Chapter 4 Eponential and Logarithmic Functions Prerequisite Skills for the chapter Eponential and Logarithmic Functions. The domain of the function is.. The range of the function is.. The inverse of the function is 5 ( ). 4. Interest 5 Prt 5 ($5000)(0.0)(4) 5 $600 Total amount 5 P Prt = $5000 $600 5 $5600 5. Interest 5 Prt 5 ($00)(0.04)(5) 5 $460 Total amount 5 P Prt 5 $00 $460 5 $760 6. 7 5 7 p 7 5 49 7. (4) 5 (4)(4)(4) 5 64 8. 4 5 p p p 5 8 9. + 4 5 6 4 5 6 5 4 0. + 5 5 0 5 5 0 5 4 5. 9 + 5 7 5 7 9 5 4 Lesson 4. Graph Eponential Growth Functions Guided Practice for the lesson Graph Eponential Growth Functions.. Range: > 0 Range: > 0. Range: > f() 5 (0, 5) f() 5 (, ) f () (0, ) (, ) 4. Using the graph, ou can estimate that the number of incidents was about 50,000 during 00 (t ø 7). 5. In the eponential growth model 5 57(.9), the initial amount is 57, the growth factor is.9, and the percent increase is.9 5 0.9 or 9%. Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. Algebra Worked-Out Solution Ke 85
6. When p 5 000, r 5 0.04, t 5, and n 5 65: A 5 P r n nt A 5 000 0.04 65 (65)() 5 000 65.04 65 095 ø 54.98 The balance after ears is $54.98. Eercises for the lesson Graph Eponential Growth Functions Skill Practice. In the eponential growth model 5.4(.5), the initial amount is.4, the growth factor is.5, and the percent increase is.5 5 0.5 or 50%.. An asmptote is a line that a graph approaches more and more closel.. C; When 5 0, 5 p 0 5 ; When 5, 5 p 5 6 4. A; When 5 0, 5 p 0 5 ; When 5, 5 p 5 6 5. B; When 5 0, 5 p 0 5 ; When 5, 5 p 5 6 6. 7. 8. 0.. f() 9... g() 5 5? 4 4. h() 5. 6. 5? (0, 7) (0, 5) (0, ) (, 4 ) (, ) (, 6) 5 5? 4 5? (, 6) 5 (, 4) Domain: all real number Range < 0 Range: > 7. 8. 5 (, ) (, 4) (, ) 5 (0, ) 7, 5 (, 0) (0, ) 5, ( ) ( ) Range: > Range: > 9. (, 6) (, 5) Range: > (0, ) 5? (, ) 5? 0. 5? 4 Range: < (, 0.75) (0,.75) (0, ) (, 5) 5? 4. f() f() 5 6? (, 9) (0, 6) (, 6) (, ) f() 5 6? Range: > Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 86 Algebra Worked-Out Solution Ke
Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved... g() 5 5? (0, 5) (, ) 7 (, ) g() 5 5? 4 g() 5 (, ) h() 5? 5 (0, 0.6) h() 5? 5 (, 0.4) (0, ) 4 (, ) Range: > 4 h() Range: < 4. B; When 5 0: (.5) 0 5 () 5 (0, ) When 5 : (.5) 5 5 4 (, 4) 5. D; Using the eponential growth model 5 a( r) t with r 5 0. and a 5 0, the model is 5 0(.) t. 6. The -intercept should be at (0, ), not (0, ). 7. The graph for 5 should have been translated units right and units up rather than units left and units up. 8. 5 number of monk parakeets t 5 number of ears since 99 5 a( r) t 5 9( 0.) t 5 9(.) t 9. A 5 amount in account t 5 ears since deposit A 5 p r n nt A 5 800 0.0 65 65t 0. 5 value of table t 5 ears since purchase 5 a( r) t 5 450( 0.06) t 5 450(.06) t. a. The balance after 7 ears is $844.8. b. You have to press enter 8 times, so it takes 8 ears.. Sample answer: 5 p 4. 5 ab h k 5 p 5 4 a. If a changes to, there will be a vertical shrink. b. If b changes to 4, the graph will increase slower. c. If h changes to, there will be a horizontal shift (of 7 units right) d. If k changes to, there will be a vertical shift (of 4 units down). 4. a. f() 5 ab ; f ( ) 5 ab f ( ) 5 ab f () ab 5 b b 5 b ( ) 5 b b. Check to see if the ratios of consecutive pairs of f ( ) -values satisf the equation 5 b, where b >. f () f () f (0) 5 4 4 5 f () f () 5 8 4 5 f () f () 5 4 8 5 f (4) f () 5 7 4 5 The ratios are not all equal and are not all greater than. So, there is no eponential function f () 5 ab whose graph passes through the points. Problem Solving 5. a. initial amount 5 0.4 million or $40,000; growth factor 5.47; annual percent increase 5.47 5.47 or 47% b. Using the graph, ou can estimate that the number of DVD plaers sold in 00 was about 6 million. Number of DVD plaers sold (millions) n 8 4 0 6 8 4 0 0 4 5 t Years since 997 6. a. Initial amount 5 500; growth factor 5.50; annual percent increase 5.5 5 0.5 or 50% b. Domain: 0,,,, 4, 5 Range: 500, 750, 565, 848,,656, 8,984 Using the graph, ou can estimate that the number of referrals in March of 00 was about,000. Number of referrals 6,000 4,000,000 0,000 8000 6000 4000 000 0 0 4 5 t Years since 998 Algebra Worked-Out Solution Ke 87
7. A 5 00 r n n(4) ; P 5 00; t 5 4 a. r 5 0.0, n 5 4 A 5 00 0.0 4 4 p 4 5 00(.0075) 6 ø 479.8 The balance after 4 ears is $479.8. b. r 5 0.05, n 5 A 5 00 0.05 p 4 5 00(.00875) 48 ø 406.98 The balance after 4 ears is $406.98. c. r 5 0.0, n 5 65 A 5 00 0.0 65 65 p 4 5 00(.0000547945) 460 ø 8. The balance after 4 ears is $8.. 8. a. When A 5 000, r 5 0.05, n 5 4, and t 5 : A 5 p n r nt 000 5 p 0.05 4 4 p 000 5 p (.00565) 804.7 ø p b. When A 5 000, r 5 0.05, n 5, and t 5 : A 5 p n r nt 000 5 p 0.05 p 000 5 p.05 6 70.9 ø p You should deposit $70.9 to have $000 in our account after ears. c. When A 5 000, r 5 0.04, n 5, and t 5 : A 5 p n r nt 000 5 p 0.04 p 000 5 p (.04) 666.99 ø p You should deposit $666.99 to have $000 in our account after ears. 9. a. Initial amount 5 494.9 thousand; percent increase 5 0.0; growth factor 5 0.0 5.0 Eponential growth model: P 5 494.9(.0) t When t 5 0: P 5 494.9(.0) 0 ø 664.84 The population in 000 was about 664,84 people. b. Domain: 0 t 0 Range: 494.9 P 664,84 Population (thousands) P 700 600 500 400 00 00 00 0 0 4 5 6 7 8 9 t Year (t 5 0; 990) c. Using the graph, ou can estimate that the population was about 590,000 during 996 (t ø 6). 40. a. Initial amount 5 50; percent increase 5 0.05; number of bids 5 n Eponential growth model: p 5 50( 0.05) n p 5 50(.05) n b. When n 5 5: p 5 50(.05) 5 ø 8.7. The price after 5 bids was $8.7. After 00 bids (n 5 00), the price will be p 5 50(.05) 00 ø $,084,40.7. No. Sample answer: This amount is unreasonable because the model is onl defined for 6 bids and 00 is out of this domain. 4. a. Initial amount 5 48.8; percent increase 5 0.06 Eponential growth model: p 5 48.8( 0.06) t p 5 48.8(.06) t b. Using the graph, ou can estimate that the price of a ticket was $60 during 00 (t ø.7). Average price (dollars) p 80 70 60 50 40 0 0 0 0 0 4 Years (t 5 0; 000) c. The graph of the model consists of the points (t, p) where 0 t 4 and p 5 48.8(.06) t. Eamine the graph to find the lowest point (which gives ou the minimum p-value) and highest point (which gives ou the maimum p-value). The lowest point occurs at (0, 48.8), so the minimum p-value is 48.8. The highest point occurs at (4, 60.95), so the maimum p-value is 60.95. 4. a. Initial amount 5 4; percent increase 5 0.089 Eponential growth model: n 5 4( 0.089) t n 5 4(.089) t t Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 88 Algebra Worked-Out Solution Ke
b. t n 0 4 44.649 48.6 5.950 4 57.66 t n 9 88.5 0 96.75 04.75 4.056 4.07 t n 7 74.686 8 90. 9 07.64 0 5.60 45.680 Lesson 4. Graph Eponential Deca Functions Guided Practice for the lesson Graph Eponential Deca Functions.. 5 6.795 4 5.6 67.545 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. c. Number of breeding pairs 6 68.8 7 74.470 8 8.097 n 0 00 70 40 0 80 50 0 90 60 0 Years since 977 5 47.0 6 60.40 0 0 6 9 5 8 4 t 9.57 4 7.88 d. Using the graph, ou can estimate that in 00 (t 5 4), there were about 5 breeding pairs of bald eagles. 4. Investing $000 at 6% annual interest: A 5 000(.06) t Investing $000 at 8% annual interest: A 5 000(.08) t Investing $6000 at 7% annual interest: A 5 6000(.07) t 4 A A 640 6870 75.8 7868.90 A 640 6869.40 750.6 7864.78 No, A A and A are not the same. After the first ear, the mone split between the 6% and 8% accounts grows at a faster rate. 44. a. Initial amount 5 500; final amount 5 9000; number of ears 5 5 9000 5 500( r) 5 45 6 5 ( r)5 Î 45 5 6 5 r 0.6 ø r The average annual growth rate was about.6% b. 5 9000( 0.6) 5 5 9000(.6) 5 ø 5,579.86 The cost will be $5,579.86 in 5 more ears.. 5. f() (, ) 4 ( 0, ) (0, 5) 5 5 ( ) 0, 5 5 ( ) ( ) 4. 6. (, 4) (0, ) (0, 5) 5 (, ) ( ) 4 5 ( 4) Range: > g() g() 5 (0, ) ( ) 5 4 4 ( ) g() 5 4 7 6, 4 (5, ) ( ) 9 (, 4 ) Range: > Range: < 4 7. Initial amount 5 400; percent decrease 5 0.0 Eponential deca model: 5 400( 0.0) t 5 400(0.80) t using the graph, ou can estimate that the value of the snowmobile will be $500 after about ears. Value (dollars) 4000 000 000 000 0 0 4 5 6 7 8 9 t Time (r) 8. Final cost 5 000; number of ears 5 ; percent decrease 5 0.07 000 5 a( 0.07) 000 5 a(0.9) 79.69 ø a The original cost of the snowmobile was $79.69. Algebra Worked-Out Solution Ke 89
Eercises for the lesson Graph Eponential Deca Functions Skill Practice. Initial amount 5 50; deca factor 5 0.85; percent decrease 5 0.85 5 0.5 or 5%. The function 5 b represents eponential growth if b >, and eponential deca if 0 < b <.. f () 5 4, b 5 Because 0 < b <, this 4 function represents eponential deca. 4. f () 5 4 5, b 5 5 Because b >, this function represents eponential growth. 5. f () 5 7 p 4, b 5 4 Because b >, this function represents eponential growth. 6. f () 5 5(0.5), b 5 0.5 Because 0 < b <, 7. 9. this function represents eponential deca. f() 8. 0. 6. 8. 0. (0, ) ( 5 (0, ) (, ) ( 4 (, ) ( 5 ( 7. 4 (0, ) ( 5 Range: > Range: < 0 7 ( 0, ) (0, ) (, ) (, ) ( 5 5 5 ( (, ) (, ) (, ) ( 5 ( ( ( ( 9. 5 ( (, ) ( (0, ) 5 ( ( 4 (4, 0) ( 5, ) Range: > Range: > (0, 6) (0, ) 5 (0.5) 5 (, 4 ) (, 4 ) 5 (0.5). (0, ) ( 5 (, ) (, ) ( 7 (, ) ( 5 (.. 5. B; 5 5. 4. h() g() When 5 0: 5 5 0 5 () 5 (0, ) When 5 : 5 5 5 5 5 6 5, 6 5. 4. Range: > Range: > f(). g() f() 5 ( 4) (0, 6) g() 5 6 5 ( ) (5, 4) (, (, ) 4 ) (0, ) (, ) (4, ) f() 5 ( 4) 5 g() 5 6 (, 4 ) ( ) Range: < 0 Range: > (0, 4) (, 4) (, ) (0, ) h() h() 5 4 ( ) ( ) h() 5 4 Range: > 0 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 90 Algebra Worked-Out Solution Ke
Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 5. 5 ab h k 5 (0.4) a. If a changes to 4, there will be a vertical stretch. b. If b changes to 0., the graph will be steeper because the deca factor is smaller. c. If h changes to 5, there will be a horizontal translation (of units right) d. If k changes to, there will be a vertical translation (of 4 units up). 6. The deca factor was written incorrectl. It should be percent decrease 5 0.0 5 0.98. 5 (inital amount)(deca factor) t 5 500(0.98) t 7. D; The graph 5 has an asmptote at the line 5. 8. You need a base between 0.5 and 0.5 and a vertical translation up between 0 and units. Sample answer: 5 (0.) 9. Yes; (4) is just another wa of saing 4 or (0.5), so f () 5 5(4) and g() 5 5(0.5) represent the same function. Problem Solving 0. a. When I 5 00 and t 5.5: A 5 I(0.7) t A 5 00(0.7) 5 ø 9.65 There is about 0 milligrams of ibuprofen remaining in the bloodstream. b. When I 5 5 and t 5.5: A 5 I(0.7) t A 5 5(0.7).5 ø 98.0 There is about 98 milligrams of ibuprofen remaining in the bloodstream. c. When I 5 400 and t 5 5: A 5 I(0.7) t A 5 400(0.7) 5 ø 7.7 There is about 7 milligrams of ibuprofen remaining in the bloodstream.. a. When r 5 0.5 and a 5 00: 5 a( r) t 5 00( 0.5) t 5 00(0.75) t When t 5 : 5 00(0.75) ø 84.8 The value of the bike after ears is about $84.8. c. Using the graph, ou can estimate that the value of the bike will be $00 after about.5 ears. Value (dollars) 00 75 50 5 00 75 50 5 0 0 4 5 6 7 8 t Years. Ratio from ear : 8 906 ø 0.96. a. Ratio from ear : 76 8 ø 0.96 Ratio from ear 4: 69 76 ø 0.96 Ratio from ear 4 5: 67 69 ø 0.96 The ratio of depreciation remains a constant 0.96 or 96%. d 5 a(0.96) 5 906 a 5 906 0.96 ø 985 An equation is d 5 985(0.96) t. Value (dollars) 8,000 4,000 0,000 6,000,000 8000 4000 0 0 4 5 6 7 8 t Years Using the graph, ou can estimate that the value of the car will be $0,000 after about 5 ears. b. When t 5 50: 5 4,000(0.845) 50 ø 5.9 The value after 50 ears is $5.9 according to the model. Sample answer: This is too low to be reasonable for the price of a car. In addition, car values usuall begin to increase once the become antiques. 4. a. P 5 00 t /570 When t 5 500: P 5 00 500/570 ø 7.90 When t 5 5000: P 5 00 5000/570 ø 54.6 When t 5 0,000: P 5 00 0,000/570 ø 9.8 After 500 ears there will be about 7.9% of the original carbon-4 remaining, after 5000 ears there will be about 54.6% remaining, and after 0,000 ears there will be about 9.8% remaining. b. P Percent 00 90 80 70 60 50 40 0 0 0 0 0 4000 8000,000 Years 6,000 c. Using the graph, ou can estimate that the bison bone is about 8000 ears old when 7% of the carbon-4 is present. t Algebra Worked-Out Solution Ke 9
5. a. The deca factor is 0.89 and the percent decrease is 0.89 5 0., or %. b. E Number of eggs produced per ear 80 70 60 50 40 0 0 0 00 0 0 0 40 60 80 00 0 40w Age of chicken (weeks) c. There are 5 weeks in a ear, so when a chicken is.5 ears old, it is.5(5) 5 0 weeks old. Using the graph, ou can estimate that the number of eggs produced b a 0 week old chicken is about 4 per ear. d. Let t represent the chicken s age in ears. Because t 5 w, ou can rewrite the equation as 5 E 5 79.(0.89) t. 6. Initial amount 5 00; resale value after t ears 5 75; number of ears 5 4 V 5 a( r) t 75 5 00( r) 4 0 p 5 ø ( r) 4 0.678 ø r The deca factor is 0.678, so an equation giving the stereo s resale value as a function of time is V 5 00(0.678) t. Etension for the lesson Graph Deca. 5 750(.0) t 5 750(.0) (/7)(7t) 5 750.0 /7 7t < 750(.008) 7t The dail percent increase in electricit is about 0.008, or about 0.8%.. A 5 n t/5 5 n /5 t < n(0.95484) t 5 n( 0.04559) t The hourl deca rate is about 0.045, or about 4.5%.. 5 400( 0.8) t 5 400(0.7) t 5 400(0.7) (/65)(65t) 5 400 0.7 /65 65t < 400(0.99900) 65t 5 400( 0.000900) 65t The dail percent decrease in the value is about 0.00090, or about 0.090%. 4. 5 n(.07) t 5 n.07 (/)t 5 n.07 / t < n(.0058) t The monthl percent increase in the value is about 0.0058, or about 0.58%. 5. A 5 P() t/0 5 P /0 t < P(.0777) t 5 P( 0.0777) t The annual interest rate needed is about 0.078, or about 7.8%. 6. A 5 P() t/ 5 P / t < P(.599) t 5 P( 0.599) t The dail percent increase in the population is about 0.599, or about 5.99%. 7. To find the annual rate of increase of the quantit, find /n, where n is the number of ears it takes for the quantit to double. Lesson 4. Use Functions Involving e Guided Practice for the lesson Use Functions Involving e. e 7 p e 4 5 e 7 4 5 e. e p 6e 5 5 e 5 5 e. 4e8 4e 5 5 6e8 5 5 6e 4. (0e 4 ) 5 0 (e 4 ) 5 000e 5 000 e 5. e /4 ø.7 6. 7. ( 0, ) (, 0.8) f() ( ) f() 5 e ( ) 0, (,.8) ( ) f() 5 e Range: > 0 Range: > Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 9 Algebra Worked-Out Solution Ke
8. 5.5e 0.5 ( ) 0, (, ) (,.9) (, 0.07) Range: > 0. e p 5e 5 5e 5 5e. e e 5 e. 4e e 4 5 4e 4 5 4e 5 4 e. Ï 8e 9 5 Ï 8 Ï e 9 5 e 9. 5.5e 0.5( ) 4. 6e4 8e 5 4 e4 5. C; (4e ) 5 4 (e ) 5 64e 6 6. D; Î 4(7e ) e 7 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. X=5 Y=.6590 Use the trace feature to determine that l ø 4 when t 5 5. The length of a 5-ear-old tiger shark is about 4 centimeters. 0. A 5 Pe rt where p 5 500 and r 5 0.05 a. When t 5 : A 5 500e 0.05() 5 500e 0. ø 76.9 The balance after ears is $76.9. b. When t 5 5: A 5 500e 0.05(5) 5 500e (0.5) ø 0.06 The balance after 5 ears is $0.06 c. When t 5 7.5: A 5 500e 0.05(7.5) 5 500e 0.75 ø 67.48 The balance after 7.5 ears is $67.48.. Amount of interest earned 5 balance principle a. The interest earned after ears is 76.9 500 5 $6.9. b. The interest earned after 5 ears is 0.06 500 5 $70.06. c. The interest earned after 7.5 ears is 67.48 500 5 $7.48. Eercises for the lesson Use Functions Involving e Skill Practice. The number e is an irrational number approimatel equal to.788.. The function f () 5 e4 is an eample of eponential growth because > 0 and 4 > 0.. e p e 4 5 e 4 5 e 7 4. e p e 6 5 e 6 5 e 4 5. (e ) 5 (e ) 5 8e 9 6. (e ) 4 5 4 (e ) 4 5 6 e8 5 e8 6 7. (e 5 ) 5 (e 5 ) 5 e5 5 e 5 8. e p e p e 4 5 e 4 5 e 4 9. Ï 9e 6 5 Ï 9 Ï e 6 5 e 5 Î 08e e 7 5 Ï 6e 7 () 5 Ï 6e 6 4 5 Ï 6 Ï e 6 Ï 4 5 6e 7. The power should have been applied to the also. (e 5 ) 5 (e 5 ) 5 9e 0 8. The term rather than should have been subtracted from 6. e 6 e 5 e6 () 5 e 8 9. e ø 0.086 0. e /4 ø 0.47. e. ø 9.05. e / ø.649. e /5 ø 0.670 4. e 4. ø 7.700 5. e 7 ø 096.6 6. e 4 ø 0.08 7. e 0. ø.48 8. 5e / ø 9.79 9. 6e.4 ø 66.9 0. 0.4e 4. ø 4.6. The function f () 5 e is an eample of eponential deca.. The function f () 5 e4 is an eample of eponential growth.. The function f () 5 e 4 is an eample of eponential deca. 4. The function f () 5 5 e is an eample of eponential growth. 5. The function f () 5 4 e5 is an eample of eponential deca. 6. The function f () 5 e is an eample of eponential growth. 7. The function f () 5 e 4 is an eample of eponential growth. 8. The function f () 5 4e is an eample of eponential deca. 9. B; When 5 0, 5 0.5e 0.5(0) 5 0.5; When 5, 5 0.5e 0.5() ø 0.8 40. C; When 5 0, 5 e 0.5(0) 5 ; when 5, 5 e 0.5() ø.0 Algebra Worked-Out Solution Ke 9
4. A; When 5 0, 5 e 0.5(0) 5 ; When 5, 5 e 0.5() ø.7 4. 44. 46. 48. 49. 4. Range: > 0 Range: > 0 45. (0, ) (0, ) (, 0.0) 5 e 5 e (, 0.90) Range: > 0 Range: > 5 ( 0, ) 5.5e 0.5 ( ) 9 0, (,.5) (,.5) 5.5e 0.5 47. 5 0.6e (,.6) ( 0, 5 ) (,.6) (, 5 ) 5 0.6e Range: > Range: > 0 f() 5 e f() 5 e ( ) f() (,.6) ( 0, ), (, 0.64) (, 4.6) (,.6) 4 g() 5 e g() 5 4 e g() 7 (, ) 4 ( 0, ) Range: > Range: > 50. 5 h() (, 0.4) (0, ) h() 5 e (, ) (0,.86) h() 5 e ( ) 5 Range: > 5. Using the table feature, ou can notice that incrementing small values of n increases the function ver slowl. Incrementing n b powers of 0 gives values that are one digit closer to the actual value of e each time. When n 5 0 0, the value of the function is.78888, which is the value of e correct to 9 decimal places. 5. No, e cannot be epressed as a ratio of two integers because it is an irrational number; it is a decimal that neither terminates nor repeats. 5. Choose a, b, r, and q, such that a > 0, b > 0, r < 0, q < 0, and r q > 0. Sample answer: f () 5 e, g () 5 e 4 f () g () 5 e e 5 4 e (4) 5 e 54. Let m 5 n r, so n 5 rm and r n 5 m. A 5 P n r nt 5 P m rmt 5 P F m m rt G As n approaches `, m approaches `, and m m approimates e. So, A 5 P n r nt approimates A 5 Pe rt as n approaches `. 55. When 5 00 997 5 5: 5.8e. 5.8e.(5) 5.8e 6.55 ø 895 About 895 million camera phones were shipped globall in 00. 56. When t 5 999 989 5 0: 5 78e 0.45t 5 78e 0.45(0) 5 78e.45 ø,47 About,47 termites were collected in 999. 57. When P 5 000, r 5 0.04, and t 5 5: A 5 Pe rt A 5 000e (0.04 p 5) 5 000e 0. ø 44.8 The balance after 5 ears is $44.8. 58. When P 5 800, r 5 0.065 and t 5.5: A 5 Pe rt A 5 800e (0.065)(.5) 5 800e 0.5 ø 4.7 The balance after.5 ears is $4.7. Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 94 Algebra Worked-Out Solution Ke
Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 59. a. When k 5 0.0: L() 5 00e k L() 5 00e 0.0 Percent of light L() 00 90 80 70 60 50 40 0 0 0 0 0 0 40 60 80 Depth below water surface (m) b. Using the graph, ou can estimate that the percent of surface light is about 45% at a depth of 40 meters. c. Using the graph, ou can estimate that the submarine can descend about 5 meters before onl 50% of surface light is available. 60. a. P(t) 5 P 0 e 0.6t ; when P 0 5 0: P(t) 5 0e 0.6t b. Population P(t) 90 80 70 60 50 40 0 0 0 0 0 4 6 8 t Hours after :00 P.M. c. Using the graph, ou can estimate that the bacteria population is about 48 at 5:00 p.m. d. P(.75) 5 0e 0.6(.75) 5 0e 0.9 ø 4 There are.75 hours between :00 p.m. and :45 p.m. so, let t 5.75. There are about 4 bacteria at :45 p.m. 6. A 5 A 0 e 0.05t where A 0 5 4 and t 5 4 A 5 4e 0.05(4) 5 4e 0.7 ø.99 The area after 4 das is about square centimeters. 6. a. X=0 Y=60 The arch is 60 feet tall at its highest point. b. The -intercepts are approimatel 5 65, so the ends of the arch are about 5 (5) 5 60 feet apart. Quiz for the lessons Graph Eponential Growth Functions, Graph Eponential Deca Functions and Use Functions Involving e... (, 6) (0, ) (, ) 5? 5? (, 6) Range: > 0 Range: > 0 8 (, ) f() 4 (, ) f() 5 ( 8) (0, ) (0, ) ( ) f() 5 8 4. e 4 p e 5 e 4 5 e 7 5. (5e ) 5 5 (e ) 5 5e 9 Range: > 6. e4 5e 5 5 e4 7. 8e5 6e 5 4e5 5 4e 8. 0. 9. Range: > 0 Range: > 0 (,.7) (0, ) 5 e (0, 0.7) (, ) 5 e. (0, 4) g() (0, 5) g() 5 4e (,.0) (, 0.0) g() 5 4e Range: > Range: > Algebra Worked-Out Solution Ke 95
. Deca factor 5 0.85; percent decrease 5 0.85 5 0.5 TVs sold (millions) n 0 0 0 0 0 4 6 8 t Years since 997 Domain: 0 t 4 Range: 6.8,.8, 9.4, 6.5, 4.0; using the graph ou can estimate that the number of black-and-white TVs sold in 999 (t 5 ) was about 9 million.. When P 5 00, r 5 0.045, and t 5 5: A 5 Pe rt A 5 00e (0.045)(5) 5 00e (0.5) ø 50.79 The balance after 5 ears is $50.79. Lesson 4.4 Evaluate Logarithms and Graph Logarithmic Functions Guided Practice for the lesson Evaluate Logarithms and Graph Logarithmic Functions. log 8 5 4. log 7 7 5 4 5 8 7 5 7. 5 0 4. log / 5 5 4 0 5 5 5 5. 5 5, so log 5 5 6. 7 / 5, so log 7 5 7. log ø.079 8. ln 0.75 ø 0.88 9. s 5 9 log d 65 5 9 log 50 65 ø 9(.76) 65 5 67.68 The wind speed near the tornado s center is about 67 miles per hour 0. 8 log 8 5. log 7 7 5. log 64 5 log 6 5 6. e ln 0 5 e log e 0 5 0 4. From the definition of logorithm, the inverse of 5 4 is 5. 5. 5 ln ( 5) 5 ln ( 5) e 5 5 e 5 5 The inverse of 5 ln ( 5) is 5 e 5. 6. Domain: > 0 Range: all real numbers 7. 8. 5 log / ( ) 8 ( ) 9, 9, ( ) 0 (, ) (4, 0) (, ) Domain: > (, 0) 5 log / Range: all real numbers f() (, 0) (0, ) (4, ) f() 5 (6, ) (5, 0) (, ) f() 5 ( ) Domain: > Range: all real numbers Eercises for the lesson Evaluate Logarithms and Graph Logarithmic Functions Skill Practice. A logarithm with base 0 is called a(n) common logarithm.. From the definition of logarithm, the inverse of 5 5 is 5 log 5, so 5 5 and 5 log 5 are inverses.. 6 5 4. log 7 4 5 4 5 6 7 5 4 5. log 6 6 5 6. log 64 5 0 6 5 64 0 5 6 7. The and 8 should be switched. Because 5 8, the equation should be log 8 5. 8. 5 5 5, so log 5 5 5. 9. 7 5 49, so log 7 49 5. 0. 6 5 6, so log 6 6 5.. 6 5 64, so log 64 5 6.. 9 0 5, so log 9 5 0.. 5 8, so log / 8 5. 4. 5 7, so log 7 5. 5. 6 / 5 4, so log 6 4 5. 6. 4 5 6, so log /4 6 5. 7. 8 5 5, so log 8 5 5. 8. 5 4 5 65, so log 5 65 5 4. 9. 5, so log 5. Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 96 Algebra Worked-Out Solution Ke
Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 0. ø.46. ln 6 ø.79. ln 0.4 ø 0.844. log 6. ø 0.79 4. log 7 ø.4 5. ln 5.8 ø.68 6. log 0.746 ø 0.7 7. ln 0 ø 4.700 8. 7 log 7 5 9. log 5 5 5 0. 0 log 0 4 5 4. 0 log 8 5 8. log 6 6 5 log 6 6 5. log 8 5 log 4 5 4 4. log 5 5 5 log 5 5 5 5. log 5 log 5 5 5 6. B; log 00 5 log 0 0 5 7. The inverse of 5 log 8 is 5 8. 8. The inverse of 5 7 is 5 log 7. 9. The inverse of 5 (0.4) is 5 log 0.4. 40. The inverse of 5 log / is 5. 4. 5 e 5 e ln 5 ln 5 The inverse of 5 e is 5 ln. 4. 5 5 5 log ( ) 5 The inverse of 5 is 5 log ( ). 4. 5 ln ( ) 5 ln ( ) e 5 e 5 The inverse of 5 ln ( ) is 5 e. 44. 5 6 log 45. 5 6 log 6 5 log 0 6 5 The inverse of 5 6 log is 5 0 6 46. Domain: > 0 Domain: > 0 Range: all real numbers Range: all real numbers 47. 49. 50. 5. 5. 5. 48. Domain: > 0 Domain: > 0 Range: all real numbers 5 log (4, ) (7, ) (, ) (5, ) (, 0) (4, 0) 5 log ( ) Domain: > Range: all real numbers 5 log 4 (9, 6) (, 5) (, 4) (, ) (9, ) (, 0) 5 log Domain: > 0 Range: all real numbers f() f() 5 (6, ) (4, ) (4, ) (, 0) (, ) (, 0) f() 5 ( ) Domain: > Range: all real numbers g() g() 5 log 6 ( 4) (40, 4) (5, ) (0, ) (6, ) (, 0) (6, ) 4 g() 5 log 6 Domain: > 4 Range: all real numbers h() h() 5 log 5 (5, ) (, 0) (4, ) (0, ) h() 5 log 5 ( ) Domain: > Range: all real numbers Range: all real numbers Algebra Worked-Out Solution Ke 97
54. log 7 9 5 55. log 8 5 7 5 9 8 5 5 5 5 5 5 5 5 5 5 56. log 5 65 5 57. 8 5 5 5 65 4 5 8 5 5 5 4 5 7 5 4 5 7 5 4 P 58. When P 5 57,000: h 5 8005 ln 0,00 h 5 8005 ln 57,000 0,00 ø 460 5 7 The altitude is about 460 meters above sea level when the air pressure is 57,000 pascals. 59. When H 5 0. : ph 5 log[h] ph 5 log[0. ] 5. The ph of lemon juice is.. 60. l Length (in.) 40 0 00 80 60 40 0 0 0 40 80 0 60 00 40 80 w Weight (lb) in. 0 ft p 5 0 in. ft Using the graph, ou can estimate that the alligator weighs about 8 pounds when it is 0 feet (0 inches) long. 6. a. When E 5.5 0 4 : M 5 0.9(ln E) 9.9 M 5 0.9 ln (.50 4 ) 9.9 ø 0.9(56.78) 9.9 ø 6.9 The energ magnitude of the earthquake was about 6.4. b. M 5 0.9(ln E ) 9.9 M 9.9 5 0.9(ln E ) M 9.9 0.9 5 ln E e (M 9.9)/0.9 5 E The inverse of the function, E 5 e (M 9.9)/0.9, represents the amount of energ released (in ergs) b an earthquake of energ magnitude M. 6. a. b. Using the graph, ou can estimate that there are about 5 different kinds of fish species when the area is 0,000 square meters. c. Using the graph, ou can estimate that the area of the lake is about 4000 square meters when there are 6 species of fish. d. As the area of the pool or lake increases, the number of fish species also increases. The answer makes sense because the larger the pool or lake, the more room there is for more varieties of fish to thrive. 6. s 5 0.59 0.8(log d ) s 0.59 5 0.8(log d ) s 0.59 0.8 5 log 0 d s 0.59 0.8 0 5 d The inverse of the function is d 5 0 (s 0.59)/0.8. 0. 0.59 When s 5 0.: d 5 0 0.8 ø. Sand particles on a beach with a slope of 0. have an average diameter of about. millimeters. Mied Review of Problem Solving for the lessons Graph Eponential Growth Functions, Graph Eponential Deca Functions, Use Functions Involving e and Evaluate Logarithms and Graph Logarithmic Functions. a. Let 5 log ( h) k. When 5 ; 5 log ( h) k 5 When 5 5; 5 log (5 h) k 5 log (5 h) log ( h) 5 log 5 h h 5 h 5 When h 5 : log ( h) k 5 log ( ) k 5 log k 5 0 k 5 k 5 Therefore, 5 log ( ). Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 98 Algebra Worked-Out Solution Ke
Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. b. 5 log ( ) 5 log ( ) 5 log ( ) log( ) 5 5 5. a. Fold number 0 4 Number of regions 4 8 6 Fractional area of each region 4 8 6 b. R(n) 5 n ; A(n) 5 n ; R(n) represents eponential growth and A(n) represents eponential deca.. Sample answer: Let r 5 0.05. Item of increasing value: 5 a( 0.05) t Item of decreasing value: 5 a( 0.05) t 00 5 a(.05) 00 5 a(0.95) 90.70 ø a 0.80 ø a An equation is An equation is 5 90.70(.05) t. 5 0.80(0.95) t. Value (dollars) 0 90 60 0 (, 00) 5 90.70(.05) t 5 0.80(0.95) t 0 0 4 6 8 t Years 4. A 5 Pe rt when P 5 000, r 5 0.04, and t 5 : A 5 000e 0.04() 5 000e 0.08 ø 66.57 Your balance after ears is $66.57. A 5 Pe rt when P 5 000, r 5 0.04, and A 5 50: 50 5 000e 0.04t.5 5 e 0.04t ln(.5) 5 0.04t 0.77 ø 0.04t.94 ø t After full ears for our balance will eceed $50. You must round up because the answer must be in full ears. 5. a. A 5 P n r nt First CD: t 5, p 5 500, r 5 0.0 and n 5 A 5 500 0.0 () 5 500.0 6 ø 59.68 Second CD: t 5 5, p 5 000, r 5 0.0, and n 5 A 5 000 0.0 (5) 5 000(.005) 60 ø. The first CD ields $59.68 at the end of its term and earns 59.68 500 5 $9.68 interest. The second CD ields $. at the end of its term and earns. 000 5 $. interest. b. The difference in the amounts of interest is $. $9.68 5 $0.55. c. Sample answer: The first CD requires a small minimum balance, and the mone can be withdrawn sooner, but the second CD earns more interest. 6. 5 0e 0.0564t when t 5 0: 5 0e 0.0564(0) 5 0e 0.564 ø 5.69 There are about 5.69 milligrams of tritium left after 0 ears. 7. a. Oil collected (billions of barrels) 60 50 40 0 0 0 0 0 500 000 500 000 500 Wells drilled b. Using the graph, ou can estimate that ou could epect to collect about 40 billion barrels after drilling 000 wells. c. Using the graph, ou can estimate that about 400 wells need to be drilled in order to collect 50 billion barrels of oil. Lesson 4.5 Appl Properties of Logarithms Guided Practice for the lesson Appl Properties of Logarithms. log 6 5 8 5 log 6 5 log 8 ø 0.898.6 5 0.6 6. log 6 40 5 log 6 (8 p 5) 5 log 6 8 log 6 5 ø.6 0.898 5.059. log 6 64 5 log 6 8 5 log 6 8 ø (.6) 5. 4. log 6 5 5 log 6 5 5 log 6 5 ø (0.898) 5.694 5. log 4 5 log log 4 5 log 4log 6. ln 4 ln ln 5 ln 4 ln ln 7. log 5 8 5 log 8 log 5 ø 0.90 0.6990 ø.9 5 ln 4 ln 7 ln 8. log 8 4 5 log 8 ø.46 0.90 ø.69 5 ln 4 p 7 08 5 ln 5 ln 9 Algebra Worked-Out Solution Ke 99
9. log 6 9 5 log 9 log 6 ø 0.954.450 ø 0.674 log 0 0. log 0 5 log ø.477.079 ø.69. Increase in loudness 5 L(I ) L(I ) 5 0 log I I 0 0 log I I 0 5 0 log I I 0 log I I 0 5 0 log log I I 0 log I I 0 5 0(log ) ø 4.8 The loudness increases b about 4.8 decibels. Eercises for the lesson Appl Properties of Logarithms Skill Practice. To condense the epression log log, ou need to use the product propert of logarithms.. You can evaluate log 7 b using the change-of-base formula with common logarithms or with natural logarithms, and then using a calculator to evaluate the result.. B; ln 6 ln 5 ln 6 5 ln 4. D; ln 6 5 ln 6 5 ln 6 5. A; 6 ln 5 ln 6 5 ln 64 6. C; ln 6 ln 5 ln 6 p 5 ln 7. log 5 log 5 log 4 ø.079 0.60 5 0.477 8. 8 5 log p 4 5 log ø.079 0.60 5.68 9. log 6 5 5 ø (0.60) 5.04 0. log 64 5 5 ø (0.60) 5.806. 4 5 log 5 log ø (.079) 5.58. log 5 log 4 5 log ø 0.60.079 5 0.477. log 5 log ø 0 0.60 5 0.60 4 4. log 5 log log ø 0.079 5.079 5. log 4 5 log 4 log 6. ln 5 5 ln 5 ln 7. log 4 5 log log 4 5 log 4 log 8. log 5 5 5 5 log 5 9. log 5 5 log log 5 0. ln 5 ln ln 5 5. 5 5. ln 4 5 ln 4 ln ln 5 ln 4 ln ln. log 7 5 z 5 log 7 5 log 7 log 7 log 7 z 5 log 7 5 log 7 log 7 log 7 z 4. log 6 6 5 log 6 6 log 6 5 log 6 6 log 6 5. ln / 5 ln ln / 5 ln ln 6. log 0 5 log 0 log 5 log 0 log 7. log Ï 5 log / 5 log 8. ln 6 4 5 ln 6 ln 4 5 ln 6 ln ln 4 5 ln 6 ln 4 ln 9. ln 4 Ï 5 ln /4 5 4 ln 0. log Ï 9 5 log Ï 5 log / 5 log log / 5 log log. When epanding the logarithmic epression, the quantities (log 5) and (log ) should have been added, not multiplied. log 5 5 log 5 log. When epanding the logarithmic epression, the should have been multiplied b ln, not ln 8. ln 8 5 ln 8 ln 5 ln 8 ln. 7 0 5 7 0 4. ln ln 4 5 ln 4 5 ln 5. log log 5 log log 5 log 6. 6 ln 4 ln 5 ln 6 ln 4 5 ln 6 4 7. 5 log 4 log 5 log 5 log 4 5 log 5 4 8. 5 7 4 5 5 7 4 5 7 4 9. ln 40 ln ln 5 ln 40 ln ln 5 ln 40 ln 4 ln 5 ln 40 4 5 ln 0 40. log 5 4 log 5 5 log 5 4 log 5 / 5 log 5 4 / 5 log 5 4 Ï 4. 6 ln 4 ln 5 ln 6 ln 4 5 ln 64 ln 4 5 ln 64 4 4. (log 0 log 4) 0.5 log 4 5 log 0 4 0.5 log 4 5 log 5 0.5 log 4 5 log 5 log 4 0.5 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 00 Algebra Worked-Out Solution Ke
Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 5 log 5 log 5 log 5 p 5 log 50 4. C; 6 5 6 5 6 44. D; log 8 log 5 log 8 log 5 log 8 log 9 5 log 8 p 9 5 log 7 log 7 Þ log 48 45. 7 5 log 7 ø 0.845 0.60 ø.404 log 46. log 5 5 log 5 ø.9 0.6990 5.594 log 5 47. log 5 5 log ø.76 0.477 ø.465 log 48. log 5 log 8 ø.44 0.90 ø.486 49. log 6 5 log 6 log ø 0.778 0.477 ø.6 50. log 5 4 5 log 5 ø.46 0.6990 ø.640 log 7 5. log 6 7 5 log 6 ø.04 0.778 ø.58 log 8 5. log 8 5 log ø.447 0.00 ø 4.808 log 9 5. log 7 9 5 log 7 ø.788 0.845 ø.5 8 54. 48 5 ø.68 0.60 ø.79 log 7 55. log 9 7 5 log 9 ø.44 0.954 ø.500 log 56. log 8 5 log 8 ø.505 0.90 ø.667 4 57. log 6 4 log 5 5 5 log 6 5 log 5 log 6.80 0.6990 ø ø 0.875 0.778 5 log 5 58. log 7 5 7 log 5 log 5 log 7 log.76 0.845 ø ø.00 0.00 9 59. log 9 log 40 5 40 log 5 log 9 0 log 0.954.60 ø ø.58 0.477 log 60. log 7 6 5 6 log 7 5 log log 6 log 7 0.477.04 ø ø 0.860 0.845 6. When using the change-of-base formula, log 7 should have been divided b log. log 7 5 log 7 log 6. a. I 5 0 4 w/m ; I 0 5 0 w/m L(I ) 5 0 log I 5 0 log I 0 04 0 5 0 log 08 5 0 p 8 5 80 The dog produces about 80 decibels of sound. b. I 5 0 0 w/m ; I 0 5 0 w/m L(I ) 5 0 log I 5 0 log I 0 08 0 5 0 log 0 5 0 p 5 0 The ambulance siren produces about 0 decibels of sound. c. I 5 0 6.5 w/m ; I 0 5 0 w/m L(I ) 5 0 log I 5 0 log I 0 06.5 0 5 0 log 05.5 5 0 p 5.5 5 55 The bee produces about 55 decibels of sound. 6. I 5 0 w/m ; I 0 5 0 w/m L(I ) 5 0 log I 5 0 log I 0 0 0 5 0 log 05 5 0 p 5 5 50 The trumpet produces about 50 decibels of sound. 64. a. Sample answer: The statement log b (M N) 5 log b M log b N is false when b 5, M 5, and N 5 4. log ( 4) 0 log log 4 log 7 0 log log 4 log 7 0 log p 4 log 7 Þ log b. Sample answer: The statement log b (M N) 5 log b M log b N is false when b 5, M 5 4, and N 5 log (4 ) 0 log 4 log log () 0 log 4 log log Þ log 4 65. Let 5 log b m and 5 log b n. Then m 5 b and n 5 b. log b mn 5 log b (b b ) 5 log b (b ) 5 5 log b m log b n 66. Let 5 log b m and 5 log b n. Then m 5 b and n 5 b. b m log b n 5 log b b 5 log b b 5 5 log b m log b n 67. Let 5 log b m. Then m 5 b and m n 5 b n. log b m n 5 log b b n 5 n 5 n log b m Algebra Worked-Out Solution Ke 0
68. Let 5 log b a, 5 log b c, and z 5 log c a. Then a 5 b, c 5 b, and a 5 c z, so that b 5 c z. 5 log b a 5 log b b 5 log b c z 5 zlog b c 5 z 5 z z 5 log c a 5 log b a log b c Problem Solving 69. I 5.4 0 5 w/m ; I 0 5 0 w/m ; n 5 L(I ) 5 0 log ni I 0 5 0 log.4 05 0 5 0,000,000 ø 0(7.65) ø 76.5 The decibel level of the combined conversations in the room is about 76 decibels. 70. I 5. 0 4 w/m ; I 0 5 0 w/m ; n 5 5 L(I ) 5 0 log ni I 0 5 0 log 5. 04 0 5 0 log (,600,000,000) ø 0(9.04) ø 9.04 All five cars in the parking garage make about 9 decibels of sound. 7. Increase in loudness 5 L(0I ) L(I ) 5 0 log 0I I 0 0 log I I 0 5 0 log 0I I 0 log I I 0 Substitute. Distributive propert 5 0 log 0 log I log I I 0 I 0 Product Propert 5 0 log 0 Simplif. 5 0 log 0 0 5 The loudness increases b 0 decibels. 7. Blue whale: I 5 0 6.8 w/m ; I 0 5 0 w/m L(I ) 5 0 log I I 0 5 0 log 06.8 0 ø 0 log (0 8.8 ) 5 0 p 8.8 5 88 Human: I 5 0 0.8 w/m ; I 0 5 0 w/m L(I ) 5 0log I I 0 5 0log 00.8 0 ø 0log(0.8 ) 5 0 p.8 5 8 A blue whale can produce sounds that are 88 8 5 60 decibels louder than those produced b a human. 7. a. s 5 log f 5 log f b. f.44.000.88 4.000 s 4 f 5.657 8.000.4 6.000 s 5 6 7 8 The amount of light that enters the camera increases b about each time. c. When s 5 9, an equation relating s and f is: 9 5 log f 9 5 log f 9/ 5 log f 9/ 5 f So, the ninth f -stop is about 9/ ø.67. 74. a. Let h 5 original altitude, so that h is the doubled altitude. Increase in speed 5 s(h) s(h) 5 ln[00(h)] ln (00h) 5 [ln (00h) ln (00h)] 5 [ln ( p 00h) ln (00h)] 5 [ln ln (00h) ln (00h)] 5 ln ø.9 The wind speed increases b about.9 knots when the altitude doubles. log (00h) b. s(h) 5 ln (00h) 5 log e 5 p log (00h) log e 5 log e (log 00 log h) 5 ( log h) log e Quiz for the lessons Evaluate Logarithms and Graph Logarithmic Functions and Appl Properties of Logarithms. 4 5 6, so 6 5. 5 0 5, so log 5 5 0. 8 5 8, so log 8 8 5 4. 5 5, so log / 5 5 5. 6. 7. (, ) 5 ln (,.69) (, 0.69) (, 0) 5 ln Domain: > 0 Domain: > 0 Range: all real numbers (9, ) (, ) (, 0) (5, ) 4 (, 0) (, ) Domain: > 4 5 log 5 log ( 4) Range: all real numbers 8. log 5 5 log 5 log 9. log 5 7 5 7log 5 Range: all real numbers 0. ln 5 5 ln 5 ln ln 5 ln 5 ln ln Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 0 Algebra Worked-Out Solution Ke
Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved.. log 6 4 8 5 log 64 log 8 5 log 6 log 4 log 8 5 log 6 4 log 8 log 5. log 5 log 0 5 log 0 5 log 4. ln 6 ln 4 5 ln 6 p 4 5 ln 4 4. log 6 5 log 6 5 log 6 5 log 6 5 log 6 5 log 6 8 5 log 6 8 p 5 5 log 6 40 5. 4 ln 5 ln 5 ln 4 ln 5 5 ln 4 5 ln 5 log 0 6. log 0 5 log ø 0.477 ø.096 7. log 7 4 5 log 7 ø.46 0.845 ø.56 8. log 5 4 5 log 5 ø.80 0.6990 ø.975 0 9. log 8 40 5 log 8 ø.60 0.90 ø.774 0. I 5 0 4 W/m ; I 0 5 0 W/m L(I ) 5 0 log I I 0 5 0 log 04 0 5 0 log 0 8 5 0 p 8 5 80 The alarm clock s loudness is 80 decibels. Graphing Calculator Activit for the lesson Appl Properties of Logarithms... 4. 5. 6. 7. 8. 9. 0.... The natural log, or ln, is equal to log e. So, if ou did not have a natural logarithm then ou could graph 5 ln b graphing 5 log log e. Lesson 4.6 Solve Eponential and Logarthmic Equations Guided Practice for the lesson Solve Eponential and Logarthmic Equations. 9 5 7. 00 7 5 000 ( ) 5 ( ) (0 ) 7 5 (0 ) 4 5 0 4 5 0 9 6 4 5 4 5 9 6 5 5 5 8. 8 5 5 6 4. 5 5 5 8 5 ( 4 ) 5 ( ) 5 6 log 5 log 5 4 5 5 6 5 log 5 4 5 5 6 5 log5 ø. log 5 6 5. 7 9 5 5 6. 4e 0. 7 5 log 7 7 9 5 log 7 5 4e 0. 5 0 9 5 log 7 5 e 0. 5 5 5 log5 9log7 ø 0.5 ln e0. 5 ln 5 7. ln (7 4) 5 ln ( ) 7 4 5 5 5 5 5 Check: ln (7 4) 5 ln ( ) ln (7 p 4) 0 ln ( p ) ln 7 5 ln 7 0. ø.6094 ø 5.6 Algebra Worked-Out Solution Ke 0
8. log ( 6) 5 5 log ( 6) 5 5 6 5 5 8 Check: log ( 6) 5 log (8 6) 5 log Because 5 5, log 5 5. 9. log 5 log ( ) 5 log [5( )] 5 0 log[5( )] 5 0 5( ) 5 00 5 5 00 5 0 0 5 0 ( 5)( 4) 5 0 5 5 or 5 4 Check: log 5 log ( ) 5 log 5 p 5 log (5 ) 0 So, 5 is a solution. log 5 0 log 00 0 5 log 5 log ( ) 5 log 5 p (4) log ((4) ) 0 log (0) log (5) 0 Because log (0) and log (5) are not defined, 4 is not a solution. 0. ( ) 5 [( )] 5 4 [( )] 5 4 ( ) 5 64 5 64 64 5 0 ( 6)( 4) 5 0 5 6 or 5 4 Check: ( ) 5 (6 ) (6) 0 (4) (6) 0 Because (4) and (6) are not defined, 6 is not a solution. ( ) 5 (4 ) 4 0 6 4 0 64 0 Because 4 5 64, 64 5. So, 4 is a solution.. M 5 5 log D When M 5 7 7 5 5 log D 5 5 5 log D 5 log D 0 5 0 log D 0 5 D The diameter is 0 millimeters. Eercises for the lesson Solve Eponential and Logarthmic Equations Skill Practice. The equation 5 5 8 is an eample of an eponential equation.. Sample answer: Logarithmic equations have etraneous solutions when ou have to factor to solve them.. 5 4 5 5 6 4. 7 4 5 49 5 4 5 (5 ) 6 7 4 5 (7 ) 5 4 5 5 7 4 5 7 4 4 5 4 5 4 8 5 5 5. 8 5 6. 7 4 5 9 8 ( ) 5 ( 5 ) ( ) 4 5 ( ) 8 5 5 0 6 6 5 5 5 0 5 6 6 7 5 6 5 9 7 5 5 9 6 7. 4 5 5 64 8. 7 5 8 ( ) 5 5 ( 6 ) 7 5 ( 4 ) 4 0 5 8 7 5 48 4 0 5 8 7 5 48 0 5 4 5 5 55 5 7 5 5 9. 6 5 5 6 0. 0 0 5 00 6 (6 ) 5 5 (6 ) 0 0 5 (0 ) 6 6 0 4 5 6 0 0 5 0 0 4 5 0 5 9 5 5 5 5 5 5 5 4 5. 5 0 8 5 5 4. 8 5 0 (5 ) 0 8 5 (5 ) 4 log 8 8 5 log 8 0 5 0 6 5 5 6 5 log 8 0 log 0 0 6 5 6 5 log 8 4 5 8 ø.44 5 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 04 Algebra Worked-Out Solution Ke
Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved.. e 5 5 4. 7 5 8 ln e 5 ln 5 log 7 7 5 log 7 8 5 ln 5 5 log 7 8 log 8 ø.6 5 log 7 ø.6 ø 0.50 5. 5 5 6. 7 6 5 log 5 5 log log 7 7 6 5 log 7 5 5 log 6 5 log 7 log 5 5 log ø 0.9 7. 4e 5 7 8. 0 4 5 9 e 5 7 4 0 5 5 log 5 6 log 7 ø 0. ln e 5 ln 7 4 log 0 5 log 5 5 ln 7 4 5 7 ln 4 5 log 5 5 log 5 ø 0.7 ø 0. 9. e 6 5 5 0. 0.5 0.5 5 4 e 5 0.5 5 4.5 e 5 log 0.5 0.5 5 log 0.5 4.5 ln e 5 ln 5 ln 5 ln ø 0.65. (6)4 5 6. 0. 5 5 7 (6)4 5 5 0. 5 5 log 0.5 4.5.5 5 log 0.5 ø.09 6 4 5 5 log 0. 5 log log 6 6 4 5 log 6 5 0. 5 log log 4 5 log 6 5 5 0. log log 5 5 4 log 6 ø 0.8 ø 5.8. 4 e 7 5 4 4 e 5 e 5 ln e 5 ln 5 ln 5 ln ø 0.0 4. log 5 (5 9) 5 log 5 6 5 9 5 6 9 5 Check: log 5 (5 9) 5 log 5 6 log 5 (5 p (9 9)) 0 log 5 6 p 9 log 5 54 5 log 5 54 5. ln (4 7) 5 ln ( ) 4 7 5 5 8 5 6 Check: ln (4 7) 5 ln ( ) ln (4 p 6 7) 0 ln (6 ) ln 7 5 ln 7 6. ln ( 9) 5 ln (7 8) 9 5 7 8 7 5 6 9 5 Check: ln ( 9) 5 ln (7 8) ln 9 9 0 ln 7 p 9 8 ln 47 5 ln 47 7. log 5 ( 7) 5 log 5 ( 9) 7 5 9 5 Check: log 5 ( 7) 5 log 5 ( 9) log 5 ( p 7) 0 log 5 ( p 9) log 5 () 0 log 5 () Because log 5 () is not defined, is not a solution. So, there is no solution. Algebra Worked-Out Solution Ke 05
8. log ( ) 5 log ( ) 5 9 5 4 5 8 Check: log ( ) 5 log ( ) log p 8 0 log p 8 The solution is 8. log 5 log 9. log (8 7) 5 log ( 8) 8 7 5 8 5 5 5 5 Check: log (8 7) 5 log ( 8) log 8 p 5 7 0 log p 5 8 The solution is 5. log 5 5 log 5 0. log 6 ( 0) 5 log 6 (4 5) 0 5 4 5 8 5 4 5 Check: log 6 ( 0) 5 log 6 (4 5) log 6 ( p 0) 0 log 6 (4 5 p ) log 6 () 0 log 6 () Because log 6 () is not defined, is not a solution. So, there is no solution.. log 8 (5 ) 5 log 8 (6 ) 5 5 6 6 5 8 5 Check: log 8 (5 ) 5 log 8 (6 ). 5 log 8 5 p 0 log8 6 p 4 5 4 5 4 Check: 5 4 log 8 5 log 8 Because 4 5 4, log 4 4 5. 5 ln 5 5 ln 5 7 e ln 5 e 7 5 e 7 ø 096.6 Check: 5 ln 5 5 ln e 7 5 5 p 7 5 5 4. log 5 5 log 5 5 6 5 log 5 5 5 6 5 5,65 5 5,65 Check: log 5 5 log 5 5,65 0 log 5 5,65 0 log 5 (5,65) / 0 5. 5. 5 6 log 5 5 5 ø.0769 4 ø 4.0769 ø 7.0 ø 5.60 Check: 5. 5 6 5. p 5.60 0 6 5. 7. 0 6 log 7. 5. 0 6 5.(.0769) 0 6 6. log ( 4) 5 6 log ( 4) 5 6 4 5 64 5 68 6.0 5 6 Check: log ( 4) 5 log (68 4) 5 log 64 Because 6 5 64, log 64 5 6. 7. log log ( ) 5 log [( )] 5 log [( )] 5 ( ) 5 8 8 5 0 ( 4)( ) 5 0 5 4 or 5 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 06 Algebra Worked-Out Solution Ke
Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. Check: log log ( ) 5 log 4 log (4 ) 0 log 4 log 0 log 8 0 log 0 5 log log ( ) 5 log () log ( ) 0 log () log (4) 0 Because log () and log (4) are not defined, is not a solution. 8. () ( 0) 5 [( 0)] 5 4 log4 [( 0)] 5 4 ( 0) 5 6 0 5 6 0 5 0 6 0 5 ( 8)( ) 5 8 or 5 Check: () ( 0) 5 ((8)) (8 0) 0 8 0 6 0 4 0 () ( 0) 5 (()) ( 0) 0 8 0 The solutions are 8 and. 9. ln ( ) ln 5 ln [( )] 5 e ln [( )] 5 e ( ) 5 e 5 e e 5 0 6 0 4 0 5 5 5 b 6 Ï b 4ac 5 6 Ï 9 4(e) 5 6 Ï 9 4e a 5 Ï 9 4e or 5 Ï 9 4e Check: ln ( ) ln 5 ln Ï 9 4e ln Ï 9 4e 0 ln Ï 9 4e ln Ï 9 4e 0 ln (0.7) ln (.7) 0 Because ln (0.7) and ln (.7) are not defined, Ï 9 4e is not a solution. ln ( ) ln 5 ln Ï 9 4e ln Ï 9 4e 0 ln Ï 9 4e ln Ï 9 4e So, the solution is Ï 9 4e. 40. 4 ln () 5 4 ln () 5 8 ln() 5 9 e ln () 5 e 9/ ø e 9/ 5 e 9/ ø 90.0 Check: 4 ln() 0 4 ln((e 9/ )) 0 4 ln(e 9/ ) 0 4 9 0 5 4. log 5 ( 4) log 5 ( ) 5 log 5 [( 4)( )] 5 5 log 5 [( 4)( )] 5 5 ( 4)( ) 5 5 5 4 5 5 5 5 0 5 56 Ï 5 4() 5 56 Ï 09 5 5 Ï 09 or 5 5 Ï 09 0 9 9 4e ln 4 0 ln (e) 0 5 Check: log 5 ( 4) log 5 ( ) 5 log 5 5 Ï 09 4 log 5 5 Ï 09 0 log 5 Ï 09 log 5 Ï 09 0 log 5 (7.44) log 5 (6.7) 0 Algebra Worked-Out Solution Ke 07
Because log 5 (7.44) and log 5 (6.7) are not defined, 5 Ï 09 is not a solution. log 5 ( 4) log 5 ( ) 5 log 5 5 Ï 09 4 log 5 5 Ï 09 0 log 5 Ï 09 log 5 Ï 09 So, 5 Ï 09 is a solution. 4. log 6 log 6 ( ) 5 log 6 [( )] 5 6 log 6 [( )] 5 6 ( ) 5 6 5 6 6 5 0 7 5 0 ( 9)( 8) 5 0 5 9 or 5 8 Check: log 6 log 6 ( ) 5 log 6 ( p 9) log 6 (9 ) 0 log 6 7 log 6 8 0 log 6 6 0 log 6 6 0 5 log 6 log 6 ( ) 5 log 6 ((8)) log 6 (8 ) 0 log 6 (4) log 6 (9) 0 0 log 5 09 9 4 0 log 5 5 0 5 Because log 6 (4) and log 6 (9) are not defined, 8 is not a solution. The solution is 9. 4. log ( 9) log ( ) 5 log [( 9)( )] 5 log [( 9)( )] 5 ( 9)( ) 5 9 7 5 9 8 5 0 5 6 Ï 44 4(8) 5 6 Ï 7 5 66 Ï 5 6 Ï or 5 6 Ï Check: log ( 9) log ( ) 5 log (6 Ï 9) log (6 Ï ) 0 log ( Ï ) log ( Ï ) 0 log (7.5) log (.4) 0 Because log (7.5) and log (.4) are not defined, 6 Ï is not a solution. log ( 9) log ( ) 5 log (6 Ï 9) log (6 Ï ) 0 log [( Ï ) ] log ( Ï ) 0 So, the solution is 6 Ï. 44. A; log 8 ( 7) 8 5 0 log 8 ( 7) 5 log 8 ( 7) 5 8 log 8 ( 7) 5 8 / 7 5 4 log [( Ï ) ] 0 5 log 0 5 or.5 5 45. The error was made in simplifing log 6 as ; log 6 ø.6, not. 5 6 log 5 log 6 5 log 6 5 log 6 ø.6 ø 0.6.58 ø 46. When eponentiating both sides, base should have been used, not base e. log 0 5 5 log 0 5 5 0 5 4 5 4. 47. Sample answer: 5 6 and log 8 (5 ) 5 48. 4 5 6 5 log 4 5 log 6 5 4 5 ( 5)(log 6) 4 5 log 6 5 log 6 4 5 log 6 5 log 6 4 5 log 6 5 ( log 6 ) 4 5 log 6 log 6 5 4 5 p log 6 log p log 6 5 log 4 5(.6) (.6) ø 5.7 ø Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 08 Algebra Worked-Out Solution Ke
Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 49. 0 8 5 5 log0 8 5 log 5 8 5 (5 ) log 8 5 5 log log ( log ) 5 8 5 log 8 log 5 log ø.88 50. log ( ) 5 log 8 log ( ) 5 log () log 8 log ( ) 5 log () log ( ) 5 log () log ( ) 5 log () ( ) 5 5 5 0 Graphing the function 5 shows that the -intercept is negative. Because the log of a negative number is not defined, there is no solution. 5. log 5 log 9 6 log 5 log 6 log 9 log 5 log 6 log 5 log 6 log 5 log 6 5 6 ( 6) 5 0 5 0 or 5 6 Check: log 5 log 9 6 log 0 0 log 9 6(0) log 0 0 log 9 0 Because log 0 and log 9 0 are not defined, 0 is not a solution. log 5 log 9 6 log 6 0 log 9 (6 p 6) log 6 0 log 9 6 log 6 log log 6 log 6 0 5 log 9 log log 6 log 5 log 6 log So, 6 is the solution. 5. p 5 0 ( 4)( 8) 5 0 5 4 or 5 8 5 log 4 or 5 log 8 5 log or 5 log 5 or 5 Check: p 5 0 () p 0 0 6 48 0 0 p 5 0 () p 0 0 64 96 0 0 0 5 0 So, the solutions are and. 5. 5 0 p 5 5 5 0 (5 5)(5 5) 5 0 0 5 0 5 5 5 or 5 5 5 5 log 5 (5) 5 log 5 5 Not defined 5 Check: 5 0 p 5 5 5 0 5 () 0 p 5 5 0 0 So, is the solution. 5 00 5 0 0 0 5 0 Problem Solving 54. T 5 00, T 0 5 00, T R 5 75, and r 5 0.054 T 5 (T 0 T R )e rt T R 00 5 (00 75)e 0.054t 75 5 5 5e 0.054t 0. 5 e 0.054t ln 0. 5 ln e 0.054t.6094 ø 0.054t 0 ø t It will take about 0 minutes to cool the beef stew. 55. T 5 7, T 0 5 75, t 5, and r 5.7 T 5 (T 0 T R )e rt T R 7 5 (75 T R )e.7() T R 7 5 75e.7 T R e.7 T R 7 75e.7 5 T R (e.7 ) 7 75e.7 e.7 5 T R 7 75(0.5) 0.5 ø T R 4 ø T R The outdoor temperature is about 48F. Algebra Worked-Out Solution Ke 09
56. A 5 P n r nt When P 5 00, r 5 0.06, and A 5 000 a. When n 5 : 000 5 00 0.06 ()t 0 5.06 t log0 5 log.06 t log.06 5 t 5 tlog.06 9.5 ø t It will take about 9.5 ears for the balance to reach $000. b. When n 5 4: 000 5 00 0.06 4 (4)t 0 5.05 4t log 0 5 log.05 4t 5 4t log.05 4 log.05 5 t 8.7 ø t It will take about 8.7 ears for the balance to reach $000. c. When n 5 65: 000 5 00 0.06 65 (65)t 0 5 65.06 65 65t log 0 5 log 65.06 65 65t 65log 65.06 5 t 65 5 65t log 65.06 65 8.4 ø t It will take about 8.4 ears for the balance to reach $000. 57. When R 5 5: R 5 00e 0.0004t 5 5 00e 0.0004t 0.05 5 e 0.0004t ln 0.05 5 ln e 0.0004t.9957 ø 0.0004t 6967 ø t It will take about 6967 ears for onl 5 grams of radium to be present. 58. C; A 5 (800) 5 400, r 5 0.05, and P 5 800 A 5 Pe rt 400 5 800e 0.05t 5 e 0.05t ln 5 ln e 0.05t.099 ø 0.05t 48.8 ø t 59. a. B changing the window settings and using the trace feature, ou can estimate the amount of energ released b each earthquake: Ocotillo Wells, CA: about 4,000 kilowatt hours Athens: about,500,000 kilowatt hours Fukuoka: about 7,000,000 kilowatt hours. b. Ocotillo Wells, CA: R 5 0.67 log (0.7E).46 4. 5 0.67 log (0.7E).46.64 5 0.67 log (0.7E).64 5 log (0.7E) 0.67 0 (.64/0.67) log (0.7E) 5 0 0 (.64/0.67) 5 0.7E 875.6 ø 0.7E,555.7 ø E The amount of energ released at Ocotillo Wells, CA, was about,556 kilowatt-hours. Athens: R 5 0.67 log (0.7E).46 5.9 5 0.67 log (0.7E).46 4.44 5 0.67 log (0.7E) 4.44 5 log (0.7E) 0.67 0 (4.44/0.67) log (0.7E) 5 0 0 (4.44/0.67) 5 0.7E 459.5 ø 0.7E,446,69.0 ø E The amount of energ released at Athens was about,446,69 kilowatt-hours. Fukuoka: R 5 0.67 log (0.7E).46 6.6 5 0.67 log (0.7E).46 5.4 5 0.67 log (0.7E) 5.4 5 log (0.7E) 0.67 0 (5.4/0.67) log (0.7E) 5 0 0 (5.4/0.67) 5 0.7E 46,950,669.66 ø 0.7E 6,89,70.8 ø E The amount of energ released at Fukuoka was about 6,89,70 kilowatt-hours. Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 0 Algebra Worked-Out Solution Ke
Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 60. a. I() 5 0.I 0 ; m 5 0.4 I() 5 I 0 e m 0.I 0 5 I 0 e 0.4 0. 5 e 0.4 ln 0. 5 ln e 0.4 ln 0. 5 0.4.0 ø 0.4.8 ø The thickness should be about.8 centimeters. b. I() 5 0.I 0 ; m 5. I() 5 I 0 e m 0.I 0 5 I 0 e. 0. 5 e. ln 0. 5 ln e. ln 0. 5..0 ø. 0.75 ø The thickness should be about 0.8 centimeters. c. I() 5 0.I 0 ; m 5 4 I() 5 I 0 e m 0.I 0 5 I 0 e 4 0. 5 e 4 ln 0. 5 ln e 4 ln 0. 5 4.0 ø 4 0.0 ø The thickness should be about 0.0 centimeters. d. Lead is a better material to use than aluminum or copper because it can be much thinner and still shield the same amount of X-ras. 56 6. When h 5 00: h(t) 5 e 0.65t 56 00 5 e 0.65t e 0.65t 5 56 00 e 0.65t 5 0.8 e 0.65t 5 0.8 ln e 0.65t ø ln 0.8 0.65t 5 ln 0.8 0.65t 5.84 t ø 5.9 It takes about 6 weeks for the seedling to reach a height of 00 centimeters. 6. 5 7 5 7 5 4 () 6 5 4 5 5 7 5 4 The solution is (4, 5). 7 5 5 5 5 6. 5 5 7 5 5 5 5 5 5 5 5 5 5 7 5 7 5 9 7 The solution is 7, 9 7. 7 5 5 7 64. 4 5 6 8 5 5 5 4(0) 5 6 5 6 The solution is (6, 0). 9 5 0 5 0 65. f() 5 5; or 0 positive real zeros f() 5 5; negative real zero Positive real zeros Negative real zeros Imaginar zeros 0 0 66. f() 5 4 6 7 8; or positive real zeros f() 5 4 6 7 8; negative real zero Positive real zeros Negative real zeros Imaginar zeros 0 67. f() 5 5 7 6 9; or 0 positve real zeros f() 5 5 7 6 9; or negative real zeros Positive real zeros Negative real zeros Imaginar zeros 0 0 0 4 Algebra Worked-Out Solution Ke
68. f() 5 7 0 6 5 4 7; or positive real zeros f() 5 7 0 6 5 4 7; or 0 negative real zeros Positive real zeros Negative real zeros Imaginar zeros 0 4 4 0 6 69. f() f() f() f(4) f(5) f(6) 9 8 7 6 5 6 9 0 0 0 0 f() 5 a b c (, 9): a b c 5 9 (, 8): 4a b c 5 8 (, 7): 9a b c 5 7 Using a calculator, the solution is a 5 5, b 5 4, and c 5 0. So, a function that fits the data is f() 5 5 4. 70. f() f() f() f(4) f(5) f(6) 0 6 80 50 0 4 44 70 8 4 0 6 6 6 6 f() 5 a b c d (, 0): a b c d 5 0 (, ): 8a 4b c d 5 (, ): 7a 9b c d 5 (4, 6): 64a 6b 4c d 5 6 Using a calculator, the solution is a 5, b 5, c 5, and d 5 0. So a function that fits the data is f() 5. Problem Solving Workshop for the lesson Solve Eponential and Logarthmic Equations. 8 e 5 4 X Y.4.598.5 -.964.6-4.099.7-8..8-4.05.9 -.76 -.7 X=.8 X=.799984 Y=-4. 7 0 5 5 9 X Y.5-4.6.6-8..7 -.95.8-8.849.9-5.589 4-4. -.94 X=.8 The solution is about.8.. e 5 8 5 5 X Y.8 5.78.9 7.487 0.89. 5.8..086. 6.5.4 57.598 X=. The solution is about.. 4..6() 4 5.6 5 6 X Y 0 7.. 6.6. 6.644. 6.08.4 5.8759.5 5.7778.6 5.746 X=. The solution is about 0.5. 5. log 5 5 X Y.4.5.9.6.585.7.8074.8.9.699.9 X=.8 The solution is 0.8. 6. log ( 7) 5 X Y -..05 -..08 - -.9.98677 -.8.97 -.7.95904 -.6.94448 X=- The solution is. 7. 4 ln 6 5 X Y 4..644 4..74 4..84 4.4.96 4.5.06 4.6.04 4.7.9 X=4.5 The solution is about 4.48. X=.79588 Y=-9 X=.09698 Y=5 X=.546788 Y=6 X=.8 Y= X=- Y= X=4.48689 Y= Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. The solution is about 0.8. Algebra Worked-Out Solution Ke
Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 8. log ( 9) 5 5 8 X Y 5.8 7.879 5.9 7.905 6 7.97 6. 7.9687 6. 8.000 6. 8.06 6.4 8.067 X=6. The solution is about 6.. 9. 5 888(.04) X Y.7 9874..8 99.9 995.9 999. 000. 0070. 009 X= X=6.9908 Y=8 X=.0850 Y=0000 The gross national product was $0 trillion in 00 ( ø ). 0. You could make the step between values even smaller to find the solution of 4 5 more precisel.. 5 5 log X Y 40.90 45.946 50.99 55 4.0 60 4.075 65 4.6 70 4.57 X=50 X=5.8864 Y=4 To reveal stars of magnitude 4, the diameter of the telescope s objective lense must be about 5 millimeters.. A 5 P n r nt, A 5 6000, P 5 5000, r 5 0.0, and n 5 4 5 5000 0.0 4 4 X Y 5.8 5946.4 5.9 5964. 6 598. 6. 6000 6. 607.9 6. 605.9 6.4 6054 X=6. X=6.0047 Y=6000 The balance will reach $6000 after about 6. ears. 0.6 7.88. 5.045 e X Y -.04 -.04 0.04.04.04.09 4.08 X=0 X=.05 Y=.04 Deep water in the South Atlantic Ocean off Antarctica has a temperature of about 0.08C. Etension for the lesson Solve Eponential and Logarithmic Equations. 0 X Y.68 8.997.69 9.07.7 9.49.7 9.64.7 9.85.7 0.07.74 0.9 X=.7 The solution is.77.. 8 > 9 X Y.75 9.84.76 9.44.77 9.07.78 9.0704.79 9.07.8 8.997.8 8.9607 X=.79 The solution is <.799.. 44(0.5) 50 X Y.48 5.595.49 5.057.5 50.5.5 49.996.5 49.474.5 48.957.54 48.446 X=.5 The solution is.5. 4. 6(0.96) < 7 X Y 0.7-7.0 0.74-7.0 0.75-7.0 0.76-7 0.77-6.98 0.78-6.97 0.79-6.96 X=0.76 The solution is < 0.76 5. 95(.6) 60 X Y 6 59.8 6.0 60. 6.0 608.9 6.0 66.5 6.04 64. 6.05 6.7 6.06 69.4 X=6.0 The solution is 6.0. X=.768 Y=0 X=.799049 Y=9 X=.509979 Y=50 X=0.75595 Y=-7 X=6.04649 Y=60 Algebra Worked-Out Solution Ke