Chapter 7: Exponential and Logarithmic F y n c t i o n s L o s s e s . /.I hrplormrj I ^ -Ait ib 1 I the C h a r a c t e r i s t i c s of txpoii-'rr.ti.jl I ijr.ct'CJrts, p a g e ; i' I : _ ... 439 4o, linear fes 40, quadratic slo. cubic !-,--;;o 1 ! — > 5 5 * { Jo, quadratic 2. b) No xontercepts; y-^intercept: y = 1 End behaviour; Qll to Ql Domain; {x | x e R}; Range; (y | y > 0, y e R) e) No x-intercepts; yontercept; y = 1 End behaviour; Qll to Ql Domain; { x | x e R}; Range; {y | y > 0, y o R} C...' cl) Number of xontercepts; 0 yontercept: y = 4 Domain: {x | x e R} flange; {y | y > 0, y e R} End Behaviour; Qll to Ql 10 y y > 0, y e R} jhaviour Ql ' - oc/ 4' . 1 IC •/ " ilii L e s s o n 7.2: R e l a t i n g t h e C h a r a c t e r i s t i c s J4.4 of a n E x p o n e n t i a l F u n c t i o n to Its -N E q u a t i o n , page 448 l!l 1.a) l - . - l - iq ji_4_p_L_ [ it: j . I • f H4IS: Term { ! L Z I I M Z I Z I i- }" .TP b) Number of xdntercepts; 0 yontercept; y = 6 Domain; {x | x o R} Range: fv'1 K-'O, y.- R} End Behoviour Ojil to 04! 28^14 Ui 2-; Foundations of Mathematics 12 Solutions Manual 1 1 Sfe'Xjtid liiffererice rii^t Oift«?rencc 7 ^ 7 io ~ VT 1" " o_h O'i \ ^ •'7 22 f ' I noticed that the difference between consecutive y-values doubles from one pair of values to the next. I know that there must be a base value of 2 being raised to an exponent and then multiplied by some constant term in the eguation. The equation looks like y = a(2'). Thus, the answer is yes, because for each unit increase in x, the value of y doubles. 1 Terrri c) Number of x-interccpts: 0 yintercept; y = 27 Domain; {x | x o R}; Range: { y ! y > 0 , yc. R} End Behaviour; Qll to Ql r j Firi,l Difference 2304 57 n 144 36 9 Second Difference 3072 - 768 1/28 768 - 192 432 192 - 4 8 108 48 12 27 12 - 3 I noticed that the difference between consecutive y-values is not constant. The function is not 7-1 4 . a) The yontercept is 5 and the function is increasing. y = 5(2)', y = 5{2f» 1 noticed each y-^value c-. ^* the previous y-value as X increases by 1, The function must be an exponential function with a base value of -1 4 being raised to an exponent and then multiplied by some constant term in the equation. Thus, the answer is yes because for each unit increase in X , the value of y is divided by 4. 2 a) Number of xontercepts; 0 yontercept; y = 4 ; Domain- {x | x o R} Range; {y | y > 0, y o R} End Behaviour; Qll to Ql b) Number of xontercepts; 0 yontercept: y = 2; Domain: {x | x o R} Range; {y | y > 0, y o R}; End Behaviour: Qll to Ql c) Number of xontercepts; 0; yonterc< Domain; { x j x e R); Range: (y | y > 0, End Behaviour; Qll to Ql d) Number of xontercepts; 0; yontercept; y = 3 Domain: { x | x e R}; Range: {y | y > 0, y G R ) End Behaviour; Qll to Ql 3, e.g. Increasing exponential functions increase as X increases, whereas decreasing exponential functions decrease as x increases. fix rt'.iMiig 1 ) 1!" _[ • - t t E --4-4-L 1 •- - 4 • - 1 , I, I E 1 'Jf The function is increasing because the base is greater than 1, b) The yontercept is 2 and the function is decreasing. y = 2(0.6)" y = 2(0.5f^ y=2(1) The function is decreasing because the base is less than 1. c) The yontercept is 10 and the function is increasing. y= 10(1.5)' y= 10(1.5)^°^ y= 10(1,5) Jo The function is increasing because the base is greater than 1. d) The yontercept is 1 and the function is decreasing y=(0.4); y= 1 The function is decreasing because the base is less than 1. '% 1 j L y=5(1) 2'"* 5. a) i) Yes, an exponential equation can be used to model the function because the rate of change in y-values doubles for each unit increase in x ii) yontercept: y = 1, the function is increasing b) i) No, y increases by 2 as x increases by 1 li) y-intercept; y = 3, the function is increasing. c| i) Yes. an exponential equation can be used to model the function because the rate of change in y-values get divided by 4 for each unit increase in X. ii) yontercept; y = 64, the function is decreasing d) i) No. e.g., y decreases, then increases then decreases again. ii) yontercept: y = 1, the function is first decreasing than increasing as it reaches the yontercept. 7-2 Chaptet 7 Exponential a n d L o i a r i t h m i c F u n c t i o n s :-i:i-=H,tH- •„ x-intercepts: 0 i.^i. u'-pi V '.; Domain: {x I X 6 R} i. pv ' i •=,]/€ R} r =.o t;.-h..-'M)Ui Oil 1'. f H f , } f y , ; , , j , : -4 - H P - . -ph. '.mrr.-.iv I-:; I t..'l C< h,ivi--tH • 41 •4p I I fo.nor V intercept; y = 25 fv'!y>0,yeR} ')! 1 1 '.fl - . 4 .P i;. • If) rpiro f.. p:.;.r.| < ; , ,ro.K..:)'-~ 0 V intercept; y = 12 f x . n . - << y i r - r r c ••Of- fj. j y > 0 , y o R} F nd Behaviour: Ql I to Ql b) Number of xontercepts; 0; yontercept: y = 4 Domain: (x | x G R ) ; Range; f^ | y > 0, y o R } End Behaviour; Qll to Ql c) Number of xontercepts: 0; y-intercept: y = 2 Df.i..:4c fx I y-- P}. Px.cgo' tv j •/ > 0, y e R} d l Number of xontercepts; 0; yontercept; y = 3.5 Domain; {x | x e R}; Range: {y | y > 0. y e R} f e d B o h w i o o r Oil to Of l i F o u n d a t i o n s of Mathematics 12 S o i u t i o n s Manual . The y-^intercept is positive and since the base is 8, it must mean the function is increasing because the base is larger than 1. b| The yontercept is positive and since the base is 0.6, it must mean the function is decreasing because the base is less than 1. c) The yontercept is positive and since the base is e. it must mean the function is increasing because e is greater than 1. 8. a) Number of xontercepts; 0; yontercept; y = 4 Domain: {x | x e R}; Range: {y | y > 0, y o R} End Behaviour: Qll to Q! b) Number of xontercepts; 0; yontercept; y = 8 Domain; {x | x e R}; Range; {y | y > 0, y o R} End Behaviour; Qll to Ql c) Number of xontercepts; 0; yontercept: y = 3 Domain; {x | x e R}; Range: {y | y > 0, y e R} End Behaviour: Qll to Ql d | Number of xontercepts; 0; yontercept; y = 10 Domain; {x | x o R}; Range; {y | y > 0, y G R} End Behaviour; Qll to Ql e) Number of xontercepts; 0; yontercept; y = 30 Domain: {x | x G R } ; Range; {y | y > 0, y G R } End Behaviour: Qll to Ql f) Number of xontercepts; 0; yontercept; y = 1 Domain; {x | x G R } ; Range; {y | y > 0, y e R } End Behaviour: Qll to Ql g) Number of xontercepts: 0; yontercept: y = 3 Domain: (x | x o R ) ; Range: {y | y > 0, y e R } End Behaviour; Qll to Ql h) Number of xontercepts; 0; yontercept: y = 45 Domain; {x | x e R } ; Range; {y i y > 0, y G R } End Behaviour: Qll to Ql 7-3 i . a) It is a d e c r e a s i n g e x p o n e n t i a l f u n c t i o n b e c a u s e it i s a f u n c t i o n w i t h a n e x p o n e n t of x a n d the base is less than 1, b ) It is n o t a d e c r e a s i n g e x p o n e n t i a l f u n c t i o n b e c a u s e a l t h o u g h it h a s a n e x p o n e n t of x , t h e base is g r e a t e r than 1, c) It is a d e c r e a s i n g e x p o n e n t i a l function because it is a f u n c t i o n with an e x p o n e n t of x and the base is less than 1. 1 1 . a) It is an increasing function with a yontercept of (0.6). It matches with iii). b) It is a decreasing function with a yontercept of (0, 4), It matches with i). c) It is a decreasing function with a yontercept of (0. 2). It matches with ii). d) It is an increasing function with a yontercept of (0, 3). It matches with iv). i ) It is not a d e c r e a s i n g e x p o n e n t i a l f u n c t i o n 1? ,»| y ii.hji. ...pl 5, because a l t h o u g h it has an base is greater than 1. lh)hhw< 10 .Jj M I . exponent it|< ,•• , r i ' j '••,ir.Jir;i. of x, the H] <j H.t RHIIOC- / i v \ v -O.yo R} b O C a U S O itS o . r , . , -s .i-i)i>^ :''tv.> " I { x \ x • ; I i•m ' j • '! ! i ' l l i • • (4 I • If' -il B — m icprxpr b| It is a decreasing function because its base is m b} yontercept; y = 7; Base: 4 Domain- {x I x < R). Range; (y | y > 0, y c R} Infjoasinq or Decreasing- increasing l e s s than 1. -A c) yontercept: y = 6; Base; | - Ik Domain; {x | x G R ) ; Range: {y | y > 0, y e R} Increasing or Decreasing: Decreasing mm c) It is an increasing function because its base is more than 1. d) It is a decreasing function because its base is less than 1. 7-4 Chaptof 7. t x p o n e n t i a l a n d L o g a r i t h m i c F u n c t i o n s 15. Yes. because all eiponentiai functions have the form y=a(bf\ Polynomial functions are characterized by having a numerical exponent while exponential functions have variable exponents. R} ncre3sir • i... 1 , . , , . . "yT'TTT >• Trtttt' 11 r ' H\ l^ ; .|__L_j.. 1 1 14 . i l l Ii 14X4 'LI 1/ •f. <1 I thmtbfjt of i 4441 441 Ittt • Hi m i l - _L_L_k. 16. i ' M'ktivftt I ydnf^nctjp-l , /Pi c, ' Y-! > • c. 1 • . j f7.«--lM/iO«o ' L«'>Pinio _ j 1-0 J , . i 1 : i 471J 0' I a-1-1, 1- r>i , : pic 1 or «x;o-j ' ' I' (a , fK'.^;.. .4.- . I O.' rcept; y=2 ^ 3/ > 0. y G R} il! > 1) or jreasing liiii) 'Moreasing re le yontercept and whether the function increases or decreases are unique to the function. IIP a;«;. • -'^o'}., -4 4 ' ? = R} a; 2 b: 0.5 1J>; IC ' O -< f'.CS...'j UjlirhO'.O ii) R a n g e 4 y I y > 0, / Y; 1: 1 ft; 3 This is an increasing function. iii) Range; {y | y > 0, y e R} a; 3 fe; 0 5 This is a decreasing function. iv) Range; {y | y > 0, y e R} a; 2 ft; 4 This is an increasing function. b) i) This function matches with the graph B. because it is a decreasing function with a yontercept of 2. ii) This function matches with grop' I ; ^ .ocause it is an increasing function with a y r m rr r.pt of 1. iii) This function matches with the graph A. because it is a decreasing function with a yontercept of 3. iv) This function matches with the graph C. because it is an increasing function with a yontercept of 2. 14. a) e.g. An example of an increasing exponential function with a yontercept of 5 would be y = 5(2)'' while an example of a decreasing exponential function with a yontercept of 5 would b e y =5(0.25)1 b) e.g. Same; number of xontercepts, ydntercepts, and behaviour, domain, and range. Different: rate of change (increasing vs. decreasing function. Foundations of Mathematics 12 Solutions Manual 18. a) Student A; yontercept; y = 80 domain; {x | x > 0, x c R} range; {y | 0 < y < 8, y o R) Student B; yontercept; y = 100 domain; {x | x > 0, x e R} range; { y | 0 < y < 100, y e R } b) e.g., Concentration of caffeine in blood naturally decreases over time as the kidneys filter it from the blood into the urine. c) Student B consumed more caffeine. Student B processed the caffeine more guickly. This can be seen from how much they initially started with and how much they consumed in four hours. Student A had 80 mg of caffeine at the start and consumed 60 mg after four hours. Student B started with 100 mg of caffeine and within the same time penod, had consumed 80 mg. d) Both Student A and B had about 20 mg of caffeine in their body after four hours. e) The y-intercepts are the values representing the initial amount of caffeine in the students' bodies. The values are different because the students drank different amounts of the same dnnk, or they drank different dnnks that did not have the same concentration of caffeine. 7-5 19 a) i Houi i " I e s s o n 7 3: Modollincj D a t a U s i n g E x p o n e n t i a l F u n c t i o n s , p a g e 461 I 'In;, d.ito :.(d docs m,l involve exponential <lo(..iy bHc.uinc differences are t-un;>Liol t! v, a tuiK.tic.ri exponential growth tur. between the V unit value o f x . StiicJonts who are TolcJ (jiowlh Ol the linssof b) ihr; thiLi -.^ii dfu-.s involve licr:<uici: il c.oritJ.Hil rnhos volu.'-s wilh oa'.h '.onooguf-ot 1 _ \ rzziiizzz izzzzizi^ I ; r ; 4--i/ " _ \ \ I I O h| ,^rj. Ath<i .-.P.hliu.io !h.-fiuri(l)f;i .4 •.Iii4f;(il , wh<i .tn; lol4 H, thai i-rnji oh.ool liu-. Oi'^..f'i4.4iuo Oouijlf.-'. Cf V I h'J fliili.llfHl I-., y (/ ;,| l.'H) 90 H;-')' finr.infn fv j x o x «o ciorjf {./) y , 3, V k; /() cJ) AflOl 'j h, /riH Sfurj,-.,-.!;, CO WoolO ho tcilO 50 40 II 1 i;^^ 30 ±t A. I • w c) This data set does not involve exponential growth or decay because the differences are not constant, or a constant ratio. d) This data set does not involve exponential growth or decay the differences are not constant, or a constant ratio. X4 1 4- " 4 - 1 ?- 2. a) Using a graphing calculator, the exponential regression function for the data is y = 10.097...(0.200...)''. 2©. a) Yes. e.g., Any data with a constant doubling time can be expressed with an exponential function. b| Y = 4 ( 1 . 2 6 ) 1 a represents the initial number of reguests. b represents the rate of growth of the number of reguests, x represents the amount of time in hours since the news broke, y represents the total number of interview reguests. c) domain: {x | x > 0, x e N} range: {y I y > 4, y e N} d) ] rime o m. jryCOajin 1 U J O ami 1 2 OOpxTj. _ 1 0 0 () m . 2 J ) 0 p rrc 3 0 0 p m. P rn. Domain: {x I x o R}; Range: { y | y > 0 , yc 40 5 0 2 4 0 R} y-intercept: y = 10.097...; End Behaviour: Qll to Ql This function is decreasing and shows exponential decay. b) Using a graphing calculator, the exponential regression function for the data is y = 2.780...{1.054...)'', Domain: {x | x e R); Range: {y | y > 0, y e R) y-intercept: y = 2 . 7 8 0 . . . ; End Behaviour: Qll to Ql This function is increasing and shows exponential growth. >'. ')0 Y+YJO U 1 L 4 4 3. a) 0 01 ^ „ 1 0 1 12 ' J6JJ 2 0 2 i 1 Years sinci^ Retirement Rent ($) IZZZiZZZI \AZ 2 3 Ratios ) 9 9 6 0 1 ------- 1 1 0 3 4 4 1 . 0 3 8 . 1 0 752 1 . 0 3 9 . . Yes, an exponential model can represent the data since the ratio of consecutive pairs of y-values are close. , 7-6 Chapter / Exponential and Logarithmic Functions b) The equation will be of the form f(x) = a{bf. Thus, We can create the formula by setting a to the first rent value and b to the average ratio of consecutive pairs of Rvalues, x will be the number of years since retirement, f(x) will be the rent after x years. c) Using graphing technology, the regression equation i s y = 9595,433...(1.038...f. f(x) = 9595.433...(1038...f f(x) = 9595.433.,.(1,038...f f(x) = 9595.433...(1.459...) f(x) = 14 000.638... 1 predict the Belands will pay about $14 000 in 10 years. 6. a) Using a graphing calculator, the exponential regression eguation for the data is y = 14.429...(1.065...)'. b) y = 14.429...(1.065,.,)™ y = 14.429...(1,879...) 4. a) Rainbow Tro'.«t l.rj.gtl, vi. >Xa%' 900 y = 27.119... cm The height of the sunflower on day 10 was about 27,1 cm. 800 c) y = 14.429...(1.065...)" 1000' 700 I c) The population is estimated to be 8.05 billion. • . c . t this number, 50 (for 2020) was substituted for X in the regression equation so that y could be determined. d) The population is expected to reach 9.50 billion 61.5 years after 1970. that is, dunng the year 2031. I plotted y = 9.50 and the regression function on a graphing calculator. The x-coordinate of the point at which these two functions intersect is the point at which the population is expected to reach 9.50 billion. y = 14.429...(6.638...) y = 95.792... cm The height of the sunflower on day 30 was about 95.8 cm. The answer does not make sense since the height was 98 0 cm on day 28. rf| I wnuld expect tho sunflower to reach a height (}i (.m oo d.ov 20 500 400 300 200 100 0 150 250 Length (mm) 50 3 b) Using a graphing calculator, the exponential regression function for the data is y = 26.934...(1.008...)! c) e.g.. 690 g; I identified the point on the curve that had an x-value of 400. d) I would expect this fish to be about 446 mm long. I identified the point on the curve that had a y-value of 1000. 5. •:.) Populatioi; bf Y*.,.r 1 h 4 4 I .4 \ 2 7. a) Using graphing technology. I input the x and y values into the spreadsheet. I then used the exponential regression function to determine the eguation that models this growth. The eguation is y=2^. b) Under this context, the domain is {x | x o N), and the range is (y | y > 0, y c N}. c) If we extend the first and last column of the table, wc get the following: Number of Possible Number of Digits Codes 1 0 '0 20 50 40 Years since 1970 ' SO b) Using a graphing calculator, the exponential regression function for the data is y = 3.911...(1.014...)T Foundations of Mathematics 12 Solutions Manual J 4 8 2 .16 .32 64 128 7-7 Based on the table, 104 codes lies in between the codes for 6 digits and the codes for 7 digits. 6 digits would not provide enough codes. Therefore, 7 digits would be needed to have 104 codes. 8. a) Here is the graph. 1000 T 900 9. a) Time (years) 0 1 2 3 4 5 Frog Population 575 460 368 294 235 188 Decrease in Population (%) 20% 20% 20.1% 20.1% 20% 800 I" 700 V 600 I 500 ^ 400 A 300 Therefore, the frog population is decreasing by 20% every year. b) Using a graphing calculator, the exponential regression function for the data is y = 575.228...(0.799^..)'. 575 c) e.g., Halfthe population is 200 100 0 2 X 4 6 8 10 12 Altitude (km) b) Using a graphing calculator, the exponential regression function for the data is y= 1050.311...(0.873...)' s15 c) y = 1050.311...(0.873...)^ y = 1050.311...(0.130...) y = 137.249... mb The pressure at 15 km is about 137.3 mb. d) Using graphing technology, the pressure reaches 500.0 mb at about 5 km. ^ T J l ^ pmnMw k tS£3 Using graphing technology, the pressure reaches 50.0 mb at about 22 km. or about 288 frogs. One possible estimate could be that the population is halved in about 3 years, since the size of the population at 3 years is 294, almost egual to halfthe initial population. Using the regression equation, we determine that the time to halve the population is about 3.1 years. 10. a) Year 1960 1965 1970 1975 1980 1985 1990 1995 2000 2005 Atmospheric CO2 (ppm) 319 322 328 333 341 348 356 364 372 381 First Differences 3 6 5 8 7 8 8 8 9 e.g., No, the rate of change of atmosphenc CO2 is increasing. b) x: The number of years since 1960. y: The atmosphenc CO2; a = 319 Solve for b using the a-value and one of the points in the table. Using (45, 381) 381 =319d^^ 1.194... =6^005 b= 1.003... Therefore, the exponential function is y = 319(1.003...)''. c) Using a graphing calculator, the exponential regression function for the data is y = 315.609...(1.004...)'. 7-8 Chapter 7: Exponential and Logarithmic Functions d) Predicted Proportion in 2010 y = 315.609...(1.004. y = 315.609...(1.225...) y = 386.609... cm The predicted proportion in 2010 is 387 ppm. Predicted Proportion in 2020 y = 315.609...(1.004...)''° y = 315.609...(1.275...) y = 402.687... cm The predicted proportion in 2020 is 403 ppm. 11. a) Using a graphing calculator, the exponential regression function for the data is y = 83.933...(0.951...)! b) It will take the sandwich 14 minutes to have an internal temperature of half its initial temperature (42 °C). b) Tnstan will have less than 3 mg of caffeine in his body after 9 hours. • -112 I •lkBul}s12t«a,,.12 IT -flO raC3f)-40.7986022432fo 747i«2W3247)'''" c) Half the initial amount of caffeine in Tnstan's body is 20 mg. Half the initial amount of caffeine in Tnstan's body remains after about 2 h. J5(^)»4Ct, 7986022432 fo, 747392903247)^ ^ ^ \ . (2-'i4S49«3S3.ao) \ c) The sandwich will need to sit for 29 minutes to reach an internal temperature of 20 °C. |28 93518SS312,20) 12. a) Using graphing technology to determine the regression function, a graphical exponential model is shown below. pml2-12a A function that models this situation algebraically is y = 40.799...(0.747...)! Foundations of Mathematics 12 Solutions Manual 13. a) Using graphing technology, the equation is y = 4120.075...(1.499...)". b) y = 4120.075...(1.499...)^° y = 4120.075...(57.659...) y = 237 559.646... There will be 237 560 bactena. c) Assuming that the obsen/ations are made every hour, the function that models the growth of these bacteria is y = 4120.075...(1.2)T The base has changed to 1.2 because the bactena are growing at a rate of 20% per hour. Since the population is growing, the value must be greater than 1. 14. a Week Pond Surface Covered by Algae (%) 25.0 27.5 30.3 33.3 36.6 40.3 Growth (%) 0 1 2 3 4 5 (10.0 + 10.2 + 9.9 + 9.9 + 10.1) ^ ^^ ^^^^^ 10.0 10.2 9.9 9.9 10.1 5 The algae are growing at a rate of about 10% every week. 7-9 b) Use the equation y = 25(1.100...f and a graphing calculator to solve this. I| mi>th12t06».I4b v -JgQ 0,2 1806123617, fl(x)-2s(l 1002)^ 16. a) y = 3000(0.9)! where x is the number of weeks that the well has operated for and y is the weekly amount of oil produced, in barrels. b) y = 3000(0.9)^° y = 3000(0.121...) y = 364.729... At the end of 20 weeks of operation, 365 barrels of oil will be extracted from the well. c) The well will be closed dunng the 61*' week. It will take 12.2 weeks for the algae to cover 80% of the pond. 15. a) i) P(26) = 200 P(26)-200 P ( 2 6 ) = 200(0.686...) P ( 2 6 ) - 1 3 7 . 3 9 5 . . . ban-els/week The expected production of this well, 26 weeks after production begins to drop is 137 barrels/week. ii) P(26) = 200 2 2 P(26)-200l I P(26) = 200(0.235...) P(26) = 47.193... ban-els/week The expected production of this well, 100 weeks after production begins to drop is 47 barrels/week. b) Production will reach this level dunng the 132nd week. 17. a) e.g., Calculate the rate of change in the decrease (b) and use 100 for a. Use software or a calculator to perform exponential regression to get those values. b) Using the method from part a) to create an algebraic model, we get our equation as so97 = 1005^ b = 0.97 y = 100(0.97)' Now substitute 7 for x: y = 100(0.97)^ y = 100(0.807...) y = 80.798...% The light intensity would be 80.80%. c) 1 0 0 - 6 0 = 40 The light intensity has to go below 40%. The minimum of gels that would be needed is 3 1 . g | T t mUhUms17e • *2{3c)-2«{i 12/ O£)c)-30 (131:374348'>2,30)^ X c) The company should cap the well dunng the 256th week. I-12 h 7-10 k ma9i12e06a,.be 18. a) Penmeter: 4(4) = 16; Inner Rows: 4(3) = 12 Inner Columns: 4(3) = 12; 16 + 12 + 12 = 40 40 toothpicks will be needed to create the next figure in the pattern. Chapter 7: Exponential and Logarithmic Functions b) Number of Toothpicks 4 12 24 40 Figure 1 2 3 4 1st Diff. 8 12 16 2nd Diff. d) Using graphing technology, the exponential regression equation i s y = 100.001 ...(0.951 ...)* It is a good model for the data because the function lies very close to the scatter plot. 1 -'J|> flmUISd* ^ € B 4 4 The second differences are constant, so the data is best modelled by a quadratic function, c) Continuing the table: Number of Toothpicks 4 12 24 40 60 84 112 Figure 1 2 3 4 5 6 7 1st Diff. 8 12 16 20 24 28 2nd Diff. 4 4 4 4 4 112 toothpicks will be needed to make the 7th figure. 19. a) Using point (9, 64.0), calculate h as shown: 1 64.0 = 100.01 - 64 ^ 1 V/i 9 100" 9 11^ 16 ^ 25" Use a graphing calculator to solve this equation. : > Ik rnatti12eC*»..«d •» The equation determined by hand and the one determined using graphing technology are very close because the difference in numbers start showing up in the thousandth decimal place. History Connection, page 468 A. For example, uranium, thonum, radon, and radium. B. The radioactive decay of an element can be modeled by a halfdife equation, which is an exponential decay function. The halfdives of the four elements listed in prompt A is given below. • U-235: 7.1 X 10^ years • U-238: 4.51 x 10^ years • Rn-220: 55 s • Rn-222: 3.823 days • Ra-226: 1600 years • Th-227: 18.5 days • Th-230: 8.0 x lO'* years • T h - 2 3 1 : 25.5 h • Th-234: 24.1 days To determine the halfdife equations for the elements shown above, I substituted each halflife into the halfdife equation. A{x) ^ A - , \IJ where A{x) is the amount of the element at time x and Ao is the onginal amount. The half-life equations for the elements are (where x represents time in years, except as noted): U-235: The value of h is 13.978... or 14 days. 13.978. b) A{t) = ( 1 ITTXIO" A x ) = A ) k m c ) a = 100 Use the same point used to calculate h in a) to calculate b for this part. Using (9, 64.0), b is calculated as shown: 64.0-1005' 16 = b' 25 5 = 0.951... The exponential regression equation is y = 100(0.951...)! Foundations of Mathematics 12 Solutions Manual u-238: A x ) = A,l 2 Rn-220: (1 V55 V2/ where x represents time in seconds 7-11 Rn-222: M i d - C h a p t e r R e v i e w , p a g e 472 { i Y3T823 A{x) = A^[l \ zJ where x represents time in days Ra-226: f -( Y-ieoo 2) Th-227: A{x) = \ 2 where x represents time in days Th-230: A{x) = A, 2 2 Th-231: l¥ where x represents time in hours Th-234: 1 12.41 where x represents time in days Applying Problem-Solving Strategies, page 469 A . - G . Answers will vary, based on the decisions that players make. H. Possible vanations: • Increase the size of the playing square, or use another shape. • If a player hits a square, she or he gets to go again immediately. • Once a function is announced, the opposing player must tell the person if the function passes above, below, or between her or his squares. • Assign points to each hit, or start out with a given number of points and deduct points for misses or questions (for example, "Does the function pass above your squares2"). 1. a) This graph does not represent an exponential function because it does not have a horizontal asymptote. In addition, its end behaviour is unlike that of an exponential equation since it extends from Qll to QIV. b) This graph does represent an exponential function because it has all of the basic charactenstics of an exponential function. It is constantly decreasing at an unsteady rate like an exponential decay function. It has a y-intercept but no x-intercept. It has a domain of {x | x e R} and a > 0 and 0 < i ) < 1. Also, it decreases from Qll to Ql. c) This graph does represent an exponential function because it has all of the basic charactenstics of an exponential function It is constantly increasing at an unsteady rate like an exponential growth function. It has a y-intercept but no x-intercept. It has a domain of {x | x o R} with a < 0 and 0 < d < 1. Also, it decreases from Qll to Ql. d) This graph does represent an exponential function because it has all of the basic charactenstics of an exponential function. It has only one asymptote that is honzontal. It is constantly decreasing at an unsteady rate like an exponential decay function. It has a y-intercept but no x-intercept. it has a domain of {x | x e R}, with a < 0 and 5 > 1. Also, it decreases from Qll to Ql. 2. a) i) 8 1 fiW-34'=y^ [I , , , IJ 1, , , ,1 T - f j •8 IS •' S -2' i# ii) y zJ 11 1} 1 1 > 1 t 1I 1 ll1 \ \ i \ t \ h- \\ 0 N 7-12 X 4 Chapter 7: Exponential and Logarithmic Functions b) Number of x-intercepts: 0; y-intercept: y = 90 domain: {x | x e R}; range: {y | y > 0, y e R} The function is decreasing. Hi) /' 1 / 11 1 1 Hi i 1U' ) 1 1 i r T 1 Ol " y7 i1) \ til J •) t 1, •\ J_ ) > r • \ \ \ 7 •\ 7 — 1r • 1 J\) A\ ) 'J il / - > 1 J) ) Z 1 \ r / [ -I \ 0 •> 3 \ _. 0 : r \ > i c) number of x-intercepts: 0 y-intercept: 8 domain: {x | x € R} range: {y\y>0,ye R} The function is increasing. b) i) number of x intercepts: 0; y-intercept: y = 3 domain: {x | x e R}; range: {y | y < 0, y o R) end behaviour: Qll to Ql ii) number of x-intercepts: 0; y-intercept: y = 8 domain: {x | x e R}; range: {y | y > 0, y € R} end behaviour: Qll to Ql iii) number of x-intercepts: 0; y-intercept: y=5 domain: {x | x o R}; range: {y | y > 0, y e R} end behaviour: Qll to Ql iv) number of x-intercepts: 0; y-intercept: y = 2 domain: {x j x e R}; range: {y | y > 0, y e R} end behaviour: Qll to Ql 3. a) Number of x-intercepts: 0; y-intercept: y = 5 domain: {x | x o R}; range: (y | y > 0, y o R} The function is increasing. 44 A' •"t 1 2 4 r—n 6 8 4. a) b) a = 2; c) a = 5; d) a = 3 b) b) It is between 0 < fe< 1 because the function is decreasing. c) It is between 0 < 5 < 1 because the function is decreasing. d) b is greater than 1 because the function is increasing. c) b) domain: {x | x o R}; range: {y I y > 0, y e R} c) domain: {x | x e R}; range: {y\y>0,ye R} d) domain: {x | x e R}; range: (y | y > 0, y e R} 5. a) e.g., It is a decreasing function because the base is less than 1. 4^ > il 1' ) \ 5> 2 4 6 1 4 S ] J n1 1t 1 i 1 \ - •,\ '\ s > \ — r- Foundations of Mathematics 12 Solutions Manual \ Ji 0 7-13 b) e.g., It is an increasing function because the base is greater than 1. 1 f< / ) 1 13 1 1'J I1 1r 11 1'. 1\ - 1J 1 r V K 0 •1 • c) e.g., It is a decreasing function because the base is less than 1. 9. a) e.g., First divide amounts in consecutive rows to determine the base, then use the initial amount and the base to write an exponential formula, b) 84H» •< -4 '2 2 4 b) i ) y = (7.628...)(1.742...)°^ y = (7.628...)(1.320) y = $10.069... The card had a value of about $10.07 6 months after purchase. ii) y = (7.628...)(1.742...)'^ y = (7.628...)(2.300...) y = $17.546... The card had a value of about $17.55 18 months after purchase. iii) y = (7.628...)(1.742...)^^ y = (7.628...)(4.008...) y = $30.575... The card had a value of about $30.58 30 months after purchase. iv) y = (7.628...)(1.742...)^^ y = (7.628...)(6.984...) y = $53.279... The card had a value of about $53.28 42 months after purchase. 6 S d) e.g., It is an increasing function because the base is greater than 1. Time (years) Amount ($) 0 1 2 3 4 1500.00 1560.00 1622.40 1687.30 1754.79 04-1 0^ ^ -6 '4 -2 2 4 6 » -oaH 6. a) Assuming that Lauren does not count as a student who has received a handout, the following is a table of values that models this situation: Round (x) 0 1 2 3 4 Number of Students Who Received Handouts (y) 7 49 343 2401 16807 b) Yes. y= 7{7f ; e.g., The ratio of consecutive numbers of students who received handouts in each round is constant. Increase in Amount (%) 4 4 4.000... 3.999... The annual interest rate is 4 % . c) The function, which we must use, is y = 1500(1.04)'. y = 1500(1.04)'° y= 1500(1.480...) y = $2220.366... Paula will have $2220.37 in her account after 10 years. 10. a ) y = 140 (0.80)° y = 140(1) y = 140 cm The ball was first dropped from a height of 140 cm. b) y = 37 cm rr 7. Using a graphing calculator, the function is y = (36.871...)(0.663...)! 8. a) Using a graphing calculator, the function is y = (7.628...)(1.742...f. 7-14 1 I r , —+—+--f - ^ , Mf,,,,j|, ,.»„.,,,._, .y. m Chapter 7: Exponential and Logarithmic Functions c ) x = 3.106... when y = 70 cm L e s s o n 7.4: C h a r a c t e r i s t i c s of L o g a r i t h m i c F u n c t i o n s w i t h B a s e 10 a n d B a s e e, p a g e 482 1. x-intercept: 1 Number of y-intercepts: 0 End Behaviour: QIV to Ql Domain: {x | x > 0, x e R} Range: {y | y e R} The height was less than half the initial drop height on the 4th bounce. 11. a) The function is decreasing. It has one y-intercept and no x-intercepts. b) The domain is time, so it must be positive: {f I f > 0, X G R} The range is all the possible temperatures from 90 °C to 21 X : {C(f) | 21 < C{t) < 90, C(f) e R} c) C(f) = 90(0.977/ C(10) = 9 0 ( 0 . 9 7 7 r C(10) = 71.316... The temperature of the coffee after 10 min is 71 °C. d) C{t) = 60 °C when f = 17 minutes. 4 - I 1 2^;. > t ma!h12c(«» .-10c • C at) = 30 °C when t = 47 minutes. 100 y 2. a) No. e.g., no x-intercept and one y-intercept whereas logarithmic functions have no y-intercepts and one x-intercept. It is an exponential function. b) No. e.g., two x-intercepts and one y-intercept whereas logarithmic functions have no y-intercepts and one x-intercept. It is a quadratic function. c) Yes. e.g., one x-intercept and no y-intercept. d) No. e.g., one x-intercept and one y-intercept whereas logarithmic functions have x-intercepts ofi. e) Yes. e.g., one x-intercept and no y-intercept. f) No. e.g., no x-intercept and one y-intercept whereas loganthmic functions have no y-intercepts and one x-intercept. 3. c: x-intercept: 1 Number of y-intercepts: 0 End Behaviour: QIV to Ql Domain: {x | x > 0, x o R} Range: {y | y e R} a: a > 0 e.g., since the graph is increasing e: x-intercept: 1 Number of y-intercepts: 0 End Behaviour: Ql to QIV Domain: {x | x > 0, x e R) Range: {y | y e R} a: a < 0 e.g., since the graph is decreasing 4. y = log x: ^rni: IB e) C(f) = 21 °C when t = 63 minutes. fllxP^O (O <>77Y 62 5428924706, 2 1 Foundations of Mathematics 12 Solutions Manual X-intercept: 1 Number of y-intercepts: 0 End Behaviour: QIV to Ql Domain: {x | x > 0, x e R} Range: {y | y e R} Increasing or decreasing: increasing y = - l o g x: 7-15 *Uni»ved 40 20 / 5 ,.fi(C"-tor~-4L-_. x-intercept: 1 Number of y-intercepts: 0 End Behaviour: Ql to QIV Domain: {x | x > 0, x e R} Range: {y | y o R} Increasing or decreasing: decreasing m tt 20 23 50 iS 40 d) x-intercept; 1 Number of y-intercepts; 0 End Behaviour; QIV to Ql Domain: {x ( x > 0, x G R } Range; {y I y e R} Increasing or decreasing; increasing hHt'Mr I ,Mi!h44 5. a) x-intercept: 1 Number of y-intercepts; 0 End Behaviour: QIV to Ql Domain: {x | x > 0, x e R} Range: {y | y e R} Increasing or decreasing: increasing e) x-intercept; 1 Number of y-intercepts; 0 End Behaviour; Ql to QIV Domain; {x | x > 0, x G R } Range: {y j y G R } Increasing or decreasing: decreasing 21- -1- 1 2 -20«b) x-intercept: 1 Number of y-intercepts: 0 End Behaviour: Ql to QIV Domain: {x | x > 0, x G R } Range: {y | y G R } Increasing or decreasing; decreasing 0.4- 0.2- r - a i 0 4 0.6.0.8 IJO t>r f 0.4- 4 2H 0 -2- <X5 X6<k^ 1.0 23 IG 3.5 4.< -4 c) X-intercept; 1 Number of y-intercepts: 0 End Behaviour; QIV to Ql Domain; {x | x > 0, x e R} Range: {y | y G R } Increasing or decreasing; increasing f) x-intercept; 1 Number of y-intercepts; 0 End Behaviour; QIV to Ql Domain; {x | x > 0, x e R} Range: {y | y G R } Increasing or decreasing: increasing 5O0H 400 300 2O0 100H -KK) too 200 300 40O -200 7-16 Chapter 7: Exponential and Logarithmic Functions 6. e.g.. one xontercept of 1. no yontercepts, I'P f\. litt; f;n<;igy " i ihr. • A-ahvI ifti.s.,>.:i%.' s, the It'.i!Ilh<':t s>[ :Jf;Olh<'f. II1! .o,;; I',<j domain; (x | x > 0, x o R } . "/ d ).g., y = 5 log x 12. a l PdLicM.--f.' • ' intercept: none. R ) . function: decreasing, p = "I ; f = o. . 0. • 12 log P ;'.f ,.,!2 loom Ofi) It would take abooi i ' < v ars for only 5% of the original amount to if • ( ' )0.422 log P )0.422 log(0-5) 230... The half-life of cesium-137 is about 30 years. 8. i) b, e.g., x-intercept is 1, no y o n t o i o c p i l , graph extends from QIV to Ql. Thus, the function is logarithmic so b and c are the only options. The function is increasing so the correct graph is b. ii) c, e.g . X intercept is 1, no yontercept, graph extends from Ql to QIV. Thus, the function is loganthmic so b and c are the only options. The function is decreasing so the correct graph is c. jjj) d. e.g., no xontercept, y-mtercept is 1, graph extends from Qll to Ql. Thus, the function is exponential so a and d are the only options. The function is increasing so the correct graph is d. iv) a. e.g., no x-intercept, y-intercept is 2, graph extends from Qll to Ql Thus, the function is exponential a and d are the only options. The function is decreasing so the correct graph is a. 9. Yes, e.g.. An exponential function has no x-intercepts, and a loganthmic function has one x-intercept. 13. a) When a < 0, each function will bo decreasing. b) The domain of each function is restncted as for each function x must be greater than 0. c) The range is unrestricted as all values of y are possible. 14 a) 10. As hydrogen ion concentration increases, pH decreases. The Tontercepl is 10 '''' and there is no A/fdntercept In context, the domain of this graph •s {71 ICr^-^ T G R } , and the range is [M \ M 0, M>: R ) . The graph is increasing, and when 7 - i, M = 5.5. b)M= log T+ 5.5 M = log(50) + 5.5 M = 7.198... An earthguake that is 50 times more intense than the Saskatchewan earthquake would have a magnitude of about 7.2. Foundations of Mathemati 'Iutions Manual 7-17 ifili'ir., t|.,,ri Uw '..r.katf tif •Well) > ;.irth(jiJ.ikc^ 1f> ,1) I lH< A inlf i;, M)(}ij . j i „ | Uit^-H I-, IK, / iiilor* f'f>t In < uii!<;yi, Ihc ilmiumi <if Ihr. 'ir.-lf.h is .A '• A ;U(J<i, A IVf Jilt* Iho IS i' I ' b) I <h / R } Ihi'. hi!uJu,u r. !fiu<..j.'.i!KJ '.H / ItHj 4 VIM 1 ' 'H> / iOf}| I I I 0 0 0 ) ' 'Ul / Xfl4 ' It Will take about 31 years for Christie's investment to reach $10 000. c) f = 58.7 log ^ ^ 2 0 4 . 1 f = 58.7 iog(6000)-^ 204,1 t= 17.677... It will take about 18 years for Christie's investment to double. b) I hf 'Ml. •. ; i o ( . . j u a t i o n that models the data I: / ;4o ()0(i i 23-957... In P P inh.-ic-pi 1 , 0 0 0 / 0 023... r llllcn.ept liuiu; I nd t)oh.ovi.-.!a «jlV lo Ql t ioi,,Hll! >P I 0 / ^ G W} H.oog«^ \i I 0 / W} I !.ii(.fini> iiK.ieaciog cH •'xlOOOO ; 28,957... In P i - :>4o 000 ^ 23,957... In (2 000 000) / / 4 i)A I The population exceeded 2 000 000 in 1974. 4. a) The independent variable is seismographic reading and the dependent variable is Richter Scale magnitude, b) • .t»-fi(j:i.)k> L e s s o n 7.5: M o d e l l i n g D a t a U s i n g Logarithmic F u n c t i o n s , page 494 4 1. a | e.g., Graph is logarithmic because graph is increasing, xontercept is 1, there is no yontercept, end behaviour is QIV to Ql. domain is { X I X > 0, X G R}, and the range is {y | y G R}. b) When the decibel level increases by 10, the relative sound intensity increases by a factor of 10. c) When the relative sound intensity doubles, the decibel level increases by a factor of 3. 2. The logarithmic regression equation that models this data is y = ^6.653... + 108.491. .In X. xontercept; 1,063... y-intercept: none End behaviour: QIV to Ql Domain: {x j x > 0, x R} Range: (y I y r- R} Function: increasing 3. a) ' i ,o,\.3^^o\J>./y,,3^^vfx? 4> ^ ^ ^ ^ ^ Seismographic reading (microns) c) The regression eguation for the data is M = -0.006... + 0.434... In r. d | M = ^0.006... + 0 , 4 3 4 . . . In r Let M = 5.7 5.7 = ^0.006... + 0.434... In n 5.706... = 0.434. .. In n 13.122 -Hi r-i .122... Let M 4 5 - - 0 006... + 0.434... In r? 4.50b - 0 4348 In f2 10.362 -In/2 r, = e''^-''''" d5,794... 7-18 Chapter 7 L x o o n e n t i a l and L o g a r i t h m i c F u n c t i o n s A magnitude 5.7 earthquake is about 15.8 times more intense than a magnitude 4.5 earthquake. 5. a) The independent vanable is pressure and the dependent variable is altitude. b) The regression equation for this data is h = 30 665.960... - 6640.436... In P. c) Pdntercept; 101.297... Ii-intercept; none End behaviour: Ql to QIV Dom? Range; {li | fi e R} Function; decreasing d) y = 30 6 6 5 . 9 6 0 . . . - 6 6 4 0 , 4 3 6 . . . In X 139 = 30 665.960... 6640.436.,. In x - 3 0 526,960... = -6640.436... In x X - 99.199... The pressure setting that Michael will need to use is 99.2 kPa. e) y = 30 6 6 5 . 9 6 0 . . . - 6 6 4 0 . 4 3 6 . . . In X 8848 = 30 665.960... - 6640.436... In x -21817,960... = -6640.436... In x 3 285... = In X X = 26.725.. The atmospheric pressure at the summit of Mt. Everest is 26.7 kPa. 6. a) Here is the graph. / a- Fhe exponential regression eguation that represents this data is 71 = 14 999.826...(1.045..,)'. b) The logarithmic regression equation that represents this data is f = -218.442... + 22.717. In A. c) Exponential: A = 14 999.826...(1.045.,.)' 25 000 = 14 999.826..,(1.045...)' 1.666... = (1.045...)' log 1.666,, = log (1.045,..)' log 1.666. - t log 1.045... togijea. logioisl = t 11 604. Logarithmic: f = .^218.442... • V 71/ In A f = ^218.442... In (25 000) f = 11.604... It will take about 12 years for the balance to egual $25 000 according to both equations, e.g., I prefer using the loganthmic because it makes the calculations simpler. 8. a) The independent variable ts distance from Earth and the dependent vanable is distance modulus. b) Here is the graph. 2 ICon !C 20 00( ttmttr 18 OCX 16 0(X 14 00( % 12 OCX ^ 10 000 < 8 000 6 000 4 OOi 2 000 10 A) J O 10 W cc 71 < HV, 190 00 Percent of carbon-14 {%) The regression eguation that represents this data i s f = 38 0 7 0 . 3 3 2 . . . - 8 2 6 6 . 9 6 1 . . . In P b) f = 38 070.332. . - 8266.961... In P t = 38 070.332... - 8266.961. . In (96.8) f = 268.437... The age of these fragments is about 268 years. c) t = 38 070.332... - 8266.961... In P f = 38 070.332 .. - 8266.961... In (50) f = 5729.790,,. The halfdife of carbon-14 is 5730 years. Foundations of Mathematics 12 Solutions Manual c) The logarithmic regression equation for this data is rn = -5.086... + 2.174... In d. d) m = - 5 . 0 8 6 . . . + 2.174. . In d m = - 5 . 0 8 6 . . . +2.174... In (2.39) r?i = - 3 . 1 9 1 . . . The distance modulus of Wolf 359 is - 3 . 1 9 . e) m =-5.086...+2.174... Ind - 1 . 6 6 = -5.086... + 2.174... In d 3.426.,. = 2.174... In d 1 575, . = In d d = 4.834... Gliese 876 is 4.8 pc from Earth. 7-19 A = 5000 1 + A 5000 In A 0.05 = 1.0125"*" = In 1.0125' 5000 l n A - l n 5 0 0 0 = 4filn1.0125 \nA 4n In 5000 Ini 0125 In 7\ In 5000 '4Tnl"oT25 iv) b) fh-: icwjanlliniic ro;jiessif)ii .;q,i;i|ion U;: tivs •'f;d,i i ' h.'58. . friO In n t'l / 3 mm ' li;..H.>:j. in n / ';..'i8{/> ^10h>;3. Iiif;'7r,i I - Ciy./30... Facebook first surpassed 275 million registered users dunng the 66th month after February 2004. which is August 2009. 10. Enter the data in my calculator, perform a loganthmic regression, graph the data points in a scatter plot and the logarithmic function on the same axes, and identify the point with the known x-value and the unknown y-value. 11-a) 1) A = P{1 + if A=5000 5000 In A 5000 1+ 0.05 365 • = 1.000136.-.*''" = In 1.000136. In7\ - In5000 = 365nIni.000 136. 365n = n= In ^ - I n 5000 In 1.000136... In 4 - I n 5000 365ln1,000136... b) e.g.. The denominator approaches In 1 = 0 as the number of compounding penods per year ^ = 5000(1 + 0,051 5000 in A 5000 = 1.05" Math in A c t i o n , p a g e 500 = In 1.05" • It looks as if you need to ask no more than 10 questions to guess a number from 1 to 1000. In 4 - I n 5000 = /? In 1.05 n= In 4 - I n 5000 In 1.05 ii) A • Let n represent the numbers remaining and let q represent the number of guestions asked. Since the secret number is from 1 to 1000, it means that there are 1000 possible numbers. Each time you do a binary search, you are dividing the set of possible numbers in half. 5000 1 ^ "I 5000 In A 5000 I n / \ - I n 5000 2n 1.025'-" In 1.025' 2:? In 1.025 In 4 - I n 5000 In 1.025 In ^ - I n 5000 2ln1.025 7-20 Chapter / '-hfoonential and Logarithmic Functions I wrote a table of values for the first few guesses. Then I substituted o = 1 and solved for q: 1 = 1 000 000 00011 1000 1000[ 2- 1 1 000 ooc 1 00 000 000 You would 0..OU :o ^c!. -0 questions to guess a number from 1 to 1 000 000 000. I noticed that the exponent on the equal to q, the number of questions asked. So, I set up an eguation expressing n in terms of q to mode! the situation. 0 = 1000 • Answers will vary. e.g.. One possibility is to allow the guesser to divide the list of remaining numbers into three groups, instead of two. Questions would now be like this; "Is your number between a and b, b and c, or c and d?" This would make the game go much more quickly, since each turn reduces the number of 2 remaining numbers by - . To guess a number from 1 to 1000, the equation that would model the situation would now be „.iooo(l . I substituted n = 1 into my eguation 1 = 10002 1 fo; 1000 (2) 1=10001.3 1000 q .log., c| = 9.965... You would need to ask 10 questions to find a number from 1 to 1000. This answer matches my expenmental answer. . To determine the number of questions needed to guess a number from 1 to one billion, I changed 1000 to 1 000 000 000 in my eguation. n = 1 000 000 000 I ^ Foundations of Mathematics 12 Solutions Manual 1 1000 g = 6.287,. C h a p t e r S e l f - T e s t , p a g e 501 1. To match the functions with the graphs, look at the X- or yontercepts and directions of the functions and of the graphs. This is because each function and each graph has a unique y-intercept. a) It is a decreasing exponential function with y-intercept of 0.2 so it must match with ii. b) It is a decreasing logarithmic function, so it must match with iv. c) It is an increasing exponential function with y-intercept of 2. so it must match with iii. d) It is an increasing loganthmic function, so it must match with i. 7-21 2. a | The exponential regression function is y= 1 2 J 2 0 . . . ( 1 0 9 4 . , . f . where y is the debt in billions of dollars and x is the number of years after 1955. N« t o d,-.,it r)..bt 4, xontercept: 1 yontercept: none End behaviour; Ql to QIV Domain: {x | x > 0, x o R} Range; {y | y e R) Function: decreasing S'jil 5 a) I .>i ^ rf.pic:.. Ill !he energy released. Let y H o'>-><-" Iho r.oigtiiPjde o f t h e earthquake. I O ' - icgo- .-.ion cgoolion is 1 I'-o t I) /;>o ..In X . h) ' : h..' / 1 hiO > :i .'P.o ..in X ' i'^'^ ' ..In (1.1 • 10^®) •/ "• 10 :h , -o^'-.i-....,.,rfh,,iMke had a magnitude of about 9.5. ',(HI 2 r.ti ,. CO • 1 too .''XI /0(, Z 150nc tq±t±t I'' M; Years since 1955 -a .|C. -ih b| 1 9 8 8 - 1 9 5 5 = 33 y = 12.620...(1.094...f^ y = 12.620,.,(19.592...) y = $247.282... billion The net federal debt was $247.28 billion. c) Assuming the same growth rate and that these figures were taken at the end of the year, the net fodoral dob! shculd have roordied $600 billion in 1998. 3. Number of xontercepts; 0 yontercept; y = 6 Domain; { x | x e R} Range: {y | y > 0, y e R} Or>o Bohavioor- OH lo Ql C h a p t e r R e v i e w , p a g e 504 1. a) e.g., The end behaviours of exponential functions extend from Qll to Ql, Polynomials have varying end behaviours, b) e.g.. The domain is always {x | x e R), the range is always {y | y > 0. y o R). and they all extend from Qll to Ql. a is the y-intercept. 2 a) domain: {x | x o R); range: {y | y > 0, y e R} y-intercept: y = 9; end behaviour: Qll to Ql This is a decreasing function. b) To make the function into an increasing exponential function, the b should be changed so Its value is greater than 1. 3. a) i) number of x-intercepts: 0 ii) y-intercept: 125 iii) end behaviour: Qll to Ql iv) domain: {x | x e R); range: {y | y > 0, y o R) v) The function decreases. b) i) number of x-intercepts: 0 ii) y-intercept: 0.12 iii) end behaviour: Qll to Ql iv) domain: { x | x e R); range: { y | y > 0, y e R) V ) The function decreases. c) i) number of x-intercepts: 0 i i | y-intercept: 1 iii) end behaviour: Qll to Ql i¥) domain: { x | x e R); range: {y | y > 0, y e R} v) The function increases. d) i) number of x-intercepts: 0 ii) y-intercept: 0.85 iii) end behaviour: Qll to Ql iv) domain: {x | x c- R); range: V ) The function increases. 4. To match the functions with the graphs, look at the y-intercepts of the functions and of the graphs. This is because each function and each graph has a unique y-intercept. a) It is an increasing exponential function with y-intercept of 5, so it must match with i. b) It is a decreasing exponential function with y-intercept of 2 so it must match with ii. 7-22 Chapter r fcxponential and Logarithmic Functions 5. a) Let x be the number of months since April 1, 1896. Let y be the population of Dawson City, a = 1000 l3 = 3 Tripling time; 3 months An exponential equation that models the population grov/th is ^ =1000 (3) , where y is the population and x is the number of quarters after April 1, 1896. b) Domain; {x | 0 < x < 3, x e N} (quarters) Range; {y | 1000 < y < 27 000, y e N) (population) c) The population of Dawson City in mid-May of 1896 was 1732. and the population of Dawson City in mid^August of 1896 was 5196. To determine the answer for mid-May. I substituted in 1.5 for X for my equation. To determine the answer for mid-August, I substituted in 4.5 for x for my eguation. These two points arc midpoints of the first two quarters, and so correspond to X = 0.5 and x = 1.5. A(8000) = 1 0 0 ' ^ | ) ' ' " /l(8000) = 100 /l(8000) = 100(0.379...) A(8000)= 39.993...% 38% of the initial carbon-14 would be present in the tools. I>) I he ago (4 ihn iools wt;ro 7400 years old. 8. x-intercept; 1 Number of yontercepts; none End Behaviour; QIV to Ql Domain; {x | x > 0, x e R) Range: {y | y e R} 6. a) .oof 350 300 9. a) x-intercept: 1 Number of y-intercepts: none End Behaviour: QIV to Ql Domain: {x | x > 0, x R) Rang^.r {y \ y R] 250 200 150 100 SO Ml 0 i ; , 4 \: ( Years after 2005 b) Using the graphing calculator, the exponential regression function is y = 200.798...(1.088...)". c) y = 400.798.,.(1.088...)'° y = 400.798...(2.330...) y = 467.971... The deer population 10 years after 2005 will be 468. d) 200 798 .(2) = 402 (rounded) Assuming the deer population numbers were recorded at the beginning of the year, I would expect the deer population to have doubled in the year 2014. Foundations of Mathematics 12 Solutions Manual b) x-interccpt: 1 Number of y-intercopts: none End Behaviour: Ql to QIV Domain: (x | x > 0, x e R) Range: {y | y e R} 7-23 c) x-intercept: 1 Number of y-intercepts: none End Behaviour: QIV to Ql Domain: {x | x > 0, x G R } Range: {y | y e R} ii) b, e.g., no x-intercept, y-intercept of 1, increasing. Thus, the function is exponential so b and c are the only possible choices. Since d > 1, the function is increasing and therefore the correct choice is b. iii) a, e.g., x-intercept of 1, no y-intercept, QIV to Ql. Thus, the function is logarithmic so a and d are the only possible choices. The function has a positive leading coefficient which means that the function is an increasing function and therefore the correct choice is a. iv) c, e.g., no x-intercept, y-intercept of 6, decreasing. Thus, the function is exponential so b and c are the only possible choices. Since d < 1, the function is decreasing and therefore the correct choice is c. 11. The logarithmic regression equation of this data is y = 151.211... - 32.836... (In X). y = 151.211... - 3 2 . 8 3 6 . . . ( I n x ) 200 = 1 5 1 . 2 1 1 . . . - 3 2 . 8 3 6 . . . ( I n x ) 48.788... = - 3 2 . 8 3 6 . . . ( I n x ) -1.485... = l n x X = e^^-'^^-X =0.226... Therefore about 0.23% of the sunlight penetrates water to a depth of 200 m. Chapter Task, page 507 A. I found that the amount of caffeine in one espresso is about 100 mg. So, a tnple espresso will have 300 mg of caffeine. d) x-intercept: 1 Number of y-intercepts; none End Behaviour; Ql to QIV Domain; {x | x > 0, x e R} Range; {y | y e R} B. I should use time as the independent vanable because it is the change in time that causes the change in the amount of caffeine. If time is the independent vanable, then the function is an exponential function. C. I chose to use 6 h as my estimate because that is the maximum possible time to reduce the amount of caffeine to a half. 1 iow^ ih) Caffeine (m g) 0 6 12 18 300 150 75 37.5 D. Using a graphing calculator, the exponential regression function for the data is y = 300(0.890...)^. <£_2 | T T ] 1^ *ma»i12c06..10c ^ IgQ 10. i) d, e.g., x-intercept of 1, no y-intercept, Ql to QIV. Thus, the function is logarithmic so a and d are the only possible choices. The function has a negative leading coefficient, which means that the function is a decreasing function and therefore the correct choice is d. 7-24 Chapter 7: Exponential and Logarithmic Functions E. To determine the amount of caffeine in Maria's body 24 h after her first triple espresso, I substituted t = 24 into my exponential equation from prompt D. Ait) =300(0.890,..)' Ai24) = 300(0.890...)^'^ 4(24) =300(0.062...)^'* 4(24) = 18.749... mg Maria will have 18.75 mg of caffeine in her system 24 h after drinking her first triple espresso. Chapter 7 Diagnostic Test, T R page 462 i o| r = 64 F. I chose to use a spreadsheet to calculate the amount of caffeine in Maria's system during the d) (5) Week Starting . .H loi^^ilh^ Week Starting amount of caffeine (mg) Amount of olleine after .' I h I) • / of the •.O.sok .05 irti .Cl ,ount of ' cdfeine (mg) ,-\o ount of rolteine after L24h i f , - c) (121)2 =V121 (121)2-11 1 |./..,. Tue. (5)'"* =0.0016 300 OOOO 318.7500 2. a) 3 • 7 7 = 3(7)^ (I o-r-.•...! of cr fl.../.c, after 24 h f r -zz —j (1 to b) 19.9219 500 c) . d) 1 1 319.S 1951 n. ,.0 0997 19.9 Sa a a (1 b = a a b . . .y.y = X X X x-^ 3. a) 27 = 3-* b) 64 = 8^ 64 = 4^ 64 = 2" c) 625 = 5'^ d) 169 = 13' 4. a) 6 ^ - 6 - " . 6 = - | ^ 20.000fs b) 20.0000 1 1 2 2 1.2500 On Saturday morning, Mana will have about 20.00 mg of caffeine in her system. If she does not drink coffee (or any other caffeinated dnnk) on Saturday, she will have only 1.25 mg of caffeine in her system by Saturday evening. 1 f i f fiT' 3 K3J K3J c) 2'-22 •2'-' = n 13 1 2^-22 2^-22-2^'^ ^ 2 ^ . 2 " ' 25.2^2.2^-2 =2^2 5J V 1 52.5^.5 5/ 5 - I -5^-5 = 5^ 5 Foundations of Mathematics 12 Solutions Manual 7-25 RS f1 0 .,1 IS c) 5 x + T ^ = 8x 5x-8x = ^ ^ 3 3 ( 1] RS 8x 0 , ^ 2 1 1 . 4 ' 33 4*-3""' 4 ^ 4'*-3-' = 432 i 9 J' 3 UJ 8 ."••1 0 6 6^-2^ 8 6 ^ . 6 - = 6''.2^^ 9 0 9 LS = RS 6-^-2^ 6 ^ = 10 368 dl ' 1 I / , a| 6 - - 3 x = 8 I - ! 3f - y « 4 1 \ — X - 3 -3x = 8 - 6 -3x = 2 13 K 2 x= — 3 6-3x X RS 8 LS 4 RS 3x -(-II 3(4) 6 +2 8 -111 LS = RS 13 12 1 12 LS = RS 7-26 Chapter 7: Exponential and Logarithmic Functions xontercept: y=4x^5 8. a) No, a constant multiple was not used to generate the numbers in this set. This can be seen in the ratio between each successive number. Nurnhct " I " 0 = 4x^^5 5 Ratio 4x 5 X ' 7 I b| I uc, u woclc^ot lOJdipio oi 2 -A'ac used Lo generate the numbers in this set. This can be seen I .P'. h»*f-.'iOHO --^cxh r y r r o c s i v o r-ismbor. riortiber • R.itio ^ e) This would match with iii) because for y = x^, the function increases when x is greater than zero. f) It would match with i) because the height of a person over time is continuous not discrete. It is a gradual increase. If the height of a person were graphed, it would be a continuous line without a break g) This matches with ii) because the function is linear since x has an exponent of 1. c) Yes, a constant multiple of — was used to generate the numbers in this set. This can be seen in the ratio between each successive number. Ratio nber m ' 1 12 3 1 A 2. a) The domain of this function is {f I f > 0 , f G R}. This is because t represents time and the ball cannot have negative seconds after it falls off a balcony, b) The domain of this function is {b\b>0,bG R). This is because b represents batches of cookies and it is impossible to bake negative batches of cookies. It is possible to bake zero or more batch of cookies. 1 1 c d) No, a constant multiple was not used to generate the numbers in this set. This can be seen in the ratio between each successive number. r _jNymt£er V 4 1 ^ _^ 10^ 23 3f. Z ^ • I Ratio ^71 ' ^ ' '2 25 _ 1 77 7 . I ~ J 562 „ 1 44 R e v i e w of T e r m s a n d C o n n e c t i o n s , T R page 464 1. a) It would match with iv) because the two coordinates are the y-intercept and x-intercept of y = 4x 5. y-intercept: y = 4(0)-5 y= 4 b) This would match with vi) because the number of books collected in a year is a discrete number. The number of books cannot be continuous; it has to ; . o df Crete unit in which each unit is separate. For example, it is impossible to have 0.5 books. It is possible to have 2 books. If this were to be graphed, it would look like a staircase because the line is not a continuous whole. c) This matches with vii) because the domain of y = x^ is all rea! numbers. d) It would match with v) because the range of y = x^' IS all values that are equal and greater than zero. 3. a) The range of this function is {11(f) I 0 < /7(f) < 3, h(t) G R} because if the ball is falling off the balcony 3 m up, that must be its maximum height. The minimum height would be 0 m, where the ball hits the ground, b) The range of this function is {n{b) I 0 < n{b), nib) e W} because it is impossible to have a negative number of cups of flour. The number of cups of flour must be zero and greater to be able to bake b batches of cookies. 4. a ) y = 3 x + 6 y-intercept: y = 3x + 6 y=3(0) + 6 y =6 5 Foundations of Mathematics 12 Solutions Manual 7-27 x-intercept: y = 3x + 6 (0) = 3x + 6 -3x = 6 x = -2 5. a) The domain is {x | x g R } and the range is {y I y e R } because it is a linear function. b) The domain is {x | x e R } and the range is {y I y < 8, y G R } because it is a quadratic function. c) The domain is {x | x € R } and the range is {y I y G R } because it is a cubic function. d) The domain is {x | x G R } and the range is {y I y e R } because it is a cubic function. The x-intercept is - 2 and the y-intercept is 6. b)y = / ^ 4 y-intercept: y = / ^ 4 y=(0f-4 x-intercepts: y=X^-4 (0) = / - 4 4 = / -2, 2 = X 6. a) The function is increasing because the slope is positive. b) The function is decreasing when x > 0 and increasing when x < 0. c) The function is decreasing because the slope is negative. d) The function is increasing because the slope is positive. 7. a) y = -2/ + 8 y-intercept: y = +8 -2x^ x-intercept; y = +8 y = ^2(0)^+8 (0) = ^2/ + 8 -8 4 = / ±2 = x The y-intercept is 8 and the x-intercepts are -2 and 2. Oa.o) •2"" * a'o)""^ The x-intercepts are - 2 and 2 while the y-intercept is 4. c)y=2x^ y-intercept: y = 2x^ y=2(0f y=0 x-intercept: y = 2x^ (0) = 2x^ 0 = x 6. $7' b)y = x' y-intercept; y = x. y - x \ (0) = x^ y=(Of 0 = x y =0 x-intercept: The y-intercept is 0 and the x-intercept is 0. c) y-intercept; 1 1 x-intercept; (0)=lx-2 y = - x ^ 2 y == 2l ( 0 ) ^ 2 4 8=x The y-intercept is 2 and the x-intercept is 8. d) y-intercept; y = ^x + 3 x-intercept; y = -4(0) + 3 -3 = ^ x y = 3 0 = -4x + 3 3 — = X to,o) -s.s? The x-intercept is 0 and the y-intercept is 0. 7-28 =-2/ y =8 I 1 -2x^ 4 The y-intercept is 3 and the x-intercept is 8. Answers will vary, e.g., y = - 2 x + 2 For a linear function, the y-intercept will give a value for b in equations such as y = mx + jb. From this, the coordinates of the x-intercept can be substituted in to calculate m. Chapter 7: Exponential and Logarithmic Functions C:f>;iptot / test ri it) I t , i , p..I . I -p.- ,-:i'c .H. ( ..J ,= ».'ii;i,.l h.notion !>-•.- . O ' 1 * tl.I'. ! > " o /•<.•!•* ; t'-.ij fi I .' 'M M pair f.i •• I;. . . ,-.fi ! • \ !Riwyt:4/| I M. y iiifiircopt j t iici Bcli.ivitjur I V R Stic : • j i l i., h'. (!. j • {y I K > 0, tjic.-c'j .:;t:) nr , b) This data docs not represent an exponential function because it does not have a constant ratio for each pair of successful y-values. x-iiYii'fi rt:'. 4 ' 4 ) I -' Increasing Oi Decreasing? ; OlihiO! y 1 / "^R. ng 2. The y4nter« The graph extends from guadrant II to guadrant I. The domain is {x | x o R) and the range is iy\y>0,ye c) This data does represent an exponential function because it has a constant ratio of 3 for '•> h ' I > V iir ( i~- ••il /- •.,nki I Ratio I R}. The function is increasing. 3. The xontercept is 1. There is no j/-intercept. The graph extends from guadrant IV to guadrant I. The domain is {x | x > 0, x e R} and the range is { y | y e R}. The function is increasing. 4. a) This function matches with ii) because it has a y-intercept of (0,4) and it is increasing. b) Function a) is increasing, because a > T The only graph that shows an increasing loganthmic function is iv). So. function a) matches graph ii). c) The graph for function i) extends from guadrant to quadrant IV and has an xontercept but no y-intercepts. These are charactenstics of a loganthmic function. Function b) matches graph i). d) This matches with iii) because it is a decreasing function with a y-intercept of (0, 4). Foundations of Mathematics 12 Solutions Manual 7. a) Using a graphing calculator, the eguation of the exponential regression function that models the data is y = 2.305...(1.181...)^ The regression equation appears to fit the data very well. The algebraic model would be y - 2.305. .(1.181...)', where y represents the number of dandelions and X represents the date in May. b) Assuming that the population growth continues at the same rate. 401 dandelions will be growing in the yard on May 3 1 . y = 2.305...(1.181...)'' y = 2.305..,(1.181...)^^ y = 401.425... 7-29 8 . Itjr- fM|u;il)<)n i)f t t m l ( K | a i ' f h n i i i : r e g r e s s i o n is y = 1.Hiy » 0 ? B 9 , . (In x) w l i i ' H j \- !o ihs^ «Mit,igv lolf.'osoMi in k J , a n d y is t h e m a < j n i l u d o ul i h o o o d h q i i o k o '/ - 1 1;-)!) » 0.280 (III x ) '» f< - 1 109 + 0,200 (In X) 10.100... 0 . 2 0 0 . . . ( I n X) 3 5 . 2 3 2 . . . - In X x = e3^^-=^32 x = 2 . 0 0 0 . . . X 10^^ A b o u t 2 . 0 X 10^*^ kJ o f e n e r g y w a s r e l e a s e d d u r i n g the e a r t h q u a k e that o c c u r r e d off the c o a s t of J a p a n in March 2 0 1 1 . 7-30 Chaptoi Exponential and Logarithmic Functions

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