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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
CHAPTER 4 LOGARITHMIC FUNCTIONS
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
CHAPTER 4 LOGARITHMIC FUNCTIONS
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
SECTION 4.1 PROPERTIES AND GRAPHS OF LOGARITHMIC FUNCTIONS
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
CHAPTER 4 LOGARITHMIC FUNCTIONS
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
SECTION 4.1 PROPERTIES AND GRAPHS OF LOGARITHMIC FUNCTIONS
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
CHAPTER 4 LOGARITHMIC FUNCTIONS
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
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CHAPTER 4 LOGARITHMIC FUNCTIONS
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
CHAPTER 4 LOGARITHMIC FUNCTIONS
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
CHAPTER 4 LOGARITHMIC FUNCTIONS
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
SECTION 4.3 EQUATIONS INVOLVING LOGARITHMS
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
CHAPTER 4 LOGARITHMIC FUNCTIONS
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
SECTION 4.3 EQUATIONS INVOLVING LOGARITHMS
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
CHAPTER 4 LOGARITHMIC FUNCTIONS
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
CHAPTER 4 LOGARITHMIC FUNCTIONS
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
CHAPTER 4 LOGARITHMIC FUNCTIONS
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
CHAPTER 4 LOGARITHMIC FUNCTIONS
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
SECTION 4.5 LOGARITHMIC TRANSFORMATIONS
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
CHAPTER 4 LOGARITHMIC FUNCTIONS
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
SECTION 4.5 LOGARITHMIC TRANSFORMATIONS
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
CHAPTER 4 LOGARITHMIC FUNCTIONS
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
SECTION 4.5 LOGARITHMIC TRANSFORMATIONS
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
CHAPTER 4 LOGARITHMIC FUNCTIONS
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
SECTION 4.5 LOGARITHMIC TRANSFORMATIONS
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
CHAPTER 4 LOGARITHMIC FUNCTIONS
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
SECTION 4.5 LOGARITHMIC TRANSFORMATIONS
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
CHAPTER 4 LOGARITHMIC FUNCTIONS
(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
SECTION 4.5 LOGARITHMIC TRANSFORMATIONS
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
CHAPTER 4 LOGARITHMIC FUNCTIONS
(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
SECTION 4.5 LOGARITHMIC TRANSFORMATIONS
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
CHAPTER 4 LOGARITHMIC FUNCTIONS
(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
CHAPTER 4 LOGARITHMIC FUNCTIONS
(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
CHAPTER 4 LOGARITHMIC FUNCTIONS
(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
CHAPTER 4 LOGARITHMIC FUNCTIONS
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
CHAPTER 4 LOGARITHMIC FUNCTIONS
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
CHAPTER 4 LOGARITHMIC FUNCTIONS
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
CHAPTER 4 LOGARITHMIC FUNCTIONS
(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.
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