Chapter 11 Exponential and Logarithmic Function

Transcription

1 Chapter Eponential and Logarithmic Function - Page 69.. Real Eponents. a m a n a mn. (a m ) n a mn. a b m a b m m, when b 0 Graphing Calculator Eploration Page 700 Check for Understanding. The quantities are not the same. When the negative is enclosed inside of the parentheses and the base is raised to an even power, the answer is positive. When the negative is not enclosed inside of the parentheses and the base is raised to an even power, the answer is negative.. If the base were negative and the denominator were even, then we would be taking an even root of a negative number, which is undefined as a real number.. Laura is correct. The negative eponent of 0 represents a fraction with a numerator of and a denominator of a positive power of 0. The product of this fraction and a number between and 0 is between 0 and (6 ) ( ) 9 8. ( ) 8 9. (a ) a a 6 a a 8a or 8 a 0. m n m n (m n ) (m n ). 8n n 7 7 8n n n 7 8 n n ( ) 7 ( ) n n 7 n n 7 n n 7 n n n m n m n m 7 n 7 or m n mn n. ( 8 ) ( ) ( 8 ) (69 ) or 69 ( ). d a b c (a b c d ). 6 b c (6b c) 6b c 7. p q 6 r (p q 6 r ) (a ) (b ) (c ) (d ) a b c d (p ) (q 6 ) (r ) p q r pq r pr 8. ( ) ( ) 7 Chapter

2 9. A r r m A ( m) (.87) (0 7 ) m (.06 0 ) m.77 0 m Pages (6) ( 6) ( ) Eercises ( 6) ( 6 ) or. ( 96. ( ) 6. 8 (9 ) ( ) 7 )(9) 7 ( ) 7 6 ( ) () (9 ) 9 6 ( ). 6 (6 ) 6 (6 ) (9 ) ) (8 (8) 7 [() 7 ] 7 ( ) 6 6. (n ) (n ) 7n 6 7. ( ) 8 ( 8 ) ( ) ( ) ( ) ( ) (7p q 6 r ) 0. [( ] [( (9 ) 7 (p ) (q 6 ) r ) ] 8 8 ( ) (p ) (q 6 ) r p q 6 pq r r 8 8 or 6 8. (6 6 ) 6 ( 6 ). b b 6 n. n n n n n n n (b n b n ) (b n ) b n Chapter 76

3 . m 7n m 7 n m ( ) n 6 m n. f 6 6g h (f 6 6 g h ) (f 6 ) (6 ) (g ) (h ) 6. 6 f 6 g h f g h or 6 f g h m n a a n m a n m a nm a m n mn a 0. () ( ) ( ) ( 6 ) z 7( ) ( 7 ) 7 (z ) 7 7 z 7. a 0 b c (a 0 ) (b ) (c ) a b c rt 80 s (r 80 ) 8 (s 6) 8 (t 7 ) 8 60r 0 s t 8 9. m 6 n (m 6 ) n m n (7a) 8 b a b 7. p q r p 6 q 6 r 6 6 p q r 9. a 7 b a b 7 a b (n m 9 ) (n ) (m 9 ) n m 9 n m mn 7 8. ) 6. ( d e f (d ) (e ) (f ) d e f d 6. a b 7 c (a ) (b 7 ) (c) ab a bc (0) ( ) ( 6 ) (.) ( ) ( ) (8) a 7 a 7 7 a (a ).68 a d km so r.97 0 km V r (.97 0 km).6 0 km 69. ; 8, 6,, 0,,, 0, 9, 7 69a. If 0 then 0. If 0 then. Since 0, 0 and. So, 0. 69b. If 0 then. If then. So,. 69c. If then. So,. 77 Chapter

4 69d. If the eponent is less than 0, the power is greater than 0 and less than. If the eponent is greater than 0 and less than, the power is greater than and less than the base. If the eponent is greater than, the power is greater than the base. An number to the zero power is. Thus, if the eponent is less than zero, the power is less than. A power of a positive number is never negative, so the power is greater than 0. An number to the zero power is and to the first power is itself. Thus, if the eponent is greater than zero and less than, the power is between and the base. An number to the first power is itself. Thus, if the eponent is greater than, the power is greater than the base. 70. r (. 0 )A If r.7 0 then.7 0 (. 0 )A A.9 A.0 A Since.0, which is the mass number of carbon, the atom is carbon. 7. () 6 ( ) ( ) () ( ) ( ) (0) ( 6) ( 6)( ) a. 7b. A -mile per hour increase in the wind speed when the wind is light has more of an effect on perceived temperature than a -mile per hour increase in the wind speed when the wind is heav. 7a. r G r Wind Speed Wind Chill Met G M t.98 0 t da 86,000 seconds ( )(.98 0 )(86,00) r,0,7. m 7b.,0,7. m,0.7 km ,870 km m factors n factors mn factors 7a. a m a n a a... a a a... a a a... a a mn n factors m factors m factors m factors 7b. (a m ) n a a... a a a... a... a a... a m n factors m factors m factors m factors a a... a a mn 7c. (ab) m ab ab... ab a a... a b b... b a m b m m factors 7d. a b m a b a b... a b a b m m 7e. a a 7. m factors m n a a... a n factors a mn a a... a 76. ( 0) ()( 0) Verte is at (0, 0) and p. The parabola opens to the right so the focus is at (0, 0) or (, 0). Since the directri is units to the left of the verte, the equation of the directri is. 77. ( i) Convert to polar form r(cos v i sin v) r v Arctan 6 Arctan 6 So, ( i) cos 6 i sin 6. Use De Moivre s Theorem. cos 6 i sin Lemniscate 6 cos 6 sin 6 cos 0 i sin. 0.i Chapter 78

5 79. Use the equation t v u sin v qt h. v u 0 g h v t(0)(sin ) ()t 6t (0 sin )t Find t when 0 (i.e., the ball is on the ground). t t 0.0,. So, the ball hits the ground after about. s. u 80. TC ( ), (6 ()), ( 6), 0, u TC ( ) (6 ()) ( 6) 8. Sample answer: tan S cos S sin S c os S co s S 8. cot v 0 v 0 sin (0 in s) 6)() ( (6) sin S Period of cot v is so cot v 0 v n, where n is an integer. 8. r 6h 0 m 0 m r 6 h r m/h 8. 90, Degree of so there are comple roots. 0 ( )( ) ;, 0, The time it takes to paint a building is inversel proportional to the number of painters. 8 k 8 k 8 8 So t 6 t The correct choice is E. - Eponential Functions Page 70 Graphing Calculator Eploration. positive reals. (0, ). For a 0. and 0.7, as, 0 as. For a and, 0 as, as.. horizontal asmptote at 0, no vertical asmptotes. Yes; the range of an eponential function is all positive reals because the value of an positive real number raised to an power is positive. 6. An nonzero number raised to the zero power is. 7. The graph of b is decreasing when 0 b because multipling b number between 0 and results in a product less than the original number. The graph of b is increasing when b because multipling b a number greater than results in a product greater than the original number. 8. There is a horizontal asmptote at 0 because a power of a positive real number is never 0 or less. As a number between 0 and is raised to greater and greater powers, its value approaches 0. As a number greater than is raised to powers approaching negative infinit, its value approaches 0. [, ] scl:0. b [, ] scl:0. abs. min; (0.7,.88) Page 708 Check for Understanding. Power; in a power function the variable is the base, in an eponential function the variable is the eponent.. Both graphs are one-to-one, have the domain of all reals, a range of positive reals, a horizontal asmptote of 0, a -intercept of (0, ), and no vertical asmptote. The graph of b is decreasing when 0 b and increasing when b.. If the base is greater than, the equation represents eponential growth. If base is between 0 and, the equation represents eponential deca.. The graphs of and are the same ecept the graph of is shifted down three units from the graph of. 79 Chapter

6 Use (0, 0) as a test point. 0? 0 0 0? 0 0 The statement is true so shade the region containing (0, 0) Use N N 0 ( r) t where N 0 70, r 0., and t. N 70 ( 0.) 70 (0.7) a. 9,,9 8,86,0 8,67 8,67 7 0,09.7. (0, 0) 0? 0 0? 0 ; no 0,09.7 8,86, % 9b. Use N N 0 ( r) t. N 8,86,0( 0.00) 0 N 9,69, ( ) Pages Eercises 7. 0 (0, 0) 0? 0 0? ( ) Chapter 80

7 8. (0, 0) 0? 0 0? b. The graph of is a reflection of the graph of across the -ais B 0.0 (0, ) (, 00) [0, 0] scl: b [0, 0] scl: c. The graph of 7 is a reflection of the graph of 7 across the -ais (, ) C A (0, 7) (0, ) ( ), 7 [0, 0] scl: b [, 9] scl: d. The graph of is a reflection of the graph of across the -ais. (, ). [, ] scl: b [, ] scl: a. The graph of is a reflection of across the -ais. The graph of is a reflection of across the -ais. b. The graph of is shifted up two units, while the graph of is shifted down two units. c. The graph of 0 increases more quickl than the graph of. The graphs are not the same because 0. a. The graph of 6 is shifted up four units from the graph of 6. a. [0, 0] scl: b [, 9] scl: b. 9.(.06) thousand $70,00 a. (0.8) [0, 0] scl: b [, 9] scl: 8 Chapter

8 b c. (0.8) 0. or % d. No; the graph has an asmptote at 0, so the percent of impurities,, will never reach 0. 6a. N 876,6( 0.007) 978,6.6 or 978,6 6b. N,6,6( 0.00),668,760.8 or,668,760 6c.,07 9,0( r) 0,07 9,0,07 9,0 ( r) 0 0 r N 9,0( r) N 7,76.7 or 7,77 6d. 9,70 68,767( r) 0 9,70 68,767 9,70 68,767 ( r) 0 0 r N 68,767( r) N, or,076 ) 7a. 00( 00( ) 700 units 7b. s hr 6.6 ft/s mi 80 ft hr mi 600 s ( ) units 8a. P n,000, n 0 or 60, i 0.07 or 0.006,000 P ( 0.006) P 86.09; $86.0 8b. P n,000, n 0 or 0, i 0.07 or ,000 P P 96.9; $96. 8c. 0 ear: I 60(86.0),000 $ ear: I 0(96.),000 $08, Chapter 8 8d. Sample answer: A borrower might choose the 0-ear mortgage in order to have a lower monthl pament. A borrower might choose the 0-ear mortgage in order to have a lower interest epense. 9a. P 000, n, i 0.07 ( 0.07) F n 000,.98; $,.9 9b. P 000, n, i 0.0 ( 0.0) F n 000 6,9.9 or $6,9. $6,9.,.9 $76, The function a is undefined when a 0 and the eponent is a fraction with an even denominator. a. Compounded once: I 000[( 0.0) ] 0; $0 Compounded twice: I 000[ 0. 0 ] 0.6; $0.6 Compounded four times: I 000[ 0. 0 ] 0.9; $0.9 Compounded twelve times:.0 I ; $.6 Compounded 6 times:. 0 I ; $.6 b. Let represent the investment. Statement savings: I [( 0.0) ] 0.0 The return is.% 6 Mone Market Savings: I The return is.7% 0.7 Super Saver: I The return is.% Mone Market Savings 6 c. ( 0.0) 6 (.0) 6 6 6[(.0) 6 ] ;.88%. () () () Use r sin v. So, r sin v.., 9, ()() (9)() ; no because the inner product does not equal 0.

9 . cos 7 8 i sin 7 8 cos i sin cos 7 8 i sin 7 8 cos 8 i sin i 6. s 7t 6t s 6t 7t s (6)(.06) 6(t..06) (s 8) 6(t.) Verte: (., 8) Maimum height: 8 feet. 7a. (7, 6) (, 8). The parent graph is translated units left. The vertical asmptote is now. The horizontal asmptote, 0, is unchanged.. C AC C AB or about 0. The correct choice is E. 7b. ( (7)), (8 6) (., 7) 8. sin A cos A cos A sin A (sin A) cos A cos A sin A ( cos A) cos A cos A sin A ( cos A cos A) cos A cos A sin A cos A cos A cos A sin A cos A sin A cos A sin A rev rev V ,7.7 or about 9,000 cm/s tan tan 69.0 about feet 00 ft - The Number e Page 7 Check for Understanding. C. If k is positive, the equation models growth. If k is negative, the equation models deca.. Amount in an account with a beginning balance of $000 and interest compounded continuousl at an annual rate of.%.. reals, positive reals. Sample answer: Continuousl compounded interest is a continuous function, but interest compounded monthl is a discrete function. 6a. growth 6b.,0 6c.,0e 0.097(60) 6,9.0 or 6,9 7. A,000e 0.06(),86. or $,86. Pages 7 77 Eercises 8. p (00 8)e 06() 8.6 or % 9a. 8e 0.() or 78.7 F 9b. too cold; After minutes, his coffee will be about 90 F. 0a. [, ] scl: b [, 0] scl: 0b. smmetric about -ais Sample answer: Chapter

10 a. Annuall: I 00[( 0.08) ] 80 $80.00; 8% 08 Semi-annuall: I $8.60; 8.6% 08 Quarterl: I $8.; 8.%.08 Monthl: I $8.00; 8.%. 08 Dail: I $8.8; 8.8% Continuousl: I 000(e 0.08() ) 8.87 $8.98; 8.9% 6 Effective Interest Annual Compounded Interest Yield Annuall $ % Semi-annuall $ % Quarterl $8. 8.% Monthl $ % Dail $ % Continuousl $ % b. continuousl c. E n r n d. E e r a. ( e 0.08() ).097 people b. after about 6h [0, 00] scl:0 b [0, 0] scl:0 a. P e 6(0.) % b. 0.0; about 0.0 h 0.0 h 60 min h. b. decimal places; decimal places; 6 decimal places c. alwas greater a. das: P e 0.07() % 0 das: P e 0.07(6) % 90 das: P e 0.07(90) % 0.9%; 60.9% 98.% b. about 9 das c. Sample answer: The probabilit that a person who is going to respond has responded approaches 00% as t approaches infinit. New ads ma be introduced after a high percentage of those who will respond have responded. The graph appears to level off after about 0 das. So, new ads can be introduced after an ad has run about 0 das. 6a. all reals 6b. 0 f() 6c. c shifts the graph to the right or left 7. 0,000 P ( ) 8 0 0,000 P(9.0687) P,7.97 $, z z 9. 6 v sin cos 6( cos sin ) r () ) ( v Arctan 6. 6(cos. i sin.) [0, ] scl:0. b [0, ] scl:0. about. min a. For 0: ( 0) 0 ( 0) For 00: ( 00) 00 ( 00) For 000: ( 000) 000 ( 000) ;.7808;.7880 Chapter 8

11 . (0, 0) d 8 u u d, F 0, 0 cos 8 0 sin cos 8 0 sin d 8.89,.697 W u F u d W 0, , ft-lb..(0).() (, ) J(9, 6), K(6, 8), L(6, ), M(9, ); The dilated image has sides that are times the length of the original figure ( ) 6 (6) (0, 6) 7. {,, }; {, 7}; es 8. The correct choice is D. Page 77 Mid-Chapter Quiz ( ) ( ) ((7 ) ) w 6 z9 8z9 7w66 ( ) ( ) (z 9 ) ( ) (w 6 ) ( 6 ) z w. a 6 b (a b ) (a 6 ) (b ) or w z a b. (a b ) b a a b b a A A A (A ). A. 0 mm 7.,786,69,67,89 ( r) 8, 786, 69, 67, 89, 786, 69, 67, 89 8, 786, 69, 67, 89 8 ( r) 8 r [( r) 8 ] 8 r 0.0 Store the eact value in our calculator s memor. N,67,89 ( 0.0) Use the stored,6,6.979 value for r.,6,7 8. A ()(.) 00(.0) 9.78 $9.7 9a. 6.7e ,9 ft 0. ears: n 8. ears: n ears: n ; 8; e 00.() 8. 9b. 6.7e 0.607,60,7 ft 00 e0.() 00 e0.(60) 8 Chapter

12 - Pages 7 7 Logarithmic Functions Check for Understanding. and are similar in that the are both continuous, one-toone, increasing and inverses. and are not similar in that the are inverses. The domain of one is the range of another and the range of one is the domain of the other. has a -intercept and a horizontal asmptote whereas has a -intercept and a vertical asmptote.. Let b m, then b m. (b ) p m p b p m p b b p b m p p b m p p b m b m p. Log is an increasing function and is a decreasing function.. Sean is correct. The product propert states that b mn b m b n.. In half-life applications r. So, ( r) becomes or. Thus, the formula N N 0 ( r) t becomes N N 0 t n n 7 8 n 8 n. 6 ( ) t t 6( 0) 0 t 6( ) 0 t 6 h 0 Pages 7 7 Eercises e Chapter 86

13 w w w 6 6w 8 ( ) or or ( ) 8 6 ( )(8) 6 ( )(8) ( ) ( ) ( )( ) ( )( ) 6 9. ( 7 7 8) 7 6 ( 7 8) (8) 7 6 (8) ( ) 6 ( ) 6 ( ) w Chapter

14 ( ) 0 ( ) 6a r 0 7 r 0 6b. r 0 0 r 0.07 r r 6.99% 6c. r 8 8 r.0 r 0.00 r 0.0% 6a. n n n n 0 6b. p 6. Let a, so a. a b b a b b a b l 6a. a l l ogb a ogb ogb a P P.7.7 p 8 p 8 p ( ) 0.0h h less light; 8 9. Use N N 0 ( r) t ; r since the rate of growth is 00% ever t time periods. 6, ( ) t 6 t 6 t 6 t t 90 min 60. All powers of are, so the inverse of is not a function. 6. Let b m and b n. So, b m and b n. b m n b b m n b b m n b m n b m b n P 6b () P P.7 (7 0.0 ). psi P 6c (6.8) P.7( ) 6.8 psi t t t t t t about 9 das Chapter 88

15 ,000 P 0..0 P 89.6 $ ellipse ( ) ( ) 9 6 9( ) ( ) 6 ( ) ( ) r, v ( cos t, sin t) 7. AB ( ()) () 06) ( 6 BC ( ()) (0 ()) AC ( ()) (0 ) 7. cos i sin cos i sin cos i sin 7 0cos i sin cos 0i sin i cos i sin, i 7. ( j)( 7j) 6 j 8j 8j 6 7j 6 7j volts 7. C 0 A 0 7. cos (A B) cos A cos B sin A sin B cos A cos B 7 7 So, sin A So, sin B 7 cos (A B) S c P A sin (kt c) h 6 A 90 h 90 c N Q b 6 k 77 k sin t c 77 6 sin () c 77 c c sin sin sin () c. c So, sin k c (6.) (.8) (6.)(.8) cos 0. c cos 0. c c 9. (6.) (.8) (9.) (.8)(9.) cos A cos A cos A 0.78 A 8. or 8 0 B 80 ( ) 6 c 9., A 8 0, B M msmn mqnm alternate interior angles b c Eterior Angle Theorem The correct choice is E. B D Both vectors have the same direction. 0 south of east. Therefore, AB u and CD u are parallel. 89 Chapter

16 Common Logarithms Page 70 Check for Understanding. 0 means 0 0. So, 0. 0 means 0 0. So, Write the number in scientific notation. The eponent of the power of 0 is the characteristic.. anti ,000 (0,000 8) (0.00 ) (0.000 ) 0 (0.77) , ( ) og 8. 8 l.6 og. 8 l ( ). 9. og 9. ( ) l ( ) ( 6) og 76. l ( ) ( ) ( ) ( ).8 [0, 0] scl: b [0, 00] scl:0.80 8a. R b. 0 times; According to the definition of arithms, R in the equation R a T B is an eponent of the base of the arithm, 0. 0 is ten times greater than 0. Pages 70 7 Eercises ,000 (00,000 ) 00, ( ) (0. ) (0.60).8. 6 ( 9) ,000 (000 9) Chapter 90

17 (0.000 ) ( ) (0.60) (0.60) og 8. 8 l. 6 lo l og og g 6 og 6. 6 l l l ( ) ( ) () 0.6 (8 ) ( ) ( ) 9 ( ) og ( ) ( 7) 7 7 ( ) l og 6 lo g When, 0, which means l og 6 is undefined. When, l og 6 is negative, which is not greater than. So, must also be greater than. Therefore, ( ) ( ) ( ) ( ) 0. ( ) 0. ( 0.) ( 0. 0.) Change inequalit sign because ( 0. 0.) is negative [0, 0] scl: b [, ] scl: lo g ( ) anti 9 anti lo g [, 0] scl: b [, ] scl: 9 Chapter

18 . [0, ] scl: b [, 0] scl:. 0.0 [, ] scl: b [0, 0] scl: [, ] scl: b [, 0] scl: 7. [, ] scl: b [, 0] scl: 8a. h mi P 8b P P P 6.0 P; 6 psi 9a. M b P 8. P.66 P 0.09 P 60a. q $9,6 60b t t t 0.9 t t ears 6. Sample answer: is between and because 7 is between 00 and 000, and 00 and a. L ( (.0 0 )) 0 db 6b. 0.0 I0 I (.0 0 ) I 0 I 0 0 I; 0 0 W/m 6. Use N N 0 t. N 60 micrograms 6 0 gram N 0 milligram.0 0 gram 6. 0 (.0 0 ) t t t r 6. a a P a q a r a p a q a r a p r a q p r q increasing from to 67. (a b ) c (a ) (b ) (c ) a ab c 68. () (0) D() E(0) F 0 D F 0 () () D() E() F 0 D E F 0 () () D() E() F 0 D E F 0 Chapter 9

19 D 0E F 0 () D E F 0 D E 0 0 D E F 0 () D E F 0 D E 0 0 D E 0 0 (D E 0 0) 0D 60 0 D 6 (6) E 0 0 E 0 E (6) 0() F 0 F 0 F 6 0 ( 6 9) ( ) 9 ( ) ( ), (, ) 70. r 6 r u 7. AB (6 ), ( 6), u AB (6 ) ( ) a 6 v cos 6 ạ 6 b sin 6. 6 a.6 cos 6 b.6 sin Use A ap, where P 0(.).. A (.99)(.).678 or.68 cm cm b 7. f() f() () () Test f(). f() 0 ( ) is a factor. 0 0 ( )( ) 0 So, the factors are ( )( )( ). 7. Neither; the graph of the function is not smmetric with respect to either the origin or the -ais ,000,000 The correct choice is D. -6 Natural Logarithms Page 7 Check for Understanding. ln e is the same as e e. And e e. So, ln e.. The two arithms have different bases or. ln 7 e 7 or.8. ln 6.89 ln ln.86 ln 6 ln The two equations represent the same thing, A Pe rt is a special case of the equation N N 0 e kt and is used primaril for computations involving mone ln ln.09 n l ln e ln 8 ln e ln Chapter

20 . 0 e k e k ln k ln e ln k 0.86 k. e 00 e ln e ln.86.. e 0.0t ln. 0.0t ln e l n. 0.0 t 8.86 t..7 [0, 0] scl: b [, 0] scl: [, 0] scl: b [0, 0] scl:0 7a. p 760e 0.(.) 760e torrs 7b e 0.a ln ln e 0.a 0.a ln e a.96 a;. km Pages Eercise n 6 n l ln. 6 l ln n 8. 8 l ln ln ln ln ln ln ln 6 ln 7 ln ln 7 n 7 n 7 l ln 6 l ln ( ) ln 9 ln 7. ln 9 ln 9 ln 7. ln 9 ln 7. ln 9 ln 7. ln 9 ln ln ln ln ln ln ln e 0.t ln t ln e ln t 0.99 t. 6.e 0.6t e t ln t ln e ln (t ) ln e ln 6. ln 0.6t ln 6. ln 0.6 t 0.76 t. ( e ) e ln e ln e e 0.07 ln 0.07 ln e ln 0.07 ln e 000 e 0 ln e ln Chapter 9

21 . 76 ln ln 7 ln 6 ln 7 ln 6 ln ( ) ln ln ln ln ln ln ln ln (ln ln ) ln ln ln ln (7.6) [, ] scl: b [0, 700] scl: [70, 0] scl: b [, 0] scl: 0. t. [0, 0] scl:0 b [00, 000] scl:00.. [6, 6] scl: b [, ] scl:..7 [0, 0] scl: b [0, 7] scl:. 0.7 [, ] scl: b [, ] scl: e ln 0.6 0,000( 0 ) ln 0.6 t t s..8 9 t ln.8 9 t ln t.68 t h 6a. ln 80 7 k(0) c.68 c 6b. ln 0 7 k().68 ln k t 0,000(0 ) t 0,000( 0 ) 0.67 k 6c. ln 00 7 (0.6)t.68 ln t 8. t about 6. min 7. e e 0 (e )(e ) 0 e 0 e 0 e e ln e ln ln e ln or a. e 0.06t ln 0.06t ln e ln t ln.8 9 ln.00 t about ears 8b. See students work. ln e 9 Chapter

22 ln r ln r antiln r r; about 70% 60a. e k(6) ln 6 k ln e k k 60b..7 e (0.0007)(t) ln. 7. ln t ln e t t about 708 r 6. is a arithmic function of. The pattern in the table can be determined b which can be epressed as ( ) 8 0 ln 6 ()( ) , ,..7, N m 6 cm m 00 c cm c; N m cos 0. sin 0. 0 (0., 0) 67. a u,,,, 9, AB ) ( p 0.6 units 9 sin 9 cos tan tan ; 9 0.6; cos v 70. d 800 (0 ) 0 The correct answer is 0. -6B Graphing Calculator Eploration: Natural Logarithms and Area Pages ; It is the same value as found in Eercise epressed to 0 decimal places. a. The result is the opposite of the result in Eercise. b. Sample answer: a negative value a b c..869 d , ,.8696 e. The value for each area is the same as the value of each natural arithm ; ; ; These values are equal to the value of ln 0.6, ln 0., and ln If k, then the area of the region is equal to ln k. If 0 k, then the opposite of the area is equal to ln k. 7. The value of a should be equal to or ver close to, and the value of b should be ver close to e. This prediction is confirmed when ou displa the actual regression equation. 8. Sample answer: Define ln k for k 0 to be the area between the graph of, the -ais, and the vertical lines and k if k and to be the opposite of this area if 0 k. Define e to be the value of k for which the area of the region is equal to. Chapter 96

23 -7 Modeling Real-World Data with Eponential and Logarithmic Functions Page 7 Check for Understanding. Replace N b N 0 in the equation N N 0 e kt, where N 0 is the amount invested and k is the interest rate. Then solve for t.. The data should be modeled with an eponential function. The points in the scatter plot approach a horizontal asmptote. Eponential functions have horizontal asmptotes, but arithmic functions do not.. e (ln ) or e.86 ; ln ln (ln ) or ln n ln. t 0ḷ 07. t r 8.66 r 6a (0.970) 6b (0.970) 0.070(e ln ) 0.070e (ln 0.970) 0.070e c e 0.00 ln ln Pages t 0ḷ r ln 9. t ;.08 min n 7 Eercises ln 8. t r eponential; the graph has a horizontal asmptote. arithmic; the graph has a vertical asmptote. arithmic; the graph has a vertical asmptote. eponential; the graph has a horizontal asmptote a..788(.7687) b..788(.7687).788(e ln.7687 ).788e (ln.7687).788e 0.70 c. Use t ln k ; k t 0ḷ 70. hr a..009(0.980) b..009(0.980).009(e ln ).009e (ln 0.980).009e n c e ln ln min 6a. 7.9(.0) 6b. 7.9(.0) 7.9(e ln.0 ) 7.9e (ln.0) 7.9e c e r ln r ln r 0.00 r;.% ln 8a ln 8b. The ear 960 would correspond to 0 and ln 0 is undefined. 9. Take the square root of each side. c c c 0a ( r).00 r 0.00 r;.0% 0b (.00) 0c (.00) (e ln.00 ) e (ln.00) e d e r a. ln r r;.99% ln b. ln c. ln e d. e 0.06() persons per square mile a. The graph appears to have a horizontal asmptote at, so ou must subtract from each -value before a calculator can perform eponential regression. b..000(.70) a. ln is a linear function of ln. c a ln ln(c a ) ln ln c ln a ln ln c a ln 97 Chapter

24 b. The result of part a indicates that we should take the natural arithms of both the - and -values. c. ln 0.99 ln.90 d. ln 0.99 ln.90 e ln e0.99ln.90 e 0.99ln e.90 (e ln ) e k(8) ln 8k k e 0.008t ln 0.008t 0 t 0.0 min or about h (7) ( 6) a. (00 0( )) (60 0) (.) 6 0(.) verte at (., 6), maimum at. $.0 7b. At maimum, 6. $6 8. cos 6 i sin 6 i i 60 b 9. sin sin 8 60 ft ln ln sin 8 b sin b 09.6 about 09.6 ft Discriminant: (8) ()() 76 The discriminant is negative, so there are imaginar roots. 8 0 b i 0 or i. units left and 8 units down. (, ) (, ): f() () 8() 0 (, ): f() () 8() 0 6 (, 8): f() () 8() 0 8 (, ): f() () 8() 0 0 0;. Circle X contains the regions a, b, d, and e. Circle Z contains the regions d, e, f, and g. Si regions are contained in one or both of circles X and Z. The correct choice is C. Chapter Stud Guide and Assessment Page 79 Understanding the Vocabular. common arithm. eponential growth. arithmic function. scientific notation. mantissa 6. natural arithm 7. linearizing data 8. eponential function 9. nonlinear regression 0. eponential equation Pages 70 7 Skills and Concepts.. (6). (7) ( ) (, ) 8 ) 9. () ( 7. ( ) 8 (, ) 6 8. (w ) (w ) w w 6w (a) (a b) (a) (a b) (a)(a b) a b 0. ( ) 9 (, ). ( ) 6 (6) ( ) a 6 a 6a Chapter 98

25 ( ) A 00e 0.06(0) ; $ A 6000e 0.07(0),88.866; $, A,000e 0.09(0),67.860, $, (8) ,000 (00,000 ) 00, (0.000 ) (0 ) (0.00 ) ( ) ( 6) (0. ) ( ) Chapter

26 . 6 ( ) ( 6) 6 ( ) 6 6 Change the inequalit because 6 is negative ( 8) 0. ( ) ( 0. 7) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) , g 00 og lo 9 6. l ln ln ln ln ( ) ln 6 ln 0 l l n 0 n 6 n 0 n 6 l l ( ) ln ln ln ln ln ln ln ln (ln ln ) ln ln ln ln ln 9 ( ) ln ln 9 ln ln ln 9 ln ln ( ln 9 ln ) ln ln ln 9 ln e 69. e 00 ln e 00 ln ln [, ] scl: b [0, 60] scl: [, ] scl: b [, 0] scl: [, ] scl: b [, ] scl: og og 60. l 6. 8 l Chapter 00

27 7. t 0 l n t 0.0 ln ln k k ln ;.8% Page t 0.6 t 0.6 t Applications and Problem solving 0.6 t or 6 r. 76a b c ,000,000e 0.0t 00 7 ln 00 7 ln e 0.0t 0.0t t 0.6 db 9. db 99. db. t a. N 6 0e 0.0().89; words per minute 78b. N 6 0e 0.0() 6.0; 6 words per minute 78c e 0.0t ln ln 0.0 Page 7 e 0.0t 0.0t t.7 t;. weeks pen-ended Assessment. Sample answer: (n ) (m). Sample answer: ( ) 6 Chapter SAT & ACT Preparation Page 7 SAT and ACT Practice. To find the greatest possible value, the other values must be as small as possible. Since the are distinct positive integers, the must be,, and. The sum of all integers is () or. The sum of the smallest is or 6, so the fourth integer cannot be more than 6 or 8. The correct choice is B.. Since one root is,,, and 0. Similarl for the root that is,,, and 0. To find the quadratic equation, multipl these two factors and let the product equal zero. ( )( ) The correct choice is E.. The result of dividing T b 6 is less than the correct average. T 6 correct answer T 6 correct average The correct average is the total divided b the number of scores,. correct average T T 6 T The correct choice is E.. A tan A Find the area of ABC. A bh Simplif the ratio. area of ABC tan A The correct choice is E. 0 B (, ). 0 0 The correct choice is B. C(, 0) 0 Chapter

28 6. C D D D and, therefore, r. Now use this value for the radius to calculate half of the area. A r 9 8 The correct choice is A. 7. The average of 8 numbers is 0. 0 sum of eight numbers 60 The average of of the numbers is. sum of eight numbers 8 sum of five numbers sum of five numbers 70 The sum of the other three numbers must be or 90. Calculate the average of these three numbers. sum of three numbers average The correct choice is D. 8. The sum of the angles in a triangle is 80. Since B is a right angle, it is 90. So the sum of the other two angles is 90. Write and solve an equation using the epressions for the two angles The question asks for the measure of A. A (8) 6 The correct choice is C. 9. A is the arithmetic mean of three consecutive ( ) ( ) positive even integers, so A 6, where is a positive even integer. Then A is also a positive even integer. Since A is even, when A is divided b 6, the remainder must also be an even integer. The possible even remainders are 0,, and. The correct choice is C. 0. First notice that b must be a prime integer. Net notice that b is greater than 0. So b could be, since (). (b cannot be.) Check to be sure that fits the rest of the inequalit. () () So is one possible answer. You can check to see that 7 and are also valid answers. The correct answer is, 7, or. Chapter 0