(8 )] log[3(8)]xx The power property for logarithms would only work in this case if both the 3 and the 8 were raised to the x power. The first step the student should have made was to divide both sides of the equation by 3. 3(8 ) 7783xx== Then take the log of both sides. Homework 5.4SSM: Intermediate Algebra 122 73. ( )( )( )log loglog logloglogxxxxab cab ccbacbacx bacaxb=== = = = 75. ( )( )( )( )log loglog logloglogloglogkxkxkxkxab d cab c dc dbac dbac dkx bac dakxbc daxk b+ == = = = = = 77. (4)(4) 4 256 f = =79. ( ) ( )( ) ( )( )( )4 3log 4 log 3log 4 log 3log 3log 40.7925xxxxx==== 81. 2(8) log (8) 3 g = =83. 2log ( ) 5 a = 5232aa== 85. 0log ( 5 ) 0 log (5 )b bk k The power property for logarithms would only work in this case if both the 5 and the k were raised to the 0 power. 87.Answers may vary. Example: When solving some equation for x, where x is in the exponent of some base b, first isolate the base b on one side of the equation. Take the log of both sides of the equation. Use the power property of logarithms to bring the exponent x in front of the log(b). Divide both sides by log(b). Isolate x on one side of the equation. Homework 5.4 1.a.( ) V f t =We know we can model the situation well by using an exponential model( )tf t ab = . The V-intercept is (0, 2000), so a = 2000 and ( ) 2000tf t b = . Since the interest rate is 5%, the base must be 1.05, so( ) 2000(1.05)tf t = . b.The V-intercept is (0, 2000). The original (t = 0) value in the account is $2000. c. 2000(1.05) 3000t= 1.05 1.5log(1.05 ) log(1.5)log(1.05) log(1.5)log(1.5)log(1.05)8.3104ttttt==== The balance will be $3000 after 8.31 years. SSM: Intermediate AlgebraHomework 5.4 123 3.a( ) 1.197(1.0163)tf t = 10 1.197(1.0163)101.01631.19710log(1.0163 ) log1.19710log(1.0163) log1.19710log1.197log(1.0163)131.2896tttttt== = = = b. ft ( )t1.19730.37Billions of People200Yea rs ft ( )t1.19730.37Billions of People200Yea rs10 There is model breakdown for the years after 2030. 5. a. a = 1.46. There was one Starbucks store in 1980. b.No. From 1971 (one store) to 1987 (17 stores) is a difference of both 16 years and 16 stores. This means that during that time the number of Starbucks stores was increasing at a rate of one per year, which is linear growth, not exponential. c.b = 1.48. The number of stores is growing exponentially by 48% per year. d.1000 stores per state = 50,000 stores 1.46(1.48) 50, 00050, 0001.481.4650, 000log(1.48 ) log1.4650, 000log(1.48) log1.4650, 000log1.46log(1.48)26.63tttttt== = = = The model predicts there will be 50,000 stores in 2007. e.1.46(1.48) 105, 000, 000t= 105, 000, 0001.481.46105, 000, 000log(1.48 ) log1.46105, 000, 000log(1.48) log1.46105, 000, 000log1.46log(1.48)46.1456ttttt= = = = The model predicts there will be 105 million stores in 2026. 7.a. Exponential function. The data points appear to fit an exponential curve better than a line. Homework 5.4SSM: Intermediate Algebra 124 b.Use exponential regression. ( ) 1.60(1.018)tf t =c. 100(100) 1.60(1.018) 9.53 f = =In 1900 the percentage of food expenditures devoted to food away from home was 9.5%. d. 1(100) f 100 1.60(1.018)1.018 62.5log(1.018 ) log(62.5)log(1.018) log(62.5)log(62.5)log(1.018)231.79tttttt===== In 2032, all food expenditures will be devoted to food away from home. e.Solution is the same as that in part d. 231.79 t . In 2032, all food expenditures will be devoted to food away from home. 9.a. Use exponential regression. ( ) 0.36(1.0036)( ) 0.037(1.0049)ssEsRs== b. 1425(1425) 0.36(1.0036) 60.3% E = 1425(1425) 0.037(1.0049) 39.2% R = c.Half = 50% Early decision applicants: ( )( )( )( ) 500.36(1.0036) 50501.00360.3650log 1.0036 log0.3650log 1.0036 log0.3650log0.36log 1.00361373sssEssss=== = = = Regular decision applicants: ( )( )( )( ) 500.037(1.0049) 50501.00490.03750log 1.0049 log0.03750log 1.0049 log0.03750log0.037log 1.00491475sssRssss=== = = = d.1475 1373 = 102 points e. (1757.58, 199.19); Students who score 1758 points on their SAT have the same change (199%) of being selected from the early SSM: Intermediate AlgebraHomework 5.4 125 decision and regular decision systems. Model breakdown has occurred.11.a.We know we can model the situation well by using an exponential model( )tf t ab . Let ( ) p f t , the total number of people. The p-intercept is (0, 30), and the base is 2. ( ) 30(2)tf t b.( ) 287, 000, 000 30 2t ( )( )( )( )287, 000, 000230287, 000, 000log 2 log30287, 000, 000log 2 log30287, 000, 000log30log 223.19ttttt _ , _ , _ , The model predicts that all Americans will have heard the rumor after 24 days. It is likely that model breakdown wil occur at f(t) approaches 271 million.13.a.Use an exponential decay model ( )dT f d ab . The T-intercept is (0, 8) so a = 8. For every increase of 5 decibels d, the exposure time is halved so the base is 12 and the exponent is 5d. 51( ) 82df d _ , b. 2451(24) 8 0.28722f _ , The model predicts that at 114 decibels, the bands could play for 0.29 hours (about 17.4 minutes!) without the fans experiencing hearing loss. The average rock concert lasts longer than 18 minutes, so the model predicts that these fans experience hearing loss. c. 1(3) f 55513 821 32 81 3log log2 81 3log log5 2 83log81 5log235log81log27.08dddddtt _ , _ , _ _ , , _ _ , , _ , _ , _ , _ , To play for 3 hours, the rock bands should play at 97 decibels. 15.Use an exponential decay model ( )tf t ab . Let ( ) P f t be the percentage of carbon-14 remaining in the wood. The P-intercept is (0, 100). Every 5730 years the amount of carbon-14 is halved so the base is 12 and the exponent is 5730t. 15730 5730 1 1( ) 100 100 100(0.999879)2 2tttf t 1 _ _ 1 1 , 1 , ]90% of the carbon-14 remains, so P = 90 = f(t) ( )( )( )90 100(0.999879)0.999879 0.9log(0.999879) log 0.9log(0.999879) log 0.9log 0.9log(0.999879)870.70tttttt The age of the wood is approximately 871 years.Homework 5.5SSM: Intermediate Algebra 126 17.Use an exponential decay model ( ) .tf t ab Let ( ) C f t be the concentration of alcohol in the blood (in milligrams per milliliter). The C-intercept is (0, 0.28). Every 2 hours, the amount of alcohol is halved so the base is 12 and the exponent is 2t 12 2 1 1( ) 0.28 0.28 0.28(0.7071)2 2tttf t 1 _ _ 1 1 , 1 , ] Find 1(0.08) f 0.08 0.28(0.7071)0.080.70710.280.08log(0.7071) log0.280.08log(0.7071) log0.280.08log0.28log(0.7071)3.6146tttttt _ , _ , _ , She must wait 3.6 hours before driving. 19.a.Use an exponential decay model ( ) .tf t ab Let( ) P f t be the percentage of radium that remains in the tank after t years. The P-intercept is (0, 100). Every 1600 years the amount of radium is halved so the base is 12 and the exponent is1600t. 16001( ) 1002tf t _ , b. 160016001600110 100210.121log log(0.1)21log 11600 211 1600log216001log25315.08ttttttt _ , _ , _ , _ , _ , _ , 10% of the radium will remain after 5315 years. 21.Answers may vary. Example: 1.Look at a scattergram for the data or how the data changes to determine what type of equation to use to model the data. 2.Find an equation for the model f. 3.Verify the results of f agree with the data 4.Use f to make estimations or predictions.If the model is exponential, the power property would be used in step 4 when calculating 1f Homework 5.5 1.log (2 ) log ( )b bx x + 2log [2 ( )]log (2 )bbxxx SSM: Intermediate AlgebraHomework 5.5 127 3.log (10 ) log (2)b bx 10log2log (5 )bbxx = = 5. 3log ( ) log (3 )b bx x + 334log ( ) log (3 )log [ (3 )]log (3 )b bbbx xx xx= +== 7.4log (2 ) 7log ( )b bx x + 4 74 711log (2 ) log ( )log [(2 ) ( )]log (16 )b bbbx xx xx= +== 9.log (8 ) 4log (2 )b bx x 4443log (8 ) log (2 )8log(2 )8log161log2b bbbbx xxxxxx= = = = 11. 3 2log ( ) 3log ( )b bx x + 3 2 33 2 33 69log ( ) log ( )log [( )( ) ]log ( )log ( )b bbbbx xx xxxx= +=== 13. 2 32log (3 ) 4log ( )b bx x 2 2 3 42 23 44128log (3 ) log ( )(3 )log( )9log9logb bbbbx xxxxxx= = = = 15. 6 6log ( ) log (4 ) 2 x x + = 6262 2212log [ (4 )] 2log (4 ) 24 636493x xxxxxx====== 17. 24 4log ( ) log ( ) 3 x x + = 24343 3133log [( )( )] 3log ( ) 34(4 )4x xxxxx===== 19. 2 37 73log ( ) log ( ) 3 x x + = 2 3 37 72 3 37979 313913log ( ) log ( ) 3log [( ) ( )] 3log ( ) 37(7 )71.9129x xx xxxxxx+ ====== Homework 5.5SSM: Intermediate Algebra 128 21.3log(5 ) 4log(2 ) 3 x x + = 3 43 43 477 3717log(5 ) log(2 ) 3log[(5 ) (2 ) ] 3log[(125 )(16 )] 3log(2000 ) 32000 1010002000120.9057x xx xx xxxxxx+ ====== = 23. 2 34 45log (2 ) 2log (2 ) 3 x x = ( )2 5 3 24 42 54 3 2104 6444 3414log (2 ) log (2 ) 3(2 )log 3(2 )32log 34log [8 ] 38 464881.6818x xxxxxxxxxx = = = ==== 25. 4 32 2log ( ) log ( ) x x + 4 3272log [( )( )]log ( )x xx== 27. 4 32 2log ( ) log ( ) 4 x x + = 4 32727 4147log [( )( )] 4log ( ) 42(2 )1.4860x xxxxx==== 29. 5 49 94log (2 ) 2log ( ) 2 x x + = 5 4 4 29 95 4 4 2920 8928928 228128log (2 ) log ( ) 2log [(2 ) ( ) ] 2log [(16 )( )] 2log (16 ) 216 9811681161.0596x xx xx xxxxxx+ ====== = 31. 5 49 94log (2 ) 2log ( ) x x + 5 4 4 29 95 4 4 2920 89289log (2 ) log ( )log [(2 ) ( ) ]log [(16 )( )]log (16 )x xx xx xx= +=== 33.( )3log(7)log 7 1.7712log(3)= 35.( )6log(1000)log 1000 3.8553log(6)= 37.( )5log(41.2)log 41.2 2.3104log(5)= 39. 81log1 70log 2.043170 log(8) = 41. 272log ( )log ( )log (7)xx 43. log ( )log ( )log ( )bsbrrs45.All three students did the problem correctly. SSM: Intermediate AlgebraHomework 5.6 129 47. 5(13) log (13) f log(13)log(5)1.5937 49. 51 1log2 2f _ _ , , 1log2log(5)0.4307 _ , 51.( ) ( )1217 log 17 g ( ) log 17log(12)1.1402 53.( ) ( )1237 log 37 g ( ) log 37log(12)1.4531 55. 2log ( ) 2bb 624log log ( ) 2b bbbb _ , 6log ( ) 6bb 66 442log ( ) log ( ) loglog ( )2b b bbbb bbb _ , 64log ( ) 61.54 log ( )bbbb The expressions 2log ( )bb , 64logbbb _ ,, 2, and 6 4log ( ) log ( )b bb b are all equal. 57.1log log ( )b bxxy y 1 _ _ 1 , , ] Use the property of logarithms. 11log ( ) loglog ( ) log ( )b bb bxyx y _ + , + Use the power property for logarithms. ( )( )log ( ) ( 1) loglog ( ) logb bb bx yx y + This is the quotient property for logarithms. ( ) log log ( ) logb b bxx yy _ , 59.a. 3 52 2log ( ) log ( ) x x + 3 5282log [( )( )]log ( )x xx b. 3 52 2log ( ) log ( ) 7 x x + ( )3 52828 7178log [( )( )] 7log ( ) 7221.8340x xxxxx c.Simplifying an expression involves combining separate logarithms into one logarithm. Solving the equation involved simplifying and the using the definition of logarithm to modify the statement into an exponential equation to solve. d.Simplifying an expression is the first step towards being able to solve an equation. Homework 5.6 1.ln(7) 1.9459 3.ln(54.8) 4.0037 Homework 5.6SSM: Intermediate Algebra 130 5. ln(0.8) 0.2231 7. 1ln 0.69312 9. 4ln( ) 4 e =11. 1ln( ) ln( ) 1 e e = =13. 11ln ln( ) 1 ee = = 15. 61 1ln( ) (6) 32 2e = =17.ln( ) 2 x = 2x e =19.20xe = ln(20)2.9957xx= 21.ln( 5) 3 x + = 335515.0855x ex ex+ == 23. 271xe = ln(71) 2ln(71)22.1313xxx== 25.7 44xe = 44744ln71.8383xexx= = 27.5ln(3 ) 5 x = 1ln(3 ) 1330.9061xx eexx=== 29. 3 14 68xe= 3 1173 1 ln(17)3 ln(17) 1ln(17) 131.2777xexxxx= == ++= 31.4 90x= ln(4 ) ln(90)ln(4) ln(90)ln(90)ln(4)3.2459xxxx=== 33.3.1 49.8x= ln(3.1 ) ln(49.8)ln(3.1) ln(49.8)ln(49.8)ln(3.1)3.4541xxxx=== 35.3(6 ) 1 97x = 3(6 ) 98986398ln(6 ) ln398ln(6) ln398ln3ln(6)1.9458xxxxxx== = = = SSM: Intermediate AlgebraHomework 5.6 131 37.ln(4 ) ln(3 ) x x + 2ln[(4 )(3 )]ln(12 )x xx== 39. 4 3ln(25 ) ln(5 ) x x 4325ln5ln(5 )xxx = = 41. 42ln(3 ) 3ln(2 ) x x + 4 2 38 38 311ln(3 ) ln(2 )ln(9 ) ln(8 )ln[(9 )(8 )]ln(72 )x xx xx xx= += +== 43. 3 23ln(3 ) 2ln(3 ) x x 3 3 2 23 9 2 43 92 45ln(3 ) ln(3 )ln(3 ) ln(3 )3ln3ln(3 )x xx xxxx= = = = 45.ln(2 1) ln(3) x + + ln[3(2 1)]ln(6 3)xx= += + 47.ln(3 ) ln( ) 4 x x + = 22 442142ln[(3 )( )] 4ln(3 ) 43334.2661x xxx eexexx==== = 49. 5 2ln(4 ) 2ln( ) 5 x x = 5 2 25 45455ln(4 ) ln( ) 5ln(4 ) ln( ) 54ln 5ln(4 ) 54437.1033x xx xxxxx eexx = = = === 51. 4 35ln( ) 2ln(4 ) 8 x x + = 4 5 3 220 620 62626 88261826ln( ) ln(4 ) 8ln( ) ln(16 ) 8ln[( )(16 )] 8ln(16 ) 81616161.2227x xx xx xxx eexexx+ =+ ===== = 53. 8 3ln( ) ln( ) x x ( )835lnlnxxx = = 55. 8 3ln( ) ln( ) 4 x x = ( )( )8355 4145ln 4ln 42.2255xxxx ex ex = === 57. ln( ) 1 e =because 1e e =Homework 5.6SSM: Intermediate Algebra 132 60. bxae c =lnlnbxceacbxacaxb= = = 61.3ln( ) 3ln( ) x x = 77 4 34ln( ) ln( ) ln ln( ) 3ln( )xx x x xx = = = 4774ln( )ln ( )ln( )xxxx= 22ln( ) ln( ) 2[ln( )] x x x = 3ln( ) 3ln( ) x x = ln(3 ) ln(3 ) x x =3ln( ) x , 7 4ln( ) ln( ) x x , and 3ln( ) xare all the equal. 63. a.The time when the person bought it is t = 0. 0.66(0) 070 137 70 13770 137207e e+ = += += The temperature was 207F when the coffee was purchased. b. 0.06180 70 137te= + ( )0.060.060.06110 137110137110ln ln1371100.06 ln( ) ln1371100.06 ln137110ln1370.063.6583ttteeet ettt== = = = = He will be able to drink the coffee in 3.66 minutes. c. Looking at the graph of y we can see that as t gets larger, y approaches 70. So the temperature of the store is 70F. 65. a.The poles are at x = 10 or x = -10 0.03(10) 0.03(10)(10) 10( )10(1.34986 0.740818)10(2.090677)20.90677h e e= += +== The poles are 20.9 feet high. b. 0.03(6) 0.03(6)(6) 10( ) h e e= + (6) 10(1.197217 0.835702)10(2.0324876)20.32488h = +==SSM: Intermediate AlgebraChapter 5 Review Exercises 133 The cable is 20.32 feet high when it is 6 feet away from the center. c.The shortest height happens in the center, where x= 0. 0.03(0) 0.03(0)(0) 10( )10(1 1)10(2)20h e e= += +== The least height of the cable is 20 feet. 67. Answers may vary. Example: When solving some equation for x, where x is in the exponent of some base b, first isolate the base b on one side of the equation. Take the log of both sides of the equation. Use the power property of logarithms to bring the exponent x in front of the log(b). Divide both sides by log(b). Isolate x on one side of the equation. Chapter 5 Review Exercises 1.( ) 3 f x x = 111Replace( ) with:3Solve for:13Replace with( ) :1( )3Write in terms of:1( )3f x yy xxx yx f yf y yxf x x==== 2. 4 7( )8xgx= 111Replace( ) with:4 78Solve for:4 7 84 8 78 74Replace with( ) :7( ) 24Write in terms of:7( ) 24gx yxyxx yx yyxx g yg y yxg x x= == ++== += + 3.( ) 3xhx = 13( ) log ( ) h x x=4. 4( ) log ( ) f x x =( ) 4xf x =5.( ) 10xgx = 1( ) log( ) g x x=6.( ) log( ) hx x = 1( ) 10xh x=7.(0) 1 f =8.(2) 4 f =9. 1(0) 4 f=10. 1(2) 1 f= Chapter 5 Review ExercisesSSM: Intermediate Algebra 134 11. 12. 13. 25log (25) 2,since 5 25 = =14. 5log(100, 000) 5,since 10 100, 000 = =15. 231 1log 2,since 39 9 = = 16. 331ln ln( ) 3 ee = = 17. ( )1261log 6 ,since 6 62= =18. ( )13 3341log 4 ,since 4 43= =19. 3log(0.001) 3,since 10 0.001= =20. 3log(7)log (7) 1.7712log(3)= 21.ln(5) 1.6094 22. 1log ( ) 1,since bb b b = =23. 7 7 7log ( ) 7,since bb b b = =24. 0log (1) 0,since1bb = =25.log ( )df t =26. ry w =27. 28.6(2) 30x= 2 5log(2 ) log(5)log(2) log(5)log(5)log(2)2.3219xxxxx==== 29. 3log ( ) 4 x = 4381xx== 30.4 75xe = 75475ln( ) ln475ln42.9312xxeexx= = = SSM: Intermediate AlgebraChapter 5 Review Exercises 135 31.4.3(9.8) 3.3 8.2x = 4.3(9.8) 11.511.59.84.311.5log(9.8) log4.311.5log(9.8) log4.311.5log4.3log(9.8)2.8333xxxxxx== = = = 32.log(83) 6b= 61683832.0886bbb== 33.3ln( ) 7 1 x + = 23ln( ) 6ln( ) 2xxx e = == 34. 22 2log( ) log(2) 4 x x + = ln(2.4) ln(63.5)ln(2.4) ln(63.5)ln(63.5)ln(2.4)4.7415xxxx=== 35. 5log( ) 3.17 x = 3.175164.3368xx= 36. 3 75(4) 40x= 3 73 74 8log(4 ) log(8)(3 7) log(4) log(8)log(8)3 7log(4)log(8)3 7log(4)log(8)7log(4)32.8333xxxxxxx== = == ++= 37. 3 812xe= 3 8 ln(12)3 ln(12) 8ln(12) 833.4950xxxx == ++= 38.log(32) 5b= 51532322bbb=== 39. 6 65log(2) 3log(4) 2 x x = 5 36 656 356 32622212log(2) log(4) 2(2)log 2(4)32log 264log 226272728.4853x xxxxxxxxxx = = = = === Chapter 5 Review ExercisesSSM: Intermediate Algebra 136 40.ln(3 1) 2 x + = 2223 13 1132.1297x ex eexx+ == = 41. 3 32log (5 ) 4log ( ) 5 x x + = 2 43 32 43636 5616log (5 ) log ( ) 5log [(5 ) ( ) ] 5log [25 ] 525 324325243251.4609x xx xxxxxx+ ===== = 42. 8 7ln(4 ) ln(2 ) 5 x x = 87554ln 52ln(2 ) 52274.2066xxxx eexx = === 43. 2 2log (log ( )) 2 x = 2224log ( ) 2log ( ) 4216xxxx==== 44. 645485xx= 225 48log(5 ) log(48)2 log(5) log(48)log(48)2log(5)log(48)2log(5)1.2027xxxxxx===== 45.log ( ) log (6 ) log (2 )b b bx x x + ( )22log [ (6 )] log (2 )log (6 ) log (2 )6log2log 3b bb bbbx x xx xxxx= = = = 46.3log (2 ) 2log (3 )b bx x + 3 23 25log (2 ) log (3 )log [(2 ) (3 ) ]log [72 ]b bbbx xx xx= +== 47. 3 53ln(4 ) 3ln(2 ) x x + 3 3 5 39 159 1524ln(4 ) ln(2 )ln(64 ) ln(8 )ln[(64 )(8 )]ln(512 )x xx xx xx= += +== 48. 2 54log (3 ) 2log (9 )b bx x 2 4 5 22 45 28102log (3 ) log (9 )(3 )log(9 )81log811logb bbbbx xxxxxx= = = = SSM: Intermediate AlgebraChapter 5 Review Exercises 137 49. 7 3ln(2 ) 4ln(2 ) x x 7 3 47 127125ln(2 ) ln(2 )ln(2 ) ln(16 )2ln161ln8x xx xxxx= = = = 50. log ( )log ( )log ( )bybwwy=51. 5 2log ( ) log ( ) 5 2 3b bb b = = 52log ( ) 52 log ( )bbbb= 3log ( ) 3bb = 5log ( ) 5bb = ( )532log log 3b bbbb = = 52.a.Use an exponential model( )tf t ab = . Let ( ) V f t = , where V is the value of the account. The V-intercept is (0, 8), so a = 8. Since the interest rate is 5%, the base is b = 1.05. ( ) 8(1.05)tf t = . b. 9(9) 8(1.05) 12.41063 f = The balance in the account after 9 years is $12, 410.63. c. The balance will have doubled when it is $16,000. Find 1(16) f 16 8(1.05)1.05 2log(1.05 ) log(2)log(1.05) log(2)log(2)log(1.05)14.2067tttttt===== The balance will be doubled in 14.2 years. 53.a.Use an exponential model( )tf t ab = . Let ( ) n f t = be the number of leaves. The n-intercept is (0, 30), so a = 30. The number of leaves doubles every week so the base, b = 2. ( ) 30(2)tf t = . b. 8(8) 30(2) 7680 f = = There are 7680 leaves on the tree 8 weeks after April 1. c.1(5000) f 5000 30(2)5000230500log(2 ) log3500log(2) log3500log3log(2)7.38tttttt== = = = There are 5000 leaves on the tree 7.38 weeks after April 1. 54.a. Use an exponential model because the data points curve upwards. b.Using exponential regression: ( ) 0.083(2.65)tf t =c.2.649 b =Every year the sales of Nantucket Nectar Juice Drinks goes up by 165%.Chapter 5 TestSSM: Intermediate Algebra 138 To find the percent growth, take the growthfrom one year to the next and divide by thesales from the previous year..1 .3 .5 511.515.1 .2 .5 19 15 301.5116 15 30= = = == = = From this the greatest percent growth can be seen to have happened in 1994 when there was a 500% increase. d.1(1000) f 1000 0.083(2.65)10002.650.0831000log(2.65 ) log0.0831000log(2.65) log0.0831000log0.083log(2.65)9.65tttttt== = = = Sales will be $1 billion in 2000. 55.a. Using exponential regression: ( ) 9.33(1.31)nf n =b.1.31; As each cassette is added to the bag the length increases by 31%c. 9.33; The initial length of the rubber band is 9.33 inches. d. 8(8) 9.33(1.31) 80.92 f = The rubber band is stretched to 80.92 inches with 8 cassettes.There are two scenarios which might cause model breakdown. The rubber band reaches a point where it can stretch no farther, or the rubber band breaks.e. 1(139) f 139 9.33(1.31)1391.319.33139log(1.31 ) log9.33139log(1.31) log9.33139log9.33log(1.31)10.00nnnnnn== = = = It would take 10 cassettes to stretch the rubber band to 139 inches. If model breakdown occurs with 8 cassettes, then it definitely occurs with 10. Either the rubber band has stopped stretching with the addition of the last two cassettes, or the rubber band is broken. Chapter 5 Test 1.( ) 2 9 gx x = 111Replace( ) with:2 9Solve for:2 992Replace with( ) :1 9( )2 2Write in terms of:1 9( )2 2gx yy xxx yyxx g yg y yxg x x= = ++== += + 2.( ) 4xhx =SSM: Intermediate AlgebraChapter 5 Test 139 14( ) log ( ) h x x=3. 5( ) log ( ) f x x = 1( ) 5xf x=4. 5. 42log (16) 4,since 2 16 = =6. 341 1log 3,since 464 64 = = 7. 7log(10)log (10) 1.1833log(7)= 8. 1log(0.1) 1,since 10 0.1= =9. 2 2log (log(10, 000)) log (4) 2 = = 4 2Since 10 10, 000 and 2 4 = =10. 0log (1) 0,since1bb = =11. 3 3 3log ( ) 3,since bb b b = =12. 121log ( ) ,since 2bb b b = =13. 22 21 1ln 2,since ee e = = 14.log (log ( )) log (1) 0b b bb = = 15.( ) log 50 4b=( )41450502.6591bbb== 16.6(2) 9 23x = 6(2) 32322632log(2 ) log632log(2) log632log6log(2)2.4150xxxxxx== = = = 17.( ) ( )3 3log log 2 5 x x + = 3232 5212log [ (2 )] 5log (2 ) 52 32432243211.0227x xxxxxx==== = 18. 3 12 54xe= 3 13 127ln( ) ln(27)(3 1) ln ln(27)3 1 ln(27)3 ln(27) 1ln(27) 131.4319xxeex exxxx== = == ++= Chapter 5 TestSSM: Intermediate Algebra 140 19. 5 24 42log (3 ) 3log ( ) 3 x x = ( )5 2 2 34 410 64 4104 6444 3414log (3 ) log ( ) 3log (9 ) log ( ) 39log 3log 9 39 46496491.6330x xx xxxxxxxx = = = === = 20.7ln( 2) 1 4 x = 57577ln( 2) 55ln( 2)7224.0427xxx ex ex = = == + 21.log ( )sw t =22. pe k =23. 3log ( ) log (5 )b bx x + 34log [ (5 )]log (5 )bbx xx== 24. 3 24ln( ) 5ln(2 ) x x 3 4 2 512 1012102ln( ) ln(2 )ln( ) ln(32 )ln32ln32x xx xxxx= = = = 25.a.Use an exponential model( )tf t ab = . Let ( ) P f t = be percentage of carbon-14 remaining in the wood. The P-intercept is (0, 100). The base is and the exponent is 5730t.57301( ) 1002tf t = . b.Find 1(40) f 573057305730140 100210.421log log(0.4)21log log(0.4)5730 25730log(0.4)1log27574.65tttttt = = = = = The wood is 7575 years old. 26. a. Use exponential regression. ( ) 0.37(1.77)tf t =b.0.37; In 1990 there were 370 arrests. c.b = 1.77; The number of arrests increases by 77% each year. SSM: Intermediate AlgebraCumulative Review Chapters 1-5 141 d.Find 1(1000) f 1000 0.37(1.77)10001.770.371000log(1.77 ) log0.371000log(1.77) log0.371000log0.37log(1.77)13.84tttttt== = = = According to the model, there will be 1 million arrests in 2004. This seems unlikely, model breakdown has probably occurred. Cumulative Review Chapters 1-5 1. 5 12(4) 17x= 5 15 117(4)217log(4 ) log217(5 1) log(4) log217log25 1log(4)17log25 1log(4)17log21log(4)50.5087xxxxxxx= = = = = + += 2. 3log ( 5) 4 x = 45 381 586xxx == += 3. 73 18 7 b = 77173 252532531.3538bbbb== = 4. 7 2 512 3 4x x = 7 2 7 5 712 3 12 4 122 83 123 2 2 32 3 3 21x x x xxxx = = = = 5.8 2 15xe + = 2 7727ln( ) ln27ln21.2528xxxeeexx== = = 6.( ) log 93 4b=( )41493933.1054bbb== Cumulative Review Chapters 1-5 SSM: Intermediate Algebra 142 7. 2 45 54log (3 ) 3log (6 ) 3 x x + = 2 4 4 35 58 125 58 12520520 320120log (3 ) log (6 ) 3log (81 ) log (216 ) 3log [(81 )(216 )] 3log (17496 ) 317496 512517496125174960.7811x xx xx xxxxxx+ =+ ===== = 8.7 3(4 2) 5 1 x x = + 7 12 6 5 113 12 5 112 5 1217 1212170.7059x xx xx xxxx + = + = + = = = 9. 9 32ln(12 ) ln(3 ) 5 x x = 9 2 318 31831515 55151515ln(12 ) ln(3 ) 5ln(144 ) ln(3 ) 5144ln 53ln(48 ) 54848481.0782x xx xxxxx eexexx = = = === = 10.Substitute2 5 x y = for x in4 5 14 x y = 4(2 5) 5 148 20 5 143 62y yy yyy = = == 11.Simplify3(2 4 ) 10 2 x y = : 6 4 10 24 2 16x yx y = + = In order to eliminate x, multiply2 3 8 x y = by 2. This gives4 6 16 x y = . Add the two equations: 4 2 164 6 16x yx y + = = This yields4 32 y = , so y = 8. Solve for x: 4 6(8) 324 48 324 164xxxx = = == 12.In order to eliminate x, multiply 1 3122 4x y = by 53. This gives 5 5 606 4 3x y+ = . Add the equations: 5 5 606 4 35 116 3x yx y+ =+ = This yields 5 1 6014 3 3y y + = . The common denominator is 12, so the equation changes to 15 4 240 1212 12 12 12y y + = . So, 191912 y =and y = 12. Solve for x when y = 12. 1 3(12) 122 419 1221326xxxx = = = = SSM: Intermediate AlgebraCumulative Review Chapters 1-5 143 13.8 3 3(4 5) x x 8 3 12 1520 181820910x xxxx + )9,10

14. 3 2 3 7 1 2(4 ) (5 ) b c b c 9 6 14 29 14 6 223 4423(64 )(25 )(64 25)( )( )16001600b c b cb b c cb ccb 15. 1 13 21 32 486bcb c 1 11 33 22 42 3 2 36 6 4 45 56 4565443434343b cb cbcbc _ ,+ 16. 35 82 723b cb c _ , 35 ( 2) 8 733 131339232323827b cb ccbcb _

, _

, _

, 17. ( ) ( )7 24log 2 2log 7b bx x ( ) ( )( ) ( )4 27 228 428424log 2 log 7log 16 log 4916log4916log49b bb bbbx xx xxxx _ , _ , 18. 6 23ln(4 ) 4ln(3 ) x x + 6 3 2 418 818 826ln(4 ) ln(3 )ln(64 ) ln(81 )ln[(64 )(81 )]ln(5184 )x xx xx xx + + 19.( ) 5(3)xf x 20.( ) 3 25 gx x +21.7 22.4023.824. 1(5) 0,since(0) 5 f f 0 x 910 Cumulative Review Chapters 1-5 SSM: Intermediate Algebra 144 25.( ) kxis an exponential function( )xkx ab =with base b = . The y-intercept is (0, a). To find a, substitute a point from the table for x and y and solve. 311602116081280aaa = = = So the y-intercept is (0, 1280). 26. 27. 28. 29. 30. 31. First find the slope: 3 7 105 4 9m = = + So 109y x b = + . Substitute a point, say (-4. 7) to solve for b. 107 ( 4)94079407963 409 9239bbbbb= += + = == So the equation is 10 239 9y x = + . SSM: Intermediate AlgebraCumulative Review Chapters 1-5 145 32. The y-intercept is (0, 8.1). So the equation is 8.1xy b = . Substitute (20, 719.3) to find b. 2020120719.3 8719.38719.381.25bbbb== = So the equation is8.1(1.25)xy =33. Both points satisfy the equationxy ab = so we have the system: 731385abab== Combining the equations yields: 734141385138513850.6254ababbbb== = Substitute this value of the base into the equation(0.6254)xy a = . Substitute (3, 85) to find a. 3385 (0.6254)85(0.6254)347.49aaa== So the equation is347.49(0.63)xy =34.4( 4) 2(3) f =35. 2(3) 16x= 3 8log(3 ) log(8)log(3) log(8)log(8)log(3)1.8928xxxxx==== 36. (0, 2) because a = 2 37. 38. 39.1(35) f 2(3) 35353235log(3 ) log235log(3) log235log2log(3)2.6053xxxxxx== = = = 40. 43 41 1 1log 4,since 381 81 3 = = = 41.( ) ( )3 2 3log log 8 log (3) 1 = = 3 1since 2 8 and 3 3 = =42. ( )17 7 71log ,since 7bb b b = =Cumulative Review Chapters 1-5 SSM: Intermediate Algebra 146 43. 8log(73)log (73) 2.0633log(8)= 44.( ) 1 1 f =45.The y-intercept is (0, 2). So the equation is 2xy b = . Substitute (-1, 1) to find b. 1111 212122bbbb== = = So the equation is( ) 2(2)xf x =46. 47. 1(2) f ( )2 2(2)2 1log(2 ) log 1log(2) 00log(2)0xxxxxx====== 48.( ) 4 6 f x x = 111Replace( ) with:4 6Solve for:6 46464 4Replace with( ) :1 3( )4 2Replace with:1 3( )4 2f x yy xxy xyxyxx f yf y yy xf x x= + =+== += += + 49. 18( ) log ( ) g x x=50.Answers may vary. Example: 0 21 32 42 52 63 7x y The relation is not a function because the input x = 2 has three different outputs y = 4, 5, and 6. SSM: Intermediate AlgebraCumulative Review Chapters 1-5 147 51.a.The y-intercept for both functions is (0, 2). b.For f, as the value of x increases by 2, the value of y increases by 3. For g, as the value of x increases by 2, the value of y is multiplied by 3. c.g;Raising 3 to a large power (as happens in g for large x values) will yield a larger number than multiplying that number by 3 power (as happens in f). d. 52.a.First find the slope: 8 3 57 4 3m= = So 53y x b = + . Substitute (4. 3) to solve for b. 53 (4)3203320339 203 3113bbbbb= += + = == So the equation is 5 113 3y x = b. Both points satisfy the equation xy ab = so we have the system: 7483abab== Combining the equations yields: 743138383831.3867ababbbb== = Substitute this value of the base into the equation (1.3867)xy a = . Substitute (4, 3) to find a. 443 (1.3867)3(1.3867)0.81aaa== So the equation is0.81(1.39)xy = c. Cumulative Review Chapters 1-5 SSM: Intermediate Algebra 148 53.a.(2) 3(2) 6 f = = 2(2) 3 9 g = =b.Finding 1f 111( ) 3Replace( ) with:3Solve for:13Replace with( ) :1( )3Replace with:1( )3f x xf x yy xxx yx f yf y yy xf x x===== 13( ) log ( ) g x x=c. 11(81) (81) 273f= = 1 43(81) log (81) 4,since 3 81 g= = =54.a.( ) 0.69 19.95 Ux x = +( ) 0.45 29.95 Bx x = +b.Slope of U is 0.69; U-Haul charges $0.69 per mile Slope of B is 0.45; Budget charges $0.69 per mile c.0.69 19.95 0.45 29.95 x x + = + 0.24 101041.67 miles0.24xx== d.( ) ( ) Ux Bx < 0.69 19.95 0.45 29.950.24 1041.67x xxx+ < +