CHAPTER 4 LOGARITHMIC FUNCTIONS - Browser Expressresources.npl.edu.hk/~hflau/MS-Sol-EE-C04.pdf · CHAPTER 4 LOGARITHMIC FUNCTIONS EXERCISE 4.1 Section 4.1 Properties and graphs of

CHAPTER 4 LOGARITHMIC FUNCTIONS

EXERCISE 4.1 Section 4.1 Properties and graphs of logarithmic functions (page 95)

1. 53 = 125 ∴ log5125 = 3 2. 34 = 81 ∴ log3 81 = 4

3. 6−2 = 136

∴ log6 136

= −2

4. 10−4 = 0.000 1 ∴ log10 0.000 1 = −4 5. a7 = 9 ∴ = 7 log a 9 6. pr = q ∴ = r log p q 7. log10100 = 2 ∴ 102 = 100 8. log5 625 = 4 ∴ 54 = 625

9. 3431log7 = −3

∴ 7−3 = 1

343

10. log 1

3

243 = −5

∴ 5

31 −

⎟⎠⎞

⎜⎝⎛ = 243

69


11. log e e1

= −12

∴ e−

12 =

1e

12. = t log b c ∴ = c bt

13. Let y = log2 32. Then 2y = 32 2y = 25

∴ y = 5 14. Let y = log9 3. Then 9y = 3 ∴ = 3 32 y

2y = 1

y = 12

15. Let y = log8 0.125. Then 8y = 0.125

8y = 18

∴ y = −1 16. Let y = log 1

6

216 .

Then y

⎟⎠⎞

⎜⎝⎛

61 = 216

= 66− y 3

∴ y = −3 17. Let y = lo . g a 1 Then a y = 1 ∴ y = 0 18. Let y = log . a a Then a y = a ∴ y = 1 19. Let y = logb b3. Then by = b3

∴ y = 3

70

SECTION 4.1 PROPERTIES AND GRAPHS OF LOGARITHMIC FUNCTIONS

20. Let y = log e e15

.

Then e y = 15e

∴ y = −5 21. Let y = 10 . 10 7log

Then log10 7 = log10 y ∴ y = 7 22. Let y = . 5logaa Then loga 5 = loga y ∴ y = 5 23. log2 x = 5 x = 25

= 32 24. log5 x = 3 x = 53

= 125

25. log 1

3

x = −4

x = 4

31 −

⎟⎠⎞

⎜⎝⎛

= 81

26. log10 x = −12

x = 21

10−

= 110

27. log16 x = 0.25 x = 160.25

= 2 28. logx 10 = 1 10 = x1

x = 10 29. logx 81 = 2 81 = x2

x = 81 = 9 Note: The base x is positive.

71


30. log x 4 = 32

32

x = 4

x = 432

= 8

31. logx e1 = −1

e1 = x−1

x = e 32. log x e3 = −6 x−6 = e3

x = 63 −

e

= 1e

33. x 0.5 1 2 4 8 16

y = log4 x −0.5 0 0.5 1 1.5 2

72


34. x 0.5 1 5 10 15 20

y = log e x −0.693 0 1.609 2.303 2.708 2.996

35. x 0.25 0.5 1 2 4 8

y = x21log 2 1 0 −1 −2 −3

73


36. x

15

1 5 25 125

y = x51log 1 0 −1 −2 −3

37.

x 31 1 3 9

y = −log3 x 1 0 −1 −2

38. x 1.5 2 4 6 8 10

y = loge (x − 1) −0.693 0 1.099 1.609 1.946 2.197

74


39. x

31 1 3 9

y = 4 − log31 x 3 4 5 6

40. x −2.5 −2 −1 0 2 5

y = loge (x + 3) − 2 −2.693 −2 −1.307 −0.901 −0.391 0.079

41.

Note: The graph of y = is the mirror image of the graph of y = about the x-axis. xalog− xalog

75


42.

Note: The graph of y = is the mirror image of the graph of y = about the y-axis. )(log xa − xalog

43.

Note: The graph of y = can be obtained by shifting the graph of y = a units

to the right. ) (log axa − xalog

44.

Note: The graph of y = − a can be obtained by shifting the graph of y = a units

downward.xalog xalog

76

SECTION 4.2 LAWS OF LOGARITHMS

45.

Note: Since yy

aa

x −⎟⎠⎞

⎜⎝⎛ = 1 = , xy

a1log = is equivalent to y = − . xalog

46.

Note: The graph of ) + (log = 1 axy

a

can be obtained by shifting the graph of xya1log = a units

to the left.

EXERCISE 4.2 Section 4.2 Laws of logarithms (page 101)

1. log10 4 + log10 25 = log10 (4 × 25) = log10 100 = 2 2. log6 42 − log6 7 = log6

427

= log6 6 = 1 3. log9 54 − log9 18 = log9

5418

= log9 3

= log9

129

= 12

77


4. loga a11 111

+ log = ⎟⎠⎞

⎜⎝⎛ ×

111 11 loga

= 1log a

= 0

5. log2 24 − log2 60 + log2 10 = log2 ⎟⎠⎞

⎜⎝⎛ ×

601024

= log2 4 = 2

6. log5 256 − log5 2 7

6 − log5 1021 = log5 ⎟

⎠⎞

⎜⎝⎛ ÷÷

2110

762

256

= log5 1251

= −3

7. 43

3

loglog

ee =

e

e

3

3

log 4

log 21

= 4

21

= 81

8. log

log7

7

32

8 2 = log

log

75

7

72

2

2

= 5 log 7 272

27log

= 107

9. ln ln 3

xx

3

= ln

ln

x

x

3

13

= 3 ln

ln

x

x13

= 9

10. 51

log 2 1010 =

2

10 51log

10⎟⎠

⎞⎜⎝

⎛

= 10 10125log

= 125

78

SECTION 4.2 LAWS OF LOGARITHMS

11. = 3ln 4e43ln e

= 81ln e = 81

12. 35log 42log

125log 12log + 18log

aa

aaa

−−

= ⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛ ×

3542 log

12512 18 log

a

a

= ⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

56 log

56 log

3

a

a

= ⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

56 log

56 log 3

a

a

= 3

13. 21 log6 16 + 2 log6 3 = log6 16 2

1

+ log6 32

= log6 (4 × 9) = log6 36 = 2

14. 3 log + 4 log 2 log2 2

98

1615

2425

− 2

= 2

2

4

2

3

2 2524 log

1516 log +

89 log ⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

= ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛÷⎟

⎠⎞

⎜⎝⎛×⎟

⎠⎞

⎜⎝⎛

243

2 2524

1516

89 log

= log2 2 = 1 15. log10 24 = log10 (23 × 3) = 3 log10 2 + log10 3 = 3p + q

16. log10 250 = log101 000

4

= log10 1 000 − log10 4 = log10 103 − log10 22

= 3 − 2 log10 2 = 3 − 2p

79


17. 2716log10 = 3

4

10 32log

= log10 24 − log10 33

= 4 log10 2 − 3 log10 3 = 4p − 3q

18. log10 0.6 = ⎟⎠⎞

⎜⎝⎛ ×

103 2 log10

= log10 2 + log10 3 − log10 10 = p + q − 1

19. log .103 13 5 =

31

10 227 log ⎟

⎠⎞

⎜⎝⎛

= log10 13

3

2

= 31

1010 2log 3log −

= pq31 −

20. log10 (7 + 4 ) + log10 (7 − 4 ) = log10 [(7 + 4 ) (7 − 4 )] = log10 [72 − ( 4 )2] = log10 45 = log10 (32 × 5) = log10 32 + log10 5 = log10 32 + log10

102

= 2 log10 3 + log10 10 − log10 2 = 2q + 1 − p 21. ln 75 = ln (3 × 52) = ln 3 + ln 52

= ln 3 + 2 ln 5 = a + 2b

22. ln 1549 = ln ⎟⎟

⎠

⎞⎜⎜⎝

⎛

×537 2

= ln 72 − (ln 3 + ln 5) = 2 ln 7 − ln 3 − ln 5 = 2c − a − b

23. ln 0.84 = ln ⎟⎠⎞

⎜⎝⎛ ×

2573

= ln 3 + ln 7 − ln 52

= ln 3 + ln 7 − 2 ln 5 = a + c – 2b = a − 2b + c

80

SECTION 4.3 EQUATIONS INVOLVING LOGARITHMS

24. 21 ln 21 −

31 ln 25 +

41 ln 63 =

21 ln (3 × 7) −

31 ln 52 +

41 ln (32 × 7)

= 21 ln 3 +

21 ln 7 −

31 × 2 ln 5 +

41 × 2 ln 3 +

41 ln 7

= ln 3 − 32 ln 5 +

43 ln 7

= a − 32 b +

43 c

25. ln x5 − 3 ln x = ln x5 − ln x3

= ln 3

5

xx

= ln x2

26. 7 ln x + 12

ln (3x − 5) = ln + ln (3 5)12x x7 −

= ln 5)

12

x

x

7

3( −

27. ln ( 2 ln ( 3 ln x y x y xy

3 2 2) )− − = 3

2223 ln )(ln )(ln ⎟⎟⎠

⎞⎜⎜⎝

⎛−−

yxyxyx

= 3

22

23

)(

ln

⎟⎟⎠

⎞⎜⎜⎝

⎛yxyx

yx

= ln yx

3

4

28. 4 ln 3 ln + 14

ln yx

y x2

6 3− y = 41

36342

)(ln + ln ln yxyx

y−⎟⎟

⎠

⎞⎜⎜⎝

⎛

= ⎥⎥

⎦

⎤

⎢⎢

⎣

⎡⋅⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛41

363

42

)( 1 ln yxyx

y

= ln y

x

234

52

EXERCISE 4.3 Section 4.3 Equations involving logarithms (page 105)

1. log8 (x − 3) + log8 (x − 5) = 1 log8 (x − 3) (x − 5) = log8 8 ∴ x2 − 8x + 15 = 8 x2 − 8x + 7 = 0 (x − 7) (x − 1) = 0 x = 7 or 1 (rejected) ∴ x = 7

81


2. log10 (x2 + 4) − 2 log10 x = 1

log10

2

2

xx + 4

= log10 10

∴ x

x

2

2

+ 4 = 10

x2 + 4 = 10x2

x2 = 49

x = 23

or −23

(rejected)

∴ x = 23

3. log7 (4x + 5) = 2 + log7 x log7 (4x + 5) = log7 72 + log7 x ∴ 4x + 5 = 49x 45x = 5

∴ x = 19

4. log6 3x + log6 (5x − 2) = 3 log6 [3x(5x − 2)] = log6 63

∴ 15x2 − 6x = 216 5x2 − 2x − 72 = 0 (x − 4) (5x + 18) = 0

x = 4 or −185

(rejected)

∴ x = 4 5. log15 (x + 2) + log15 (3x + 4) = 2 log15 (x + 2) (3x + 4) = log15 225 ∴ 3x2 + 10x + 8 = 225 3x2 + 10x − 217 = 0 (x − 7) (3x + 31) = 0

x = 7 or −331 (rejected)

∴ x = 7 6. = 0 52 x x + 5 6− = 0 (5 x x 2) (5 + 3)− = 2 or −3 (rejected) 5x

∴ x log 5 = log 2

x = log 2log 5

82


7. 22 x x 5(2− 1+ ) + 21 = 0 2 = 0 2 x x 5(2 2 + 21− × ) 2 = 0 2 x x 10(2 + 21− ) (2 = 0 x x 3) (2 7)− − = 3 or 7 2 x

∴ x = log 3log 2

or log 7log 2

8. = 0 e ex x2 7 18− − ( = 0 e ex x 9) ( + 2)− = 9 or −2 (rejected) e x

∴ x = ln 9 9. 6 2e ex x + 13 5− = 0 (2e ex x + 5) (3 1)− = 0

= e x 13

or −52

(rejected)

∴ x = ln 13

10. 2ex − 7 2x

e − 15 = 0

( 2x

e – 5) ( 22x

e + 3) = 0

e 2x

= 5 or −23 (rejected)

∴ 2x = ln 5

x = 2 ln 5 11. 23x = 7x–1

3x log 2 = (x − 1) log 7 x(3 log 2 − log 7) = −log 7

∴ x = 2 log 37 log

7 log−

12. = 5 32 1x+ x

(2x + 1) log 3 = x log 5 x (2 log 3 − log 5) = −log 3

∴ x = −

−

log 32 log 3 log 5

13. = 56 23 2 1− ⋅x x 7 +

+ = log (2log ( ) 723 2 1− ⋅x x 3 × 7) ∴ (3 − 2x) log 2 + (x + 1) log 7 = 3 log 2 + log 7 x log 7 − 2x log 2 = 0 x(log 7 − 2 log 2) = 0 log 7 − 2 log 2 ≠ 0 ∵ ∴ x = 0

83


14. = 3 4 8 2x+ 2( )x

= 3 [(2 ( )23 x+2 22 ) ]x

= 3 2 23 6x+ 4( )x

(3x + 6) log 2 = log 3 + 4x log 2 x log 2 = 6 log 2 − log 3

x = 6 log 2 log 3

log 2−

15. e x

x

3

3 =

2log64log

10

10

x

e⎟⎟⎠

⎞⎜⎜⎝

⎛

3

3

= log

log10

6

10

2

2

x

e⎟⎟⎠

⎞⎜⎜⎝

⎛

3

3

= 6

⎟⎟⎠

⎞⎜⎜⎝

⎛

3ln

3ex = ln 6

x = ln 6ln ln 3e3 −

= ln 63 ln 3−

16. = 13 . . . . . . . . . . . . . . . . . . . . . . . . . . (1) yx 7 + 6 = 5 . . . . . . . . . . . . . . . . . . . . . . . . . . (2) yx 7 6 − (1) + (2), 2(6x) = 18 6x = 9

∴ x = 6 log9 log

(1) − (2), 2(7y) = 8 7y = 4

∴ y = 7 log4 log

17. = 13 . . . . . . . . . . . . . . . . . . . . (1) yx 3 + 2 = 82 . . . . . . . . . . . . . . . . . . . . (2) 21 3 + 2 ++ yx

From (2), = 82 . . . . . . . . . . . . . . . . . . . . (3) )9(3 + )2(2 yx

(3) − (1) × 2, = 56 )3(7 y

= 8 y3

∴ y = log 8log 3

(1) × 9 − (3), = 35 7 2( x ) = 5 2 x

∴ x = log 5log 2

84


18. = 7 . . . . . . . . . . . . . . . . . (1) yx 5 4 − = 9 . . . . . . . . . . . . . . . . . (2) 21 5 + 4 −− yx

From (2), 14

4 5( ) ( )x y + 125

= 9

= 900 . . . . . . . . . . . . . . . (3) 25 4( ) )x + 4(5 y

) (1) × 4 + (3), = 928 29 4( x

= 32 4 x

∴ x = 52

(3) − (1) × 25, = 725 29 5( )y

= 25 5y

∴ y = 2 19. 2x + y = 6 . . . . . . . . . . . . . . . . . . . . . . . . . . (1) 3 2x y− = 10 . . . . . . . . . . . . . . . . . . . . . . . . . . (2)

From (2), x − 2y = log 10log 3

x − 2y = 1

log 3 . . . . . . . . . . . . . . . . . . . . . (3)

(1) × 2 + (3), 5x = 12 + 1

log 3

∴ x = 12 log 3 + 1

5 log 3

(1) − (3) × 2, 5y = 6 2

log 3−

y = 6 log 3 2

5 log 3−

20. = 7 y . . . . . . . . . . . . . . . . . . . . (1) 7 2( x )

)

x2 − y + 1 = 0 . . . . . . . . . . . . . . . . . . . . (2) From (2), y = x2 + 1 . . . . . . . . . . . . . . . . . . (3) Putting (3) into (1), = 7 7 2( x 2 1x +

= 7 2 x 2x

∴ x log 2 = x2 log 7 x(x log 7 − log 2) = 0

x = 0 or log 2log 7

. . . . . . . . . . . . (4)

Putting (4) into (3), when x = 0, y = 02 + 1 = 1

when x = log 2log 7

, y = 2

222

7) (log7) (log + 2) (log = 1 +

7 log2 log⎟⎟⎠

⎞⎜⎜⎝

⎛

85


EXERCISE 4.4 Section 4.4 Applications of logarithmic functions (page 110)

1. (a) 36 000 = 30 000t

⎟⎠⎞

⎜⎝⎛ +

10081

1.2 = (1.08)t

∴ t = 1.08 log1.2 log

= 2.369 i.e. It will take 2.4 years.

(b) 36 000 = 30 000t4

410081 ⎟

⎠⎞

⎜⎝⎛

×+

1.2 = (1.02)4t

∴ 4t = 1.02 log1.2 log

t = 2.302 i.e. It will take 2.3 years.

(c) 36 000 = 30 000 e 1008t

1.2 = e 1008t

∴ 1008t = ln 1.2

t = 2.279 i.e. It will take 2.3 years.

2. (a) 21 000 = 18 00012

121001 ⎟

⎠⎞

⎜⎝⎛

×+

r

67 =

12

200 11 ⎟

⎠⎞

⎜⎝⎛ +

r

121

67⎟⎠⎞

⎜⎝⎛ = 1 +

200 1r

r = 1 200⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−⎟

⎠⎞

⎜⎝⎛ 1

67 12

1

= 15.514 5 i.e. The interest rate is 15.51% per annum.

(b) 21 000 = 18 000 e 1001×r

67 = e 100

r

∴ 100

r = ln 67

r = 15.415 1 i.e. The interest rate is 15.42% per annum.

86

SECTION 4.4 APPLICATIONS OF LOGARITHMIC FUNCTIONS

3. (a) P = 6.5 et

1003

(b) 8 = 6.5 et

1003

5.6

8 = et

1003

∴ 100

3 t = ln 5.6

8

t = 6.92 i.e. The population will become 8 million after 7 years. 4. (a) 1 800 × 50.04t = 1 300 × 50.08t

300 1800 1 = t

t

04.0

08.0

55

1318 = 50.08

t − 0.04

t

∴ 0.04t = 5 log

1318 log

t = 5.05 ∴ The populations will be equal after 5 months.

(b) 2(1 800 × 50.04t ) = 1 300 × 50.08t

300 1

800 12 × = t

t

04.0

08.0

55

1336 = 50.08

t − 0.04

t

∴ 0.04t = 5 log

1336 log ⎟

⎠⎞

⎜⎝⎛

t = 15.82 ∴ The population of the black ants will be twice that of the red ants after 16 months.

5. (a) 2

0N = N0 e–k(1 590)

21 = e–1 590k

∴ −1 590k = ln 21

k = 0.000 436 (3 sig. fig.)

(b) N = 10 e–0.000 436(2 000)

= 4.181 1 i.e. There will be 4.18 grams remaining.

(c) 5 = 6 e–0.000 436t

∴ −0.000 436t = ln 65

t = 418.17 i.e. It takes 418 years.

87


6. (a) Initially, the time t = 0. ∴ W = 48 e–0.028(0)

= 48 i.e. The initial weight of the piece is 48 grams.

(b) After 10 years, W = 48 e–0.028(10)

= 36.277 6 i.e. There will be 36.28 grams remaining after 10 years.

(c) 24 = 48 e–0.028t

∴ −0.028t = ln 21

t = 24.76 i.e. The half-life is 25 years.

(d) 9 = 48 e–0.028t

∴ −0.028t = ln 489

t = 59.78 i.e. The piece will decay to 9 grams after 60 years. 7. (a) Let the required equation be N(t) = N0 e–kt.

Then 21 N0 = N0 e–k(5 750)

∴ −5 750k = ln 21

k = 0.000 121 Hence, the required equation is N(t) = N0 e–0.000 121t.

(b) 0.3 = e–0.000 121t

∴ −0.000 121t = ln 0.3 t = 9 950 (to the nearest ten) i.e. The age of the skeleton is 9 950 years old.

8. (a) 2

0N = N0 e–0.000 125t

21 = e–0.000 125t

∴ −0.000 125t = ln 21

t = 5 545.18 i.e. The half-life of carbon-14 is 5 545 years.

(b) 1 − 0.6 = e–0.000 125t

∴ −0.000 125 t = ln 0.4 t = 7 330.33 i.e. The piece of wood is 7 330 years old.

88

SECTION 4.4 APPLICATIONS OF LOGARITHMIC FUNCTIONS

9. (a) When I = I0,

D = 10 log10 ⎟⎟⎠

⎞⎜⎜⎝

⎛

0

0

II

= 10 log10 1 = 0 i.e. The sound level of the threshold sound is 0 dB.

(b) (i) When I = 10I0,

D = 10 log10 ⎟⎟⎠

⎞⎜⎜⎝

⎛

0

010II

= 10 log10 10 = 10 i.e. The sound level of rustle of leaves is 10 dB.

(ii) When I = 109.5I0,

D = 10 log10 ⎟⎟⎠

⎞⎜⎜⎝

⎛

0

05.910

II

= 10 log10 109.5

= 95 i.e. The sound level of drilling is 95 dB.

(c) For printer,

50 = 10 log10 ⎟⎟⎠

⎞⎜⎜⎝

⎛

0II

∴ 0II = 10 10

50

I = 105I0 The ratio of the intensity of the sound of drilling to that of the printer

= 0

50

5.9

1010

II

= 104.5

∴ The sound of drilling is 104.5 times as intense as the printing noise.

(d) Let D′ and I′ be the new sound level and intensity respectively. Then D′ = D + 10 and I′ = kI

10 log10 ⎟⎟⎠

⎞⎜⎜⎝

⎛

0II + 10 = 10 log10 ⎟⎟

⎠

⎞⎜⎜⎝

⎛

0IkI

log10 ⎟⎟⎠

⎞⎜⎜⎝

⎛

0II + 1 = log10 ⎟⎟

⎠

⎞⎜⎜⎝

⎛

0IkI

log10 ⎟⎟⎠

⎞⎜⎜⎝

⎛÷

00

II

IkI = 1

∴ log10 k = 1 k = 10 i.e. We have to multiply the intensity by 10.

89


10. (a) pH = log10 41016.31

−×

= 3.500 3 i.e. The pH value of the orange juice is 3.50.

(b) (i) 7 = log10 ]OH[

1

3+

∴ ]OH[

1

3+ = 107

[H3O+] = 10–7

i.e. The hydronium ion concentration of distilled water is 10–7 moles per litre.

(ii) 1 = log10 ]OH[

1

3+

∴ [H3O+] = 10−1

i.e. The hydronium ion concentration of hydrochloric acid is 10–1 moles per litre.

(iii) 12 = log10 ]OH[

1

3+

∴ [H3O+] = 10–12

i.e. The hydronium ion concentration of household ammonia is 10–12 moles per litre.

(iv) 5.5 = log10 ]OH[

1

3+

∴ [H3O+] = 10–5.5

i.e. The hydronium ion concentration of face cleaning solution is 10–5.5 moles per litre.

90

SECTION 4.5 LOGARITHMIC TRANSFORMATIONS

EXERCISE 4.5 Section 4.5 Logarithmic transformations (page 123)

1. (a) x 2 4 8 10 12

y 10.6 7.3 5.4 4.8 4.2

log10 x 0.301 0.602 0.903 1 1.079

log10 y 1.025 0.863 0.732 0.681 0.623

(b) Since the points plotted in (a) lie approximately in a straight line, thus x and y are related by y k . xn =

(c) log10 k = 1.18 k = 15.1 (3 sig. fig.)

n = 118

0. 0.82

0.7−

−

= −0.514 (3 sig. fig.)

91


2. (a) t 5 10 15 20 25

s 120 490 1 100 1 965 3 070

log10 t 0.699 1 1.176 1.301 1.398

log10 s 2.079 2.690 3.041 3.293 3.487

(b) log10 p = 0.65 ∴ p = 4.47 (3 sig. fig.)

n = 2.690

10 0.65 0−

−.

= 2.04 (3 sig. fig.)

92


3. (a) d 108 150 778 1 427 2 875 4 497

T 0.6 1.0 11.9 29.5 84.0 164.8

log10 d 2.03 2.18 2.89 3.15 3.46 3.65

log10 T −0.22 0 1.08 1.47 1.92 2.22

(b) Since the points plotted in (a) are approximately on a straight line, the data are consistent with a relation of the form T = kd . n

(c) n = 2.7 0.54 2.5

−

−

= 1.47 (3 sig. fig.)

2.7

4

log

010−

−

k =

2.7 0.54 2.5

−

− (same slope)

∴ log10 k = −3.17 k = 6.76 × 10−4 (3 sig. fig.)

93


4. (a) d 5 12 15 20 28

F 7.2 1.3 0.8 0.5 0.2

log10 d 0.70 1.08 1.18 1.30 1.45

log10 F 0.86 0.11 −0.10 −0.30 −0.70

From the graph of log10 F against log10 d, we see that the points plotted are approximately on a straight line. It reveals that the data support the claim.

(b) log10 k = 2.3 ∴ k = 200 (3 sig. fig.)

n = 2.3 −

−

1.50 0.4

= −2

(c) From the result of (b), F = 200d − 2

When d = 3, F = 200(3−2) = 22.2

94


5. (a) x 6 15 22 30 43

y 4.0 2.6 2.1 1.8 1.6

ln x 1.79 2.71 3.09 3.40 3.76

ln y 1.39 0.96 0.74 0.59 0.47

(b) Let y = kx n . Then ln k = 2.25 k = 9.49 (3 sig. fig.)

and n = 2.25

0−

−

0.8 3

= −0.483 (3 sig. fig.) ∴ y = 9.49 0 483x − .

95


6. (a) t 4 8 12 16 20 24

S 55 157 290 450 620 830

ln t 1.39 2.08 2.48 2.77 3.00 3.18

ln S 4.01 5.06 5.67 6.11 6.43 6.72

(b) Let S = kt . n

Then ln k = 1.9 k = 6.69 (3 sig. fig.)

and n = 5 2. 1.92.2 0

−

−

= 1.5 ∴ S = 6 6 9 1. .5t (c) When t = 10, S = 6.69 (101.5) = 212

96


7. (a) P = a ekt

∴ ln P = ln (a ekt) = ln a + ln ekt

= kt + ln a (b) t 1 2 3 4 5 P 160 276 441 743 1 210 ln P 5.075 5.620 6.089 6.611 7.098

(c) ln a = y-intercept of the line in (b) = 4.6 ∴ a = e4.6

= 99.5 (3 sig. fig.) k = slope of the line in (b)

= 05

6.41.7−−

= 0.5

(d) From the result of (c), P= 99.5 e0.5t

When t = 7, P = 99.5 e0.5(7)

= 3 294.99 i.e. The population of the bacteria is approximately 3 295 after 7 hours.

97


8. (a) H(t) = h ebt

∴ ln [H(t)] = ln (h ebt) = ln h + ln ebt

= bt + ln h (b) t 1 2 3 4 5

H(t) 2.9 3.3 4.1 4.5 5.2

ln [H(t)] 1.065 1.194 1.411 1.504 1.649

(c) b = slope of the line in (b)

= 04

9.05.1−−

= 0.15 ln h = y-intercept of the line in (b) = 0.9 ∴ h = e0.9

= 2.46 (3 sig. fig.)

(d) From the result of (c), H(t) = 2.46 e0.15t

When H(t) = 6.5, 6.5 = 2.46 e0.15t

∴ 0.15t = ln 46.25.6

t = 6.478 i.e. The tree will be 6.5 m tall after 6.5 years.

(e) There is a constraint because the tree will not grow at such a rapid rate after 10 years. Besides, a tree has a limiting height and it cannot grow at this rate forever.

98


9. (a) t 0 2 4 6 8 10 M 8.0 7.5 7.0 6.7 6.3 5.8 ln M 2.079 2.015 1.946 1.902 1.841 1.758

(b) M = a e–kt

∴ ln M = ln (a e–kt) = ln a + ln e–kt

= ln a − kt ln a = y-intercept of the line in (a) = 2.062 5 ∴ a = 7.87 (3 sig. fig.) −k = slope of the line in (a)

= 210275.1

−−

= −0.031 3 ∴ k = 0.031 3 (3 sig. fig.)

(c) From the result of (b), M = 7.87 e–0.031 3t

When t= 15, M= 7.87 e–0.031 3(15)

= 4.921 2 i.e. The mass of the substance will be 4.92 grams.

(d) 287.7 = 7.87 e–0.031 3t

∴ −0.031 3t = ln 21

t = 22.145 i.e. The half-life of the substance is 22.1 months.

99


10. (a) t 30 60 90 120 150

Q 1 776 1 150 723 450 290

ln Q 7.482 7.048 6.583 6.109 5.670

If Q = a ebt, then ln Q = ln (a ebt) = ln a + ln ebt

= bt + ln a Since the plotted points of ln Q against t are approximately on a straight line, the above

relation is valid.

(b) ln a = y-intercept of the line in (a) = 7.9 ∴ a = 2 697 (to the nearest integer) b = slope of the line in (a)

= 01509.77.5

−−

= −0.015 (2 sig. fig.)

(c) From the result of (b), Q = 2 697 e–0.015t

When t = 0, Q = 2 697 e–0.015(0)

= 2 697 i.e. The battery can hold 2 697 C of charge.

(d) 100 = 2 697 e–0.015t

∴ −0.015t = ln 697 2

100

t = 219.65 i.e. The battery can last for 220 minutes.

100


11. (a) x 2 4 6 8 10 12

y 3.5 3.8 4.1 4.3 4.5 4.6

e y 33.1 44.7 60.3 73.7 90.0 99.5

(b) y = ln (a + bx) ∴ = a + bx e y

Hence, a = y-intercept of the line in (a) = 20 b = slope of the line in (a)

= 68 20

7 0−

−

= 6.86 (3 sig. fig.)

12. (a) x 1 3 5 7 9

y 3 34 85 160 255

yx2 3 3.78 3.4 3.27 3.15

log102

xx

0 0.053 0.028 0.017 0.012

101


(b) y = ax2 + b log10 x

∴ y

x 2 = 2

10log + x

xba ⋅

Hence, a = y-intercept of the line in (a) = 3 b = slope of the line in (a)

= 3 3 . −

−

30.02 0

= 15

13. (a) ln ⎥⎦

⎤⎢⎣

⎡−1

)(500

tP = ln ⎥

⎦

⎤⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛+

÷−

1 1

500500btea

= ln (1 + a e–bt − 1) = ln (a e–bt) = ln a + ln e−bt

= ln a − bt

(b) Draw the graph of ln ⎥⎦

⎤⎢⎣

⎡− 1

)(500

tPagainst t.

t 1 2 3 4 5 P(t) 120 142 170 209 236

ln ⎥⎦

⎤⎢⎣

⎡−1

)(500

tP 1.153 0.925 0.663 0.331 0.112

102


ln a = y-intercept of the line in the figure = 1.475 ∴ a = 4.37 (3 sig. fig.) −b = slope of the line in the figure

= 15

2.11.0−−

= −0.275 ∴ b = 0.275

(c) From the result of (b),

P(t) = te 275.0 37.41

500−+

(i) When t = 0,

P(t)= 0 275.0 37.41

500×−+ e

= 93.11 i.e. The population of frogs is 93.

(ii) When t = 12,

P(t) = 12 275.0 37.41500

×−+ e

= 430.60 i.e. The population of frogs is 431. (d) When t tends to infinity, 4.37 e–0.275t tends to zero.

∴ P(t) = 01

500+

= 500 i.e. The population of frogs in the pond will eventually be 500. 14. (a) y = 50 + 30 (1 − p e–qx)

30

50−y = 1 − p e–qx

1 − 30

50−y = p e–qx

∴ ln ⎟⎠⎞

⎜⎝⎛ −−

30501 y = ln (p e–qx)

= ln p + ln e–qx

= ln p − qx

Hence, ln ⎟⎠⎞

⎜⎝⎛ −−

30501 y and x are linearly related.

103


(b) Draw the graph of ln ⎟⎠⎞

⎜⎝⎛ −−

30501 y against x.

x 1 2 3 4 5 y 55 62 64 68 69

ln ⎟⎠⎞

⎜⎝⎛ −−

30501 y −0.182 −0.511 −0.629 −0.916 −1.003

ln p = y-intercept of the line above = 0 ∴ p = 1 −q = slope of the line above

= 03

063.0−−−

= −0.21 ∴ q = 0.21

(c) From the result of (b), y = 50 + 30(1 − e–0.21x) 75 = 50 + 30(1 − e–0.21x)

e–0.21x = 61

∴ −0.21x = ln 61

x = 8.532 i.e. 8.5 kg per m2 of fertilizer should be applied.

(d) (i) When x = 0, y = 50 + 30[1 − e–0.21(0)] = 50 i.e. The yield will be 50 kg per m2. (ii) When x is very large, e–0.21x ≈ 0 ∴ y = 50 + 30(1 − 0) = 80 i.e. The yield will be 80 kg per m2.

104


15. (a) N = atbe–t

ln N = ln (atbe–t) = ln a + ln tb + ln e–t

= ln a + b ln t − t ∴ ln N + t = ln a + b ln t Hence, ln N + t and ln t are linearly related.

(b) Draw the graph of ln N + t against ln t. t 1 2 3 4 5

N 55 160 200 172 128

ln t 0 0.693 1.099 1.386 1.609

ln N + t 5.007 7.075 8.298 9.147 9.852

ln a = y-intercept of the line above = 5 ∴ a = 148.4 (1 d.p.) b = slope of the line above

= 0 15 8

−−

= 3

105


(c) From the result of (b), N = 148.4 t3 e–t

When t = 8, N = 148.4(83)e–8

= 25.49 i.e. 25 chickens are infected.

REVISION EXERCISE 4 Revision exercise 4 (page 128)

1. (a) log2 (x2y3) = log2 [(2p)2 (8q)3] = log2 [(22p) (29q)] = log2 [22p+9q] = 2p + 9q

(b) 57

37

loglog

xx = 5

7

31

7

loglog

xx

= x

x

log 5

log 31

7

7

= 151

2. (a) 2 log10 x + log10 3 = log10 (7x − 2) log10 x2 + log10 3 = log10 (7x − 2) log10 3x2 = log10 (7x − 2) ∴ 3x2 = 7x − 2 3x2 − 7x + 2 = 0 (3x − 1) (x − 2) = 0

∴ = 31 or 2

(b) 1 + 2 log10 (x − 1) = log10 (2x − 3) + log10 (5x − 2) log10 10 + log10 (x − 1)2 = log10 (2x − 3) (5x − 2) log10 (10x2 − 20x + 10) = log10 (10x2 − 19x + 6) ∴ 10x2 − 20x + 10 = 10x2 − 19x + 6 ∴ x = 4 3. (a) 52x − 5 x+1 + 6 = 0 52x − 5(5x) + 6 = 0 (5x − 2) (5x − 3) = 0 5x = 2 or 3

∴ x = 5 log2 log or

5 log3 log

= 0.430 7 or 0.682 6

106

REVISION EXERCISE 4

(b) e0.2x − 2e0.1x − 3 = 0 (e0.1x − 3) (e0.1x + 1) = 0 e0.1x = 3 or −1 (rejected) ∴ 0.1x = ln 3 x = 10.986 1

(c) − e3log88 ln 5 = x2

x3log77− ∴ 3 − 5 = x 2 − 3x x2 − 3x + 2 = 0 (x − 1) (x − 2) = 0 ∴ x = 1 or 2 4. ln = hx + k )]([ xf = 1 ∵ f ( )0 ∴ ln 1 = h(0) + k Hence, k = 0 = e∵ f ( )2 3

∴ ln e3 = h(2) + k 3 = 2h

∴ h = 32

5. (a) The equation connecting x and y should be of the form ln y = m ln x + k where m and k are constants. m = slope of the graph

= 42−

= 21

−

k = y-intercept of the graph = 2 ∴ The required equation is

ln y = 21

− ln x + 2

ln y + ln x = 2 ln x y = 2 i.e. x y = e2

107


(b) x y = e2

∴ y = x

e2

x 1 2 3 4 5 6 7 8 9

y =x

e2

7.39 5.22 4.27 3.69 3.30 3.02 2.79 2.61 2.46

6. (a) y = kax Taking logarithms, log10 y = log10 k + x log10 a. log10 a = slope of the graph

= 1 52 4

−−

= 21

∴ a = 21

10 log10 k = y-intercept of the graph

∴ 0 1

log 2 10

−− k

= 21

log10 k = 23

∴ k = 23

10

108

REVISION EXERCISE 4

(b) y = 10 23

(10 21

)x

∴ y = 10 23

21

+x

x −3 −2 −1 0 0.5 1

y = 10 2

321

+x 1 3.16 10 31.6 56.2 100

7. (a) When x tends to negative infinity, y = a ex + b tends to b. From the graph, b = 2. The point (0, 3) is on the graph. ∴ 3 = a e0 + 2 ∴ a = 1

109


(b)

Note: The graph of y = a e−x + b is the mirror image of the graph of y = a ex + b about the

y-axis. 8. (a) Q = A ekt

Putting t = 1, Q = 96 into it, 96 = A ek(1)

∴ 96 = A ek. . . . . . . . . . . . . . . . . . . . . . . . (1) Putting t = 3, Q = 144 into it, 144 = A ek(3)

∴ 144 = A e3k. . . . . . . . . . . . . . . . . . . . . . . (2)

(2) ÷ (1), 23 = e2k

ek = 23 . . . . . . . . . . . . . . . . . . . (3)

∴ k = ln 23

= 0.202 7 Putting (3) into (1),

96 = A ⎟⎟⎠

⎞⎜⎜⎝

⎛

23

∴ A = 78.38

(b) From the result of (a), Q = 78.38 e0.202 7t

When Q = 300, 300 = 78.38 e0.202 7t

∴ 0.202 7t = ln 38.78

300

t = 6.622 i.e. The amount of polluted material will become 300 after 7 months.

9. (a) (i) When S = 21 S0,

21 S0 = S0 e–0.12t

∴ −0.12t = ln 21

t = 5.776 i.e. The sales will reduce to half after 5.8 months.

110

REVISION EXERCISE 4

(ii) When S = 30% of S0, 0.3S0 = S0 e–0.12t

0.3 = e–0.12t

∴ −0.12t = ln 0.3 t = 10.033 i.e. The sales will diminish to 30% of the original amount after 10.0 months. (b) A = the sales at the beginning of the campaign = S0 e–0.12 × 6

= S0 e–0.72

When S = S0, S0 = S0 e–0.72 × e0.08t

1 = e–0.72 + 0.08t

∴ −0.72 + 0.08t = 0 t = 9 i.e. The sales will resume to S0 after 9 months. 10. (a) 2 log2 x = y + 1 . . . . . . (1) log2 (2x) = y + 5 . . . . . . (2) (1) − (2), 2 log2 x − log2 (2x) = −4 ∴ log2 x2 − log2 (2x) = −4

log2

2

2x

x = −4

x2

= 2−4

x = 18

. . . . . . . . (3)

Putting (3) into (1), 2 log2 18

= y + 1

2 (−3) = y + 1 ∴ y = −7

(b) + = 4 e x2 xe 23 −

− 4 + = 0 e x2 xe 23 −

(ex − ) (e − ) = 0 xe− x xe−3 ∴ = or ex xe− xe−3 = 1 or 3 e x2

2x = 0 or ln 3

x = 0 or 12

ln 3

(c) log10 (9 − 1) = logx10 (3 + 1) + 1 x

1 + 31 9log10 x

x − = 1

1 + 3

1) + (3 1) 3(log10 x

xx − = 1

log10 (3 − 1) = 1 x

− 1 = 10 3x

= 11 3x

∴ x = loglog

10

10

113

= 2.183 (4 sig. fig.)

111


11. (a) (ln x)2 = 2 ln x (ln x)2 − 2 ln x = 0 ln x (ln x − 2) = 0 ln x = 0 or 2 ∴ x = 1 or e2

(b)

(c) From the graph in (b), (ln x)2 > 2 ln x when x < 1 or x > e2

(d) (i) Substitute the point (e2, 5) into y = 2 ln x + k, 5 = 2 ln e2 + k 5 = 2(2) + k ∴ k = 1 (ii)

Note: The graph of y = 2 ln x + 1 can be obtained by shifting the graph of y = 2 ln x

in (b) 1 unit upward.

112

REVISION EXERCISE 4

12. (a) m = 5 e−0.4t

When t = 0, m = 5 e−0.4(0)

= 5

(b) If the mass becomes 54

g,

54

= 5 e−0.4t

−0.4t = ln 14

t = ln40 4.

= 3.466 i.e. The required time is 3.466 hours.

(c) Let t1 and t2 be the times when the mass becomes 3 g and 2.9 g respectively. 3 = . . . . . . . . . . . . (1) 14.0 5 te−

2.9 = . . . . . . . . . . . . (2) 24.0 5 te−

From (1), 35

= 14.0 te−

−0.4t1 = ln 35

t1 = 53 ln

4.01

−

Similarly, from (2), t2 = 59.2 ln

4.01

−

t2 − t1 = ⎟⎠⎞

⎜⎝⎛ −

− 53 ln

59.2 ln

4.01

= 39.2 ln

4.01

−

= 0.084 75 i.e. The required time = 0.084 75 hour

(d) (i) m = 5 e–0.4t

ln m = ln (5 e–0.4t) = ln 5 + ln e–0.4t

= ln 5 − 0.4t ∴ ln m and t are connected by the equation in the form ln m = a + bt where a and b

are constants. Hence, the graph of ln m against t is a straight line. (ii) The slope is −0.4. The intercept on the ln m axis is the value of ln m when t = 0. i.e. ln m = ln 5 ∴ The intercept = ln 5

113


13. (a) (i) The half-life is the time at which P = 50. 50 = 100 e–0.000 12t

∴ −0.000 12t = ln 10050

t = 5 780 (3 sig. fig.) i.e. The half-life of carbon-14 is 5 780 years. (ii) When P = 30,

30 = 100 e−0.000 12t

∴ −0.000 12t = ln 10030

t = 10 033.11 i.e. The bone has been buried for 10 033 years.

(b) (i) Suppose the man has died for t0 hours at 11 p.m. Now T0 = 16. When t = t0, T = 25. ∴ 25 − 16 = (37.5 − 16) 0kte−

9 = 21.5 . . . . . . . . . . . . . . . . . . . . . . (1) 0kte−

When t = t0 + 1, T = 20. ∴ 20 − 16 = (37.5 − 16) )1( 0 +− tke 4 = 21.5 )1( 0 +− tke

(2) ÷ (1), 94 = . . . . . . . . . . . . . . . . . . . . . (2) )( 00 ktkkte −−−−

94 = e–k. . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)

∴ −k = ln 94

k = 0.81 (2 sig. fig.) (ii) Putting (3) into (1),

9 = 21.50

94 t

⎟⎠⎞

⎜⎝⎛

∴ t0 ln ⎟⎠⎞

⎜⎝⎛

94 = ln

5.219

∴ t0 = 1.074 ∴ The man has died for 1 hour at 11 p.m. i.e. The murder was committed at 10 p.m. (iii) No, carbon-dating cannot be employed to estimate the time of death as only a very little

amount of carbon-14 would have decayed. The change in percentage of carbon-14 cannot be measured accurately enough to evaluate the murder time.

114

REVISION EXERCISE 4

14. (a) I = kV n

log10 I = log10 k + n log10 V Thus, the graph of log10 I against log10 V is a straight line.

V 160 180 200 220 240

I 52 70 93 121 152

log10 V 2.204 2.255 2.301 2.342 2.380

log10 I 1.716 1.845 1.968 2.083 2.182

n = slope of the graph

= 275.2 375.29.1 175.2

−−

= 2.75 log10 k = y-intercept of the graph

∴ 0 275.2

log 9.1 10

−− k

= 2.75

log10 k = −4.356 k = 4.4 × 10−5

(b) Thus, I = 4.4 × 10−5 V 2.75. When V = 230, I = 4.4 × 10−5 (2302.75) = 137

(c) If V increases to V1 = 2V, then I becomes I1 = 4.4 × 10−5 (2V)2.75

= 6.727 (4.4 × 10−5 V 2.75) ∴ The percentage increase in I = (6.727 − 1) × 100% = 572.7%

115


15. (a) t 1 2 3 4 5 Q 136 370 1 004 2 730 7 415 ln t 0 0.693 1 1.098 6 1.386 3 1.609 4

ln Q 4.912 7 5.913 5 6.911 7 7.912 1 8.911 3 (b) (i) (ii)

116

REVISION EXERCISE 4

(iii)

(c) From the straight line graph in (b)(ii), we can say that ln Q = ln a + bt where a and b are constants. Hence, Q = a ebt. ln a = y-intercept of the line in (b)(ii) = 3.9 ∴ a = 49.4 (3 sig. fig.) b = slope of the line in (b)(ii)

= 05

9.39.8−−

= 1 ∴ The equation connecting Q and t is Q = 49.4 et.

(d) When t = 7, Q = 49.4 e7

= 54 200 (3 sig. fig.) i.e. The amount of algae will be 54 200 after 7 days. 16. (a) (i) ln [P(t) + 5] = ln [a ebt − 5 + 5] = ln (a ebt) = ln a + ln ebt

= bt + ln a

117


(ii) t 2 4 6 9 P(t) 8.5 13.0 19.7 28.1 ln [P(t) + 5] 2.602 7 2.890 4 3.206 8 3.499 5

(iii) ln a = y-intercept of the line in (ii) = 2.3 ∴ a = 10 (2 sig. fig.) b = slope of the line in (ii)

= 08

3.25.3−−

= 0.15 (iv) Hence, the equation connecting P and t is P(t) = 10 e0.15t − 5. When t = 10, P(10) = 10 e0.15(10) − 5 = 40 i.e. The amount of product is 40 grams when t = 10.

118

REVISION EXERCISE 4

(b) (i) t 0 4 8 12 15 R(t) = 115 − e0.3t 115 111.7 104.0 78.4 25.0

(ii) 21 (115 − e0.3t) = 10 e0.15t − 5

115 − e0.3t = 20 e0.15t − 10 e0.3t + 20 e0.15t − 125 = 0 (e0.15t − 5) (e0.15t + 25) = 0 e0.15t = 5 or −25 (rejected)

∴ t = 0.15

5ln

= 10.730 ∴ The reaction will reach the equilibrium state after 10.7 minutes.

119

CHAPTER 4 LOGARITHMIC FUNCTIONS - Browser Expressresources.npl.edu.hk/~hflau/MS-Sol-EE-C04.pdf · CHAPTER 4 LOGARITHMIC FUNCTIONS EXERCISE 4.1 Section 4.1 Properties and graphs of

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