Page of 4 Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise A, Quesion Quesion: Skech he graphs of (a) y = e x + (b) y = 4e x (c) y = e x 3 (d) y = 4 e x (e) y = 6 + 0e x (f) y = 00e x + 0 Soluion: (a) y = e x + This is he normal y = e x moved up (ranslaed) uni. (b) y = 4e x
Page of 4 x = 0 y = 4 As x, y As x, y 0 This is an exponenial decay ype graph. (c) y = e x 3 x = 0 y = 3 = As x, y As x, y 0 3= 3 (d) y = 4 e x
Page 3 of 4 x = 0 y = 4 = 3 As x, y 4, i.e. y As x, y 4 0=4 (e) y = 6 + 0 x x = 0 y = 6 + 0 = 6 As x, y As x, y 6+0 0 = 6 (f) y = 00e x + 0
Page 4 of 4 x = 0 y = 00 + 0 = 0 As x, y 00 0 + 0 = 0 As x, y Pearson Educaion Ld 008
Page of Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise A, Quesion Quesion: The value of a car varies according o he formula V = 0 000e where V is he value in 's and is is age in years from new. (a) Sae is value when new. (b) Find is value (o he neares ) afer 4 years. (c) Skech he graph of V agains. Soluion: V = 0 000e (a) The new value is when = 0. V = 0 000 e = 0 000 = 0 000 New value = 0 000 (b) Value afer 4 years is given when = 4. V = 0 000 e = 0 000 e 3 = 4 330.63 Value afer 4 years is 4 33 (o neares ) (c) 0 4 Pearson Educaion Ld 008
Page of Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise A, Quesion 3 Quesion: The populaion of a counry is increasing according o he formula P = 0 + 0 e 50 where P is he populaion in housands and is he ime in years afer he year 000. (a) Sae he populaion in he year 000. (b) Use he model o predic he populaion in he year 00. (c) Skech he graph of P agains for he years 000 o 00. Soluion: P = 0 + 0e 50 (a) The year 000 corresponds o = 0. Subsiue = 0 ino P = 0 + 0 e P = 0 + 0 e 0 = 0 + 0 = 30 Populaion = 30 housand (b) The year 00 corresponds o = 0. Subsiue = 0 ino P = 0 + 0 e 0 P = 0 + 0 e 50 = 0 + 4.98 = 34.98 housand Populaion in 00 will be 34 98 50 50
Page of (c) Year 00 is = 00 P = 0 + 0 e 00 50 = 0 + 0 e = 93.89 housand Pearson Educaion Ld 008
Page of Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise A, Quesion 4 Quesion: The number of people infeced wih a disease varies according o he formula N = 300 00e 0.5 where N is he number of people infeced wih he disease and is he ime in years afer deecion. (a) How many people were firs diagnosed wih he disease? (b) Wha is he long erm predicion of how his disease will spread? (c) Graph N agains. Soluion: N = 300 00e 0.5 (a) The number firs diagnosed means when = 0. Subsiue = 0 in N = 300 00e 0.5 N = 300 00 e 0.5 0 = 300 00 = 00 (b) The long erm predicion suggess. As, e 0.5 0 So N 300 00 0 = 300 (c) Pearson Educaion Ld 008
Page of Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise A, Quesion 5 Quesion: The value of an invesmen varies according o he formula V = Ae where V is he value of he invesmen in 's, A is a consan o be found and is he ime in years afer he invesmen was made. (a) If he invesmen was worh 8000 afer 3 years find A o he neares. (b) Find he value of he invesmen afer 0 years. (c) By wha facor will he original invesmen have increased by afer 0 years? Soluion: V = Ae (a) We are given ha V = 8000 when = 3. Subsiuing gives 8000 = Ae 3 8000 = Ae 4 ( e 4 ) A = 8000 e 4 A = 8000 e 4 A = 630.4 A = 630 (o he neares ) (b) Hence V = ( 8000 e 4 ) e (use real value) Afer 0 years V = 8000 e 4 e (use laws of indices) = 8000 e = 8000e 7 0 3 0
Page of = 4 336.0 Invesmen is worh 4 336 (o neares ) afer 0 years. (c) Afer 0 years V = Ae 0 This is e imes he original amoun A = 5.9 imes. Pearson Educaion Ld 008 0
Page of Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise B, Quesion Quesion: Solve he following equaions giving exac soluions: (a) e x = 5 (b) lnx = 4 (c) e x = 7 x (d) ln = 4 (e) e x = 8 (f) ln ( x + ) = 5 (g) e x = 0 (h) ln ( x ) = 4 (i) e 4x 3 = 8 Soluion: (a) e x = 5 x = ln5 (b) lnx = 4 x = e  4 (c) e x = 7 x = ln7 x = ln 7 x (d) ln = 4 = e  4 x = e  4 (e) e x = 8 x = ln8 x = ln 8 + (f) ln ( x + ) = 5 x + = e 5 x
Page of x = e 5 x = e 5 (g) e x = 0 x = ln0 x = ln0 x = ln0 x = ln ( 0. ) (h) ln ( x ) = 4 x = e  4 = e  4 + x x = e  4 (i) e  4x 3 = 8 e  4x = e  4x = 4x = ln x = 4 ln Pearson Educaion Ld 008
Page of Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise B, Quesion Quesion: Solve he following giving your soluion in erms of ln : (a) e 3x = 8 (b) e x = 4 (c) e x + = 0.5 Soluion: (a) e 3x = 8 3x = ln 8 3x = ln 3 3x = 3 ln x = ln (b) e x = 4 x = ln 4 x = ln x = ln x = ln x = ln (c) e x + = 0.5 x + = ln ( 0.5 ) x + = ln x + = ln x = ln x = ln Pearson Educaion Ld 008
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Page of 4 Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise B, Quesion 3 Quesion: Skech he following graphs saing any asympoes and inersecions wih axes: (a) y = ln ( x + ) (b) y = lnx (c) y = ln ( x ) (d) y = ( lnx ) (e) y = ln ( 4 x ) (f) y = 3 + ln ( x + ) Soluion: (a) y = ln ( x + ) When x = 0, y = ln ( ) = 0 When x, y y wouldn' exis for values of x < When x, y (slowly) (b) y = lnx
Page of 4 When x =, y = ln ( ) = 0 When x 0, y y wouldn' exis for values of x < 0 When x, y (slowly) (c) y = ln ( x ) When x =, y = ln ( ) = 0 When x 0, y y wouldn' exis for values of x < 0 When x, y (slowly) (d) y = ( lnx )
Page 3 of 4 When x =, y = ( ln ) = 0 For 0 < x <, lnx is negaive, bu ( lnx ) is posiive. When x 0, y When x, y (e) y = ln ( 4 x ) When x = 3, y = ln = 0 When x 4, y y doesn' exis for values of x > 4 When x, y (slowly) When x = 0, y = ln4 (f) y = 3 + ln ( x + )
Page 4 of 4 When x =, y = 3 + ln = 3 + 0 = 3 When x, y y doesn' exis for values of x < When x, y slowly When x = 0, y = 3 + ln ( 0 + ) = 3 + ln When y = 0, 0 = 3 + ln ( x + ) 3 = ln ( x + ) e 3 = x + x = e 3 Pearson Educaion Ld 008
Page of Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise B, Quesion 4 Quesion: The price of a new car varies according o he formula P = 5 000e 0 where P is he price in 's and is he age in years from new. (a) Sae is new value. (b) Calculae is value afer 5 years (o he neares ). (c) Find is age when is price falls below 5 000. (d) Skech he graph showing how he price varies over ime. Is his a good model? Soluion: P = 5 000e (a) New value is when = 0 P = 5 000 e 0 = 5 000 The new value is 5 000 (b) Value afer 5 years is when = 5 P = 5 000 e 0 = 5 000e 0.5 = 9097.96 Value afer 5 years is 9 098 (o neares ) (c) Find when price is 5 000 Subsiue P = 5 000: 5 000 = 5 000e 0 ( 5 000 ) 5 000 5 000 3 ln = e = e 3 = 0ln 0 0 = 0 3 0 5
Page of = 0ln 3 = 0ln3 = 0.99 years The price falls below 5 000 afer years. (d) A fair model! Perhaps he price should be lower afer years. Pearson Educaion Ld 008
Page of Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise B, Quesion 5 Quesion: The graph below is of he funcion f ( x ) = ln ( + 3x ) { x R, x > a {. (a) Sae he value of a. (b) Find he value of s for which f ( s ) = 0. (c) Find he funcion f ( x ) saing is domain. (d) Skech he graphs f(x) and f ( x ) on he same axes saing he relaionship beween hem. Soluion: (a) x = a is he asympoe o he curve. I will be where + 3x = 0 3x = x = 3 Hence a = (b) If f ( s ) = 0 hen ln ( + 3s ) = 0 3
Page of + 3s = e 0 3s = e 0 s = e 0 3 (c) To find f ( x ), change he subjec of he formula. y = ln ( + 3x ) e y = + 3x e y = 3x x = e y 3 Therefore f ( x ) = e x 3 domain of f ( x ) = range of f(x), so x R (d) f ( x ) is a reflecion of f(x) in he line y = x. Pearson Educaion Ld 008
Page of Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise B, Quesion 6 Quesion: The graph below is of he funcion g ( x ) = e x + 4 { x R{. (a) Find he range of he funcion. (b) Find he value of p o significan figures. (c) Find g ( x ) saing is domain. (d) Skech g(x) and g ( x ) on he same se of axes saing he relaionship beween hem. Soluion: (a) g ( x ) = e x + 4 As x, g ( x ) 0 + 4 = 4 Therefore he range of g(x) is g ( x ) > 4 (b) If ( p, 0) lies on g ( x ) = e x + 4 e p + 4 = 0 e p = 6 e p = 3 p = ln3
Page of p = ln3 p = 0.55 ( s.f.) (c) g ( x ) is found by changing he subjec of he formula. Le y = e x + 4 y 4 = e x y 4 ln x = = e x ln = x Hence g ( x ) = ln Is domain is he same as he range of g(x). g ( x ) has a domain of x > 4 (d) y 4 y 4 x 4 g ( x ) is a reflecion of g(x) in he line y = x. Pearson Educaion Ld 008
Page of Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise B, Quesion 7 Quesion: The number of baceria in a culure grows according o he following equaion: N = 00 + 50 e 30 where N is he number of baceria presen and is he ime in days from he sar of he experimen. (a) Sae he number of baceria presen a he sar of he experimen. (b) Sae he number afer 0 days. (c) Sae he day on which he number firs reaches 000 000. (d) Skech he graph showing how N varies wih. Soluion: N = 00 + 50 e 30 (a) A he sar = 0 N = 00 + 50e 30 = 00 + 50 = 50 There are 50 baceria presen a he sar. (b) Afer 0 days = 0 N = 00 + 50e 30 = 00 + 50e 3 = 70 There are 70 baceria presen afer 0 days. (c) When N = 000 000 000 000 = 00 + 50e 30 ( 00 ) 999 900 = 50e 30 ( 50 ) 9 998 = e ln ( 9 998 ) = 30 30 0 0 = 30ln ( 9 998 ) = 97.0 The number of baceria reaches 000 000 on he 98h day (o he neares day).
Page of (d) Pearson Educaion Ld 008
Page of 3 Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise B, Quesion 8 Quesion: The graph below shows he funcion h ( x ) = 40 0e 3x { x > 0, x R{. (a) Sae he range of he funcion. (b) Find he exac coordinaes of A in erms of ln. (c) Find h ( x ) saing is domain. Soluion: (a) h ( x ) = 40 0 e 3x The range is he se of values ha y can ake. h ( 0 ) = 40 0 e 0 = 40 0 = 30 Hence range is h ( x ) < 30
Page of 3 (b) A is where y = 0 Solve 40 0 e 3x = 0 40 = 0e 3x ( 0 ) 4 = e 3x ln4 = 3x x = ln4 x = ln x = ln A is ln, 0 3 (c) To find h ( x ) change he subjec of he formula. Le y = 40 0e 3x 0 e 3x = 40 y e 3x = 3 3 3 3x = ln 40 y 0 40 y 0 40 y x = 3 ln 0 The domain of he inverse funcion is he same as he range of he funcion. Hence h 40 x ( x ) = 3 ln { x R, x < 30 { 0
Page 3 of 3 Pearson Educaion Ld 008
Page of 4 Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise C, Quesion Quesion: Skech he following funcions saing any asympoes and inersecions wih axes: (a) y = e x + 3 (b) y = ln ( x ) (c) y = ln ( x + ) (d) y = 3e x + 4 (e) y = e x + (f) y = 4 lnx Soluion: (a) y = e x + 3 This is he graph of y = e x moved up 3 unis. x = 0, y = e 0 + 3 = + 3 = 4 (b) y = ln ( x )
Page of 4 x =, y = ln ( ) = ln ( ) = 0 y will no exis for values of x > 0 x, y (slowly) The graph will be a reflecion of y = ln ( x ) in he y axis. (c) y = ln ( x + ) x =, y = ln ( + ) = ln ( ) = 0 y will no exis for values of x < x, y x, y (slowly) x = 0, y = ln ( 0 + ) = ln (d) y = 3e x + 4
Page 3 of 4 x = 0, y = 3e 0 + 4 = 3 + 4 = 7 x, y 3 0+4=4 x, y (e) y = e x + x =, y = e + = e 0 = x, y 0 x, y x = 0, y = e (f) y = 4 lnx
Page 4 of 4 x =, y = 4 ln ( ) = 4 x 0, y 4 ( ), so y + y will no exis for values of x < 0 y = 0 4 lnx = 0 lnx = 4 x = e  4 Pearson Educaion Ld 008
Page of Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise C, Quesion Quesion: Solve he following equaions, giving exac soluions: (a) ln ( x 5 ) = 8 (b) e  4x = 5 (c) 4 e x = 0 (d) lnx + ln ( x 3 ) = 0 (e) e x + e x = (f) ln + lnx = 4 Soluion: (a) ln ( x 5 ) = 8 (inverse of ln) x 5 = e 8 ( + 5 ) x = e 8 + 5 ( ) x = e 8 + 5 (b) e  4x = 5 (inverse of e) 4x = ln5 ( 4 ) x = (c) 4 e x = 0 ( + e x ) 4 = 0 + e x ( 0 ) 4 = e x (inverse of e) ln ( 4 ) = x ( ) ln5 4 ln ( 4 ) = x x = ln ( 4 )
Page of (d) ln ( x ) + ln ( x 3 ) = 0 ln [ x ( x 3 ) ] = 0 x ( x 3 ) = e 0 x ( x 3 ) = x 3x = 0 x = x = x = (x canno be negaive because of iniial equaion) (e) e x + e x = e x + = ( e x ) ( e x ) + = e x ( e x ) e x + = 0 ( e x ) = 0 e x = x = ln = 0 (f) ln + lnx = 4 lnx = 4 x = e  4 x = 3 ± \ 9 + 4 3 ± \ 3 3 + \ 3 e x e  4 Pearson Educaion Ld 008
Page of Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise C, Quesion 3 Quesion: The funcion c ( x ) = 3 + ln ( 4 x ) is shown below. (a) Sae he exac coordinaes of poin A. (b) Calculae he exac coordinaes of poin B. (c) Find he inverse funcion c ( x ) saing is domain. (d) Skech c(x) and c ( x ) on he same se of axes saing he relaionship beween hem. Soluion: (a) A is where x = 0 Subsiue x = 0 ino y = 3 + ln ( 4 x ) o give y = 3 + ln4 A = ( 0, 3 + ln4 ) (b) B is where y = 0 Subsiue y = 0 ino y = 3 + ln ( 4 x ) o give 0 = 3 + ln ( 4 x ) 3 = ln ( 4 x ) e 3 = 4 x x = 4 e 3 B = ( 4 e 3, 0 )
Page of (c) To find c ( x ) change he subjec of he formula. y = 3 + ln ( 4 x ) y 3 = ln ( 4 x ) e y 3 = 4 x x = 4 e y 3 The domain of he inverse funcion is he range of he funcion. Looking a graph his is all he real numbers. So c ( x ) = 4 e x 3 { x R{ (d) Pearson Educaion Ld 008
Page of Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise C, Quesion 4 Quesion: The price of a compuer sysem can be modelled by he formula P = 00 + 850e where P is he price of he sysem in s and is he age of he compuer in years afer being purchased. (a) Calculae he new price of he sysem. (b) Calculae is price afer 3 years. (c) When will i be worh less han 00? (d) Find is price as. (e) Skech he graph showing P agains. Commen on he appropriaeness of his model. Soluion: P = 00 + 850e (a) New price is when = 0. Subsiue = 0 ino P = 00 + 850e o give P = 00 + 850e ( e 0 = ) = 00 + 850 = 950 The new price is 950 (b) Afer 3 years = 3. 0 Subsiue = 3 ino P = 00 + 850e o give 3 P = 00 + 850e = 89.66 Price afer 3 years is 90 (o neares ) (c) I is worh less han 00 when P < 00 Subsiue P = 00 ino P = 00 + 850 e o give
Page of 00 = 00 + 850e 00 = 850e 00 850 ln = e = ln = = 4.8 I is worh less han 00 afer 4.8 years. (d) As, e 0 Hence P 00 + 850 0 = 00 The compuer will be worh 00 evenually. (e) 00 850 00 850 Pearson Educaion Ld 008
Page of Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise C, Quesion 5 Quesion: The funcion f is defined by f : x ln ( 5x ) x R, x >. 5 (a) Find an expression for f ( x ). (b) Wrie down he domain of f ( x ). (c) Solve, giving your answer o 3 decimal places, ln ( 5x ) =. Soluion: (a) Le y = ln ( 5x ) e y = 5x e y + = 5x e y + 5 = x The range of y = ln ( 5x ) is y R So f ( x ) = e x + 5 { x R{ (b) Domain is x R (c) ln ( 5x ) = 5x = e 5x = e + x = e + 5 Pearson Educaion Ld 008 =.8778 x =.878 (o 3d.p.)
Page of Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise C, Quesion 6 Quesion: The funcions f and g are given by f : x 3x { x R{ g : x e x { x R{ (a) Find he value of fg(4), giving your answer o decimal places. (b) Express he inverse funcion f ( x ) in he form f : x. (c) Using he same axes, skech he graphs of he funcions f and gf. Wrie on your skech he value of each funcion a x = 0. (d) Find he values of x for which f ( x ) =. 5 f ( x ) Soluion: (a) fg ( 4 ) = f ( e ) = f ( e ) = 3e =.7 (d.p.) (b) If f : x 3x { x R{ hen f x + : x 3 x R by using flow diagram mehod: 4 3x (c) gf ( x ) = g ( 3x ) = e f ( x ) = 3x
Page of 0 A x = 0, gf ( x ) = e = e and f ( x ) = 3 0 = (d) f ( x ) = x + 3 = (cross muliply) ( x + ) ( 3x ) = 5 3 3x + x = 5 3x + x 6 = 0 ( 3x + 8 ) ( x ) = 0 x =, 5 3x 8 3 5 f ( x ) Pearson Educaion Ld 008
Page of Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise C, Quesion 7 Quesion: The poins P and Q lie on he curve wih equaion y = e x. The x-coordinaes of P and Q are ln 4 and ln 6 respecively. (a) Find an equaion for he line PQ. (b) Show ha his line passes hrough he origin O. (c) Calculae he lengh, o 3 significan figures, of he line segmen PQ. Soluion: (a) Q has y coordinae e ln6 = e ln 6 = 6 = 4 P has y coordinae e ln 4 = e ln4 = 4 = change in y Gradien of he line PQ = = = = change in x 4 ln 6 ln4 ln 4 ln 6 4 Using y = mx + c he equaion of he line PQ is y = ln 4 x + c (ln 4, ) lies on line so
Page of = + c c = 0 Equaion of PQ is y = x ln 4 (b) The line passes hrough he origin as c = 0. (c) Lengh from (ln 4, ) o (ln 6, 4) is Pearson Educaion Ld 008
Page of Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise C, Quesion 8 Quesion: The funcions f and g are defined over he se of real numbers by f : x 3x 5 g : x e x (a) Sae he range of g(x). (b) Skech he graphs of he inverse funcions f and g and wrie on your skeches he coordinaes of any poins a which a graph mees he coordinae axes. (c) Sae, giving a reason, he number of roos of he equaion f ( x ) = g ( x ). (d) Evaluae fg, giving your answer o decimal places. 3 Soluion: (a) g ( x ) > 0 (b) f ( x ) = g ( x ) = x + 5 3 lnx
Page of (c) f ( x ) = g ( x ) would have roo because here is poin of inersecion. (d) fg = f e = f e 3 3 = 3 e 3 5 = 0.84 3 Pearson Educaion Ld 008
Page of Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise C, Quesion 9 Quesion: The funcion f is defined by f : x e x + k, x R and k is a posiive consan. (a) Sae he range of f(x). (b) Find f(ln k), simplifying your answer. (c) Find f, he inverse funcion of f, in he form f : x, saing is domain. (d) On he same axes, skech he curves wih equaions y = f ( x ) and y = f ( x ), giving he coordinaes of all poins where he graphs cu he axes. Soluion: (a) f : x e x + k As x, f ( x ) 0 + k = k Range of f(x) is f ( x ) > k (b) f ( lnk ) = e lnk + k = k + k = k (c) Le y = e x + k
Page of y k = e x ln ( y k ) = x Hence f : x ln ( x k ), x > k (d) Pearson Educaion Ld 008
Page of Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise C, Quesion 0 Quesion: The funcion f is given by f : x ln ( 4 x ) { x R, x < { (a) Find an expression for f ( x ). (b) Skech he curve wih equaion y = f ( x ), showing he coordinaes of he poins where he curve mees he axes. (c) Sae he range of f ( x ). The funcion g is given by g : x e x { x R{ (d) Find he value of gf(0.5). Soluion: f ( x ) = ln ( 4 x ) { x R, x < { (a) Le y = ln ( 4 x ) and change he subjec of he formula. e y = 4 x x = 4 e y x = 4 e y f : x 4 e x { x R{
Page of (b) x = 0 f ( x ) = = y = 0 4 e x x, y x, y = 4 = 0 e x = 4 x = ln4 4 0 (c) Range of f ( x ) is f ( x ) < (d) gf ( 0.5 ) = g [ ln ( 4 0.5 ) ] = g ( ln3 ) = e ln 3 = 3 3 Pearson Educaion Ld 008
Page of Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise C, Quesion Quesion: The funcion f(x) is defined by f ( x ) = 3x 3 4x 5x + (a) Show ha ( x + ) is a facor of f(x). (b) Facorise f(x) compleely. (c) Solve, giving your answers o decimal places, he equaion 3 [ ln ( x ) ] 3 4 [ ln ( x ) ] 5 ln ( x ) + = 0 x > 0 Soluion: f ( x ) = 3x 3 4x 5x + (a) f ( ) = 3 ( ) 3 4 ( ) 5 ( ) + = 3 4 + 5 + = 0 As f ( ) = 0 hen ( x + ) is a facor. (b) f ( x ) = 3x 3 4x 5x + f ( x ) = ( x + ) ( 3x 7x + ) f ( x ) = ( x + ) ( 3x ) ( x ) (by inspecion) (c) If 3 [ ln ( x ) ] 3 4 [ ln ( x ) ] 5 [ ln ( x ) ] + = 0 [ ln ( x ) + ] [ 3ln ( x ) ] [ ln ( x ) ] = 0 ln ( x ) =,, x = e, e 3, e x = e, e 3, e x = 0.8, 0.70, 3.69 3 Pearson Educaion Ld 008