Certificate Mathematics in Action Full Solutions 4A
5 Exponential and Logarithmic Functions
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Activity 5.3 (p.238)
Activity
1.
1
Activity 5.1 (p.216)
2.
No
1.
1
3.
y increases as x increases; the graph lies on the right-hand
side of the y-axis; it cuts the x-axis at (1,0).
2.
No
3.
y increases as x increases; the graph lies above the x-axis;
it cuts the y-axis at (0, 1).
4.
x
y
0.3
–0.5
0.6
–0.2
1
0
2
0.3
3
0.5
4
0.6
4.
–1
1
3
x
y
5.
0
1
2
3
1
3
9
27
Yes
5.
Activity 5.2 (p. 222)
They are both defined for x > 0 and cut the x-axis at (1,0);
y increases as x increases in both graphs; they both lie on
the right-hand side of the y-axis.
1.
M
1
10
100
N
M 
logM logN logMN log   logM2
N 
100
–1
2
1
–3
–2
100
1
1000
10
1
10
100
1
1000
100
2
2
4
0
4
3
–1
2
4
–2
–1
3
2
–2
3
1
–5
–4
log MN = log M + log N
3.
M 
log   = log M – log N
N 
4.
log M2 = 2 log M
p. 205
1.
6
1
2.
Follow-up Exercise
(3 x 3 ) 4 3 4 ( x 3 ) 4

(2 x 2 ) 3 2 3 ( x 2 ) 3
81x1 2
8x 6
81x1 26

8
81x 6

8

1
5
2.
 a 2 b 3
 2
 a b
1

  a 2( 2 ) b 31



 a 4 b 4

1
2.
1
3
x


b4
a4
x
3.

(64)
4
3
3
3
5
3
 [(4) 3 ]
1
(4) 4
1

256
1000  10

 216   6
4.
4.
∵ 11  11 = 121
112 = 121
3
2 2
 25  2  5  
     
 36 
 6  
5
 
6
125

216
 121   11
∴
∵ 4  4  4  4 = 256
44 = 256
 4 256   4
∴
 x
3
4
5.
1 1 1 1 1
   
3 3 3 3 81
∵
x
5.
1
6.
4

x

5
1
1
  
 2  32
∴
5
1
1

32 2
x
1
 x4
1 1 1 1 1 1
    
2 2 2 2 2 32
∵
3
x4
3
1 1

81 3
3
( x 4 )3

x
4
1
1
  
81
 3
∴
4
3
 4
3  
 3
3
3.

 (4) 4
∵ (–6)  (–6)  (–6) = –216
(–6)3 = –216
∴

 (4)
∵ 10  10  10 = 1000
103 = 1000
∴
2.
5
x3
p. 207
1.
1
(x5 ) 3
1
1
 a 4b 4

1

5
Exponential and Logarithmic Functions

1
4
1
1
x4
1
6.
1
4
1
a 
5
a4
a4

1
(a 4 ) 5
1

p. 210
a4
4
a5
1.
 x
3
4
1
1 4

5
 (x 3 )4
 a4
4
x3
a



11
20
1
11
a 20
2
Certificate Mathematics in Action Full Solutions 4A
(c) ∵
∴
p. 213
1.
3x + 2 – 3x = 216
– 3x = 216
3x(32 – 1) = 216
8(3x) = 216
3x = 27
3x = 33
∴
x =3
(32)(3x )
2.
3.
p. 221
52x – 24(5x) = 25
(5x)2 – 24(5x) – 25 = 0
Let y = 5x, the equation becomes
y2 – 24y – 25 = 0
(y – 25)(y + 1) = 0
y = 25
or y = –1
∴ 5x = 25 or
5x = 52
∴ x=2
1.
∵
92 = 81
∴ log9 81 = 2
2.
∵
43 = 64
∴ log4 64 = 3
3.
∵
112 = 121
∴ log11 121 = 2
4.
∵
25 = 32
∴ log2 32 = 5
5x = –1 (rejected)
3 x 1  3 y  0  (1)
 x y
 ( 2)
2  16
p. 225
From (1),
3x + 1 = 3y
∴ x+1=y
From (2),
2x + y = 24
∴ x+y=4
Consider the simultaneous equations:
 x  1  y  (3)

 x  y  4  ( 4)
By substituting (3) into (4), we have
x + (x +1) = 4
2x = 3
3
x=
2
3
By substituting x =
into (3), we have
2
3
5
y= +1=
2
2
3
5
∴ The solution is x = , y = .
2
2
1
1.
1
2.
3.
log2 8 – log2 16 = log2 23 – log2 24
= 3 log2 2 – 4 log2 2
=3–4
= 1
log 16 log 2 4

log 32 log 2 5
4 log 2

5 log 2
4

5
1
4.
(a) ∵
0.01 = 10–2
∴ log 0.01 = 2
(b) ∵
100 000 = 105
∴ log 100 000 = 5
2.
2 log 2 + log 5 = 2 log 2 2 + log 5
= log ( 2 2 )2 + log 5
= log 2 + log 5
= log (2  5)
= log 10
=1
p. 220
1.
10x = 5
x = log 5
= 0.699 (cor. to 3 sig. fig.)
(a) ∵
∴
10x = 20
x = log 20
= 1.30 (cor. to 3 sig. fig.)
(b) ∵
∴
10x = 0.8
x = log 0.8
= 0.0969 (cor. to 3 sig. fig.)
3
log 5 log 5 2

log 25 log 5 2
1
log 5
 2
2 log 5
1

4
5
5.
6.
2 log x
log x  log x

1
2
log x  log x
2 log x

1
log x  log x
2
2 log x

 1
1   log x
 2
2

1
2
4
log x 2  log x
log x 2  log x
6x = 3x + 2
log 6x = log 3x + 2
x log 6 = (x + 2) log 3
x log 6 = x log 3 + 2 log 3
x log 6 – x log 3 = 2 log 3
(log 6 – log 3) x = 2 log 3
2 log 3
x=
log 6  log 3
= 3.17 (cor. to 3 sig. fig.)
4.
log (3x – 2) = 2
log (3x – 2) = log 100
3x – 2 = 100
3x = 102
x = 34
5.
log (3x + 1) – log (x – 2) = 1
 3x  1 
log
 = log 10
 x2 
3x  1
= 10
x2
3x + 1 = 10(x – 2)
3x + 1 = 10x – 20
7x = 21
x =3
6.
Let y = log (x – 1), then the equation
[log (x – 1)]2 + 2 log (x – 1) + 1 = 0 becomes
y2 + 2y + 1 = 0
(y + 1)2 = 0
y = –1
∴ log (x – 1) = –1
1
x–1=
10
11
x=
10
1





7.
3.
2 log x
Exponential and Logarithmic Functions
log x 2  log x 2
1
log x 2  log x 2
1
2 log x  log x
2
1
2 log x  log x
2
1

 2   log x
2

1

 2   log x
2

3
2
5
2
3
5
54
6
= log 54 – log 6
= yx
log 9 = log
p. 233
p. 229
1.
2.
3x – 1 = 7
log 3x – 1 = log 7
(x – 1)log 3 = log 7
log 7
x–1=
log 3
log 7
x=
+1
log 3
= 2.77 (cor. to 3 sig. fig.)
52x = 8
log 52x = log 8
2x log 5 = log 8
log 8
2x =
log 5
log 8
x=
2
log 5
= 0.646 (cor. to 3 sig. fig.)
1.
Let I1 and I2 be the sound intensities of the concert and
that of the football match respectively.
By the definition of sound intensity level, we have
I 
95 = 10 log  1 
 I0 
I
and 80 = 10 log  2
 I0




I
∴ 95 – 80 = 10 log  1
 I0
 I
15 = 10 log 1
  I 0
I
3
= log  1
2
 I2

I
 – 10 log  2

I

 0

I
  log 2

I

 0












3
I1
= 10 2
I2
= 31.6 (cor. to 3 sig. fig.)
4
Certificate Mathematics in Action Full Solutions 4A
∴ The sound intensity of the concert is 31.6 times to
that of the football match.
Exercise
2. Let I be the original sound intensity produced by the radio,
then the sound intensity produced after adjusting the
volume is 1.25I.
∵ The original sound intensity level produced by the
radio is 30 dB.
 I 
∴ 30 = 10 log  
 I0 
Let dB be the sound intensity level produced by the
radio after adjusting the volume.
 1.25I 

∴  = 10 log 

 I0 
 1.25I
– 30 = 10 log 
 I0


 I
 – 10 log 

I

 0
Exercise 5A (p.210)
Level 1
1
1.
7
x3  (x3 ) 7
3
 x7
2.
 x   (x
8
3
8




 x3
  1.25I 
 I 
  log 
– 30 = 10 log

I 
 0 
  I 0 
– 30 = 10 log 1.25
= 30 + 10 log 1.25
1
3.
 x
5

1
( x 2 )5
1
5
x2
∴ The sound intensity level produced by the radio after
adjusting the volume is 31.0 dB.
x
4.
p.235
Let E1 and E2 be the relative energies released by the
earthquakes measured 6 and 3 respectively.
By the definition of the Richter scale, we have
6 = log E1 and 3 = log E2
i.e. E1 = 106
and E2 = 103
6
E1 10

∴
E 2 10 3
= 103
= 1000
1
5
x
2


5
2
1

1
(x2 ) 5
1
2
x5
x

2
5
3
3
5.
2 2
 49  2  7  
     
 25 
 5  
7
 
5
343

125
∴ The strength of an earthquake measured 6 is 1000
times to that measured 3.
2.
1

= 31.0 (cor. to 3 sig. fig.)
1.
1
3 )8
Let E1 and E2 be the relative energies released by the
earthquakes in Nantou and Turkey respectively.
By the definition of the Richter scale, we have
7.3 = log E1 and
6.3 = log E2
i.e. E1 = 107.3
and
E2 = 106.3
7.3
E1 10

∴
E 2 10 6.3
= 10
∴ The strength of the earthquake in Nantou is 10 times
to that in Turkey.
3
3
3
6.
4 4
 16  4  2  
     
 81 
 3  
2
 
3
8

27
5
3
5
3
3
7.
3
 
4
27

64
8.
1
2 2
 9  2  3  
     
 16 
 4  
 27 
 
 64 

14. a 3 b
1
4
2
a
1

1
3
11. (m n )





15.
m n
aq
3
m

1
2
1
a2a p

1
1
  ( 2)
 ( 2)
2
b2

3
4
 ab 1
3
1   ( 1)
4

2 1
3b 4
b4
2
a 2 p
q
a
2 p 
q
2
a
2 p 
q
2
a2
1
 a0
q
0
2
q = 2(2 + p)
when p = 2, q = 2(2 + 2) = 8.
when p = 3, q = 2(2 + 3) = 10.
∴ A possible solution is p = 2, q = 8 or p = 3, q = 10.
(or any other reasonable answers)
 2 p
 1
4  
 2
n
 ax
16.  y
a
 1
2  
 2
1
m n
3
3
1
2

4 1 3 2


a 3 2b 2 3

11 13
a6b6
1
a

 1
4  
 2
2
1 2
3
1 2

2  2

2b 3  a 3 b 4
2b 3

a

a


 a 3b 2  a 2b 3




a x  y  a1

= a3b–9  a–2b–1
= a3 – (–2)b–9 – (–1)
= a5b–8
a5
= 8
b
2
1

(a  x ) 1 a x
  y 1   y  a x  y
(a )
a

 a x

 ay


m–1n–3
 a 1( 3) b 3( 3)  a
4

(a q ) 2
aq
=

= m2 – (–1)n1 – (–3)
= m3n 4
12. (a 1b 3 ) 3  (a 4 b 2 )
a 2 p


1
1
a2a p
1
m2n
 2 3
 3 4
a b
13. 

a
a3
1

 = [(m–4 – 3)(n –3 – 4)]–1

= (m–7n –7)–1
= m7n7
1
2
3
4
1
1


 a3 b

 = (a–1+3 – 2)–1

= (a0)–1
= a0
=1
 m 4 n 3
10.  3 4
 m n
1
 a 3b
1
 3
  
 4
4

3
9.
2
 a 3b

3
 3  
    
 4  

 a 1 a 3
 2
 a
 1 1 


a 2b 2 




3
4
3
1


 27  3 

  
 64  


1
3

Exponential and Logarithmic Functions
b
x  y 1
y=1–x
When x = 2, y = 1 – 2 = –1.
When x = 3, y = 1 – 3 = –2.
∴ A possible solution is x = 2, y = –1or x = 3, y = –2.
(or any other reasonable answers)
 1
2  
 2
Level 2
17.
1
(3 x ) 2
x
3
4
 x 
4
2
(x 3 )2
x
3
4
2

x3
x
3
4
x
1
 (x2 ) 4
1
2
2 3 1
 
4 2
 x3
5
 x 12
6
Certificate Mathematics in Action Full Solutions 4A
1
18. x
2
3

1
x4

5
x
x3

2
3

1
3 5
(x )
x

2
3
x
1 1
1 1
 1 1 
2
2 2
a b
1 1
1 1
 1
 1
 1 1  1 1 
24.  m 3 n 2  m 2 n 3   mn  m 3 2 n 2 3






61
60
61
3
 p 
 ( p2 )3
1
5
( p )2
p4
p
1

5
2
 s s
3
1
26.
1

2
s
b2
3
2
1
 (27 ab) 3  a 3 b 
a
t
t
t

1
3
4
t t


2
3
22. q 
2
(q )
9
q
3
1
4
27.
2
3
2

 1
 m n  2 2 m n 2   1
 2
m
23
12
1
4
2
3

4
3
1
2

4
23
1
t 12
 22 m 4
q
q 
4
(  )
3
17
8
 4m 1 2 n 3
1
3 9
1
3
8
 4m 1 2 6 n 3
1
2
(q )
 q4 
13

1

q2
1
q3
3 1 1
 
2 3
 q4
11
 q 12
7
2
n3
 ( 2 )
1
 m6
1
3




 m 4 n 3  2 2 m 3 n 2  (m 2 )
1
3
4
2
1 2 1
  
3 3 6
 1 
 2 

m n   2m 3 n   



 m
2
3
1
4
19 1

3
4
1
1
 (t 3 ) 4  t
1
1
 (33 ab) 3  (a 3 b) 2
1
3
 9a 6 b 6
1 3
2
  (  )
2 4
3

2
b2
 9a
13
1
2
5

s5

1
2
1
1
 a 3 b 2  32 a 3 b 3  a 6 b 2
5 1 3
 
2 5
1
2
1 1

4
b2
b4
a
3
5
 s2

3
2
2a 4
1
t
5

3
2
1
s
2
3
1

 2 1 a 4 b 4
19
60
3

1
1 1
3
 (  )
4
2
 s 5  ( s 5 ) 2  ( s 2 ) 1  s 5
 4 t3 t
1
 2 1 a 2
19
60
p
1
 a 2 b 2  2a 4 b 4  a
1

1
2
1
 (ab) 2  ( 2 4 ab) 4  a


p
1
2
3
2
3 2 2
 
4 5 3
p
s5 
p
2
5
 3
ab  4 16ab   a 4 
 
25.
3


1
1
p4
2
3
(5 p ) 2
21. t

3
p4
20.
1

 m 6n 6
1
 1 1
m6n6
x 60
19.
b
2
1

1
 a2
2 1 3
  
3 4 5

1
3
5
x
x
1
1
4
x

1
1
a2
 a 2
23. a b     ab 2  a 2 b  1  ab 2
b
b2
1
2
x4


1
3
1
3
b
2 1
2 
3 2
5
1

1
2
1
1

1
2
2
3 2
3 5
3
3 5
4
4
3
28. xy  64 x y  x y  xy  (4 x y )  x y
2
1
1

1
6.
2
 xy 4  4 x 3 y 3  x 3 y 5
 4 1 x
2
1
1  (  )
3
3
2

1 1 2
 
3 5
y4
3(3x) – 3x – 1 = 216
3(3x) – 3x(3– 1) = 216
3x(3 –3–1) = 216
1
3x(3 – ) = 216
3
29
1 3  60
x y
4
2

x3
∴
29
4 y 60
2x + 2 + 2x = 10
(22)(2x ) + 2x = 10
2x(22 + 1) = 10
5(2x) = 10
2x = 2
2x = 21
∴
x =1
8.
42x = 16(2x)
(22)2x = 24(2x)
24x = 24 + x
∴ 4x = 4 + x
4
x=
3
9.
23x = 8(4x)
23x = 23(22)x
23x = 23(22x)
23x = 23 + 2x
∴ 3x = 3 + 2x
x=3
Level 1
2
5 2  125 3
1.
2
x
5 2  (53 ) 3
x
∴
5 2  52
x
=2
2
x=4
x
3
8  16
2.
x
(23 ) 3  2 4
2 x  24
∴ x=4
10.
6
3.
x
1
2
 36
x
1
2
∴
6  62
x
+1= 2
2
x=2
x
x
3
(2 )  (2 )
2
5
2
5
2x
∴
9(3x + 1) = 272x
32(3x + 1) = (33)2x
3x + 3 = 36x
∴ x + 3 = 6x
3
x=
5
Level 2
2
4 3  32 5
4.
5.
8
3x   = 216
3
3x = 81
3x = 34
x=4
7.
Exercise 5B (p.214)
x
Exponential and Logarithmic Functions
2 3  22
2x
=2
3
x =3
3x + 2 – 3x = 8
(32)(3x ) – 3x = 8
3x(32 – 1) = 8
8(3x) = 8
3x = 1
3x = 30
∴
x=0
11.
4x + 1 + 4x – 4x – 1 = 19
(4)4x + 4x – (4– 1)4x = 19
4x(4 + 1 – 4– 1) = 19
1
4x(5 – ) = 19
4
19
4x(
) = 19
4
4x = 4
∴
x =1
12. 6x + 2 – 2(6x + 1) – 12(6x) = 2
(62)6x – 2(6)6x – 12(6x) = 2
6x(36 – 12 – 12) = 2
6x(12) = 2
1
6x =
6
6x = 6–1
∴
x = 1
8
Certificate Mathematics in Action Full Solutions 4A
13.
22x – 3(2x) + 2 = 0
(2x)2 – 3(2x) + 2 = 0
Let y = 2x, the equation becomes
y2 – 3y + 2 = 0
(y – 1)(y – 2) = 0
y = 1 or y = 2
∴ 2x = 1
or
2x = 2
2x
or
or
2x = 21
x =1
∴
14.
20
=
x=0
x = –1
By substituting x = –1 into (4), we have
y = –2(–1) = 2
∴ The solution is x = –1, y = 2.
18. 3 x  y  27  (1)
 3 x y
 4  ( 2)
4
From (1),
3x + y = 33
∴ x+y=3
From (2),
43x – y = 41
∴ 3x – y = 1
Consider the simultaneous equations:
 x  y  3  (3)

3x  y  1  ( 4)
(3) + (4),
(x + y) + (3x – y) = 3 + 1
4x = 4
x=1
By substituting x = 1 into (3), we have
1+y=3
y=2
∴ The solution is x = 1, y = 2.
32x – 12(3x) + 27 = 0
(3x)2 – 12(3x) + 27 = 0
Let y = 3x, the equation becomes
y2 – 12y + 27 = 0
(y – 3)(y – 9) = 0
y = 3 or y = 9
∴ 3x = 3
or
3x = 9
3x
or
or
3x = 32
x=2
∴
31
=
x =1
15. 5(52x) – 26(5x) + 5 = 0
5(5x)2 – 26(5x) + 5 = 0
Let y = 5x, the equation becomes
5y2 – 26y + 5 = 0
(5y – 1)(y – 5) = 0
1
y=
or y = 5
5
16.
1
=
or
5
∴
5x
∴
5x = 5–1 or
x = 1 or
5x
19. 32 x  y  729  (1)
 x2 y
 3  ( 2)
9
From (1),
32x + y = 36
∴ 2x + y = 6
From (2),
(32)x – 2y = 3
32x – 4y = 31
∴ 2x – 4y = 1
Consider the simultaneous equations:
2 x  y  6  (3)

2 x  4 y  1  ( 4)
(3) – (4),
(2x + y) – (2x – 4y) = 6 – 1
5y = 5
y=1
By substituting y = 1 into (3), we have
2x + 1 = 6
5
x=
2
5
∴ The solution is x = , y = 1.
2
=5
5x = 51
x =1
24x – 20(22x) + 64 = 0
(22x)2 – 20(22x) + 64 = 0
Let y = 22x, the equation becomes
y2 – 20y + 64 = 0
(y – 4)(y – 16) = 0
y=4
or y = 16
∴ 22x = 4
or
22x = 16
=
or
2x = 2 or
x = 1 or
22x = 24
2x = 4
x=2
22x
∴
22
17. 5 x  2 y  125  (1)
 2 x y
 1  ( 2)
25
From (1),
5x + 2y = 53
∴ x + 2y = 3
From (2),
252x + y = 250
∴ 2x + y = 0
y = –2x
Consider the simultaneous equations:
 x  2 y  3  (3)

 y  2 x  ( 4)
By substituting (4) into (3), we have
x + 2(–2x) = 3
–3x = 3
20. 2 3 x 2 y  4  (1)
 x y
5  625  ( 2)
From (1),
23x – 2y = 22
∴ 3x – 2y = 2
From (2),
5x + y = 54
∴ x+y=4
y=4–x
Consider the simultaneous equations:
3x  2 y  2  (3)

 y  4  x  ( 4)
9
5
By substituting (4) into (3), we have
3x – 2(4 – x) = 2
5x = 10
x=2
By substituting x = 2 into (4), we have
y=4–2=2
∴ The solution is x = 2, y = 2.
Exercise 5C (p.226)
Level 1
1.
∵
1 000 000 = 106
∴ log 1 000 000 = 6
2.
∵
625 = 54
∴ log5 625 = 4
3.
∵
1
= 4–2
16
10.
11.
log 9 log 32

log 27 log 33
2 log 3

3 log 3
2

3
log 64 log 4 3

log 16 log 4 2
3 log 4

2 log 4
3

2
 1 
 1 
log 2   log 2  2 
25 

5 
12.

1
log 2 5
log 2 5 2

1
∴ log4   = 2
 16 
4.
 1 
∴ log 
 = 2
 100 
5.
∵
∴
10x
6.
∵
∴
10x
= 120
x = log 120
= 2.08 (cor. to 3 sig. fig.)
log 2 5  2
1
log 2 5 2
 2 log 2 5
1
log 2 5
2
 4

1
= 10–2
100
∵
Exponential and Logarithmic Functions
13. log 6 = log (2  3)
= log 2 + log 3
= x y
1
7.
8.
9.
= 88
x = log 88
= 1.94 (cor. to 3 sig. fig.)
200
2
= log 100
= log 102
= 2 log 10
=2
14. log 18 = log (2  32 ) 2
1
= log ( 2 2  3)
1
= log 2 2 + log 3
1
= log 2+ log 3
2
1
= x y
2
log 200 – log 2 = log
log 4 + log 25 = log(4  25)
= log 100
= log 102
= 2 log 10
=2
1
1
log3   + log3 108 = log3 (
 108)
12
 12 
= log3 9
= log3 32
= 2 log3 3
=2
15. Let loga 8 = x, logb 8 = y.
Then ax = 8 and by = 8.
∵ loga 8 logb 8 = 1
∴
xy = 1
1
x=
y
Let a = 2, then
2x = 8
2x = 23
x=3
1
∴ y=
3
1
∴
b3  8
b = 83
= 512
10
Certificate Mathematics in Action Full Solutions 4A
Let a =
1
1
, then
2
log 8  log 32 log 2 3  log( 2 5 ) 2
20.

log 4
log 2 2
x
1
  8
2
x
1
1
   
2
 
2
x  3
1
∴ y
3
∴

b
1
3
5
3 log 2  log 2 2

2 log 2
5
3 log 2  log 2
2

2 log 2
11
log 2
 2
2 log 2
11

4
3
8
b  83
1
512
∴ A possible solution is a = 2, b = 512
1
1
or a = , b =
. (or any other reasonable answers)
512
2

16.
2 2  log 10
21. log 5  2 log 2  1  log 5  log
2 log 4  log 0.5
log 4 2  log 0.5

log a
=3
log b
log a = 3 log b
log a = log b3
∴ a = b3
∴ A possible solution is a = 8, b = 2 or a = 27, b = 3.
(or any other reasonable answers)



Level 2

 2 
 2 
17. 3 log 5 + log   = log 53 + log  
 25 
 25 
2 

= log  53  
25 

= log 10
=1
22.
3
 2
3
19. log4 2 – log4 18 + 2 log4   = log4   + log4  
2
 18 
2


2

 1  3 2 
= log4     
 9  2  

1 9
= log4   
9 4
1
= log4  
4
= log4 4–1
= –log4 4
= 1
11



log 8  3 log 5  2 log 8  log 5 3  2 log 10

2 log 5  log 0.25
log 5 2  log 0.25
log 8  log 125  log 100

log 25  log 0.25

 100 10 

18. log (100 10 ) – log (10 10 ) = log 
 10 10 


= log 10
=1
 5  22
log 
 10
42
log
0.5
log 2
log 32
log 2
log 2 5
log 2
5 log 2
1
5
 8  125 
log 

 100 
25
log
0.25
log 10
log 100
1
log 10 2
1
2 log 10
1
2
5
23.
log x 4
4 log x

3 log x 2  4 log x 3( 2 log x)  4 log x
4 log x

6 log x  4 log x
4 log x

(6  4) log x
4

10
2

5
1
(3 log 4 + log 5)
2
1
= (3x  y )
2
=
Exercise 5D (p.229)
Level 1
1.
2x = 10
log 2x = log 10
x log 2 = 1
1
x=
log 2
= 3.32 (cor. to 3 sig. fig.)
2.
32x – 1 = 6
log 32x – 1 = log 6
(2x – 1) log 3 = log 6
log 6
2x – 1 =
log 3
log 6
2x =
+1
log 3
2 log 3 x  log 3 x 3
2 log 3 x  3 log 3 x
24. 4 log x 2  3 log x  4(2 log x)  3 log x
3
3
3
3

(2  3) log 3 x
8 log 3 x  3 log 3 x

5 log 3 x
(8  3) log 3 x
5
5
1

 x3 y 4

log
4
log x 3 y 4  log xy 4
 xy

25.
1
log 3 x
log x 3



log x 2
1
log x
3
2 log x

1
log x
3
6
 log 6 
x = 
 1  2
 log 3 
= 1.32 (cor. to 3 sig. fig.)

26.
2 log x 3  3 log x 4 2(3 log x)  3(4 log x)

log x 2 y  log xy
 x2 y 

log
 xy 
6 log x  12 log x

log x
(6  12) log x

log x
 6
3.
1
log 320 = log 320 2
1
= log 320
2
1
= log (43  5)
2
1
= (log 43 + log 5)
2
41 – 3x = 5
log 41 – 3x = log 5
(1 – 3x) log 4 = log 5
log 5
1 – 3x =
log 4
log 5
3x = 1 –
log 4
 log 5 
  3
x = 1 
 log 4 
= 0.0537 (cor. to 3 sig. fig.)
4.
27. log 200 = log (4  5  10)
= log 4 + log 5 + log 10
= x  y 1
28.
Exponential and Logarithmic Functions
5.
2(3x + 2) = 5
log [2(3x + 2)] = log 5
log 2 + log 3x + 2 = log 5
(x + 2) log 3 = log 5 – log 2
log 5  log 2
x+2=
log 3
log 5  log 2
x=
–2
log 3
= 1.17 (cor. to 3 sig. fig.)
9x + 1 = 8x
log 9x + 1 = log 8x
(x + 1) log 9 = x log 8
x log 9 + log 9 = x log 8
(log 8 – log 9) x = log 9
log 9
x=
log 8  log 9
= 18.7 (cor. to 3 sig. fig.)
12
Certificate Mathematics in Action Full Solutions 4A
6.
7.
8.
9.
42x = 63x – 1
log 42x = log 63x – 1
2x log 4 = (3x – 1) log 6
2x log 4 = 3x log 6 – log 6
(3 log 6 – 2 log 4) x = log 6
log 6
x=
3 log 6  2 log 4
= 0.688 (cor. to 3 sig. fig.)
x2 = 25
x = 5 or
x = –5 (rejected)
∴ x =5
14. log3 (4x + 9) – log3 (3x – 2) = 1
 4x  9 
log3 
 = log3 3
 3x  2 
4x  9
=3
3x  2
4x + 9 = 9x – 6
5x = 15
x =3
log (6x – 4) = 2
log (6x – 4) = log 100
6x – 4 = 100
6x = 104
1
x = 17
3
15. log (x – 1) + log (2x – 1) = 1
log [(x – 1)(2x – 1)] = log 10
log (2x2 – 3x + 1) = log 10
2x2 – 3x + 1 = 10
2x2 – 3x – 9 = 0
(x – 3)(2x + 3) = 0
log3 (4x – 1) = 3
log3 (4x – 1) = 3 log3 3
log3 (4x – 1) = log3 33
4x – 1 = 27
4x = 28
x=7
x=3
or
x=
3
(rejected)
2
∴ x =3
log (3x – 2) = 0
log (3x – 2) = log 1
3x – 2 = 1
3x = 3
x =1
16. Let y = log2 (2x – 1), then the equation
[log2 (2x – 1)]2 – 3 log2 (2x – 1) + 2 = 0 becomes
y2 – 3y + 2 = 0
(y – 1)(y – 2) = 0
y = 1 or y = 2
For y = 1, log2 (2x – 1) = 1
2x – 1 = 2
3
x=
2
For y = 2, log2 (2x – 1) = 2
2x – 1 = 4
5
x=
2
3
5
∴ x=
or
2
2
10. log2 (3x + 1) = 4
log2 (3x + 1) = 4 log2 2
log2 (3x + 1) = log2 24
3x + 1 = 16
3x = 15
x =5
Level 2
11. log (8x – 2) – log 3 = 1
log (8x – 2) = log 10 + log 3
log (8x – 2) = log (10  3)
8x – 2 = 30
8x = 32
x=4
17. Let y = log (x + 1), then the equation
[log (x + 1)]2 – log (x + 1) – 12 = 0 becomes
y2 – y – 12 = 0
(y + 3)(y – 4) = 0
y = –3 or y = 4
For y = –3, log (x + 1) = –3
1
x+1=
1000
999
x=
1000
For y = 4, log (x + 1) = 4
x + 1 = 10 000
x = 9999
999
∴ x=
or 9999
1000
12. log (2x – 3) + log 2 = –1
log [(2x – 3)  2] = –log 10
log (4x – 6) = log 10–1
4x – 6 = 10–1
1
4x =
+6
10
61
4x =
10
61
x=
40
18. Let y = log (x – 1), then the equation
[log (x – 1)]2 – 3 log (x – 1) + 2 = 0 becomes
y2 – 3y + 2 = 0
(y – 1)(y – 2) = 0
y = 1 or y = 2
For y = 1, log (x – 1) = 1
13. log4 (x + 3) + log4 (x – 3) = 2
log4 [(x + 3)(x – 3)] = 2log4 4
log4 (x2 – 9) = log4 42
x2 – 9 = 16
13
5
x – 1 = 10
x = 11
For y = 2, log (x – 1) = 2
x – 1 = 100
x = 101
∴ x = 11 or 101
3
I2
= 10 2
I1
= 31.6 (cor. to 3 sig. fig.)
∴ The sound intensity in restaurant B is 31.6 times to
that in restaurant A.
 (1)
19.  x  2 y  32

 ( 2)
log x  log y  1
From (2),
log x = log y + log 10
log x = log 10y
x = 10y
……(3)
By substituting (3) into (1), we have
10y – 2y = 32
8y = 32
y=4
By substituting y = 4 into (3), we have
x = 10(4) = 40
∴ The solution is x = 40, y = 4.
2.




I
∴ 35 – 67 = 10 log  1
 I0

I
16
= log  1
5
 I2

I
  log 2

I

 0












16

I1
= 10 5
I2
= 0.000 631 (cor. to 3 sig. fig.)
∴ The sound intensity after the party is 0.000 631 times
to that during the party.
3.
Let I1 and I2 be the sound intensities before adjustment
and after adjustment respectively.
Then the sound intensity levels before and after
I 
I 
adjustment are 10 log  1  dB and 10 log  2  dB
 I0 
 I0 
respectively.
∵ The sound intensity level increases by 5 dB after the
adjustment.
I 
I 
∴
5 = 10 log  2  – 10 log  1 
I
 0
 I0 
 I
5 = 10 log 2
  I 0
Level 1
I
1
= log  2
2
 I1
Let I1 and I2 be the sound intensities in restaurants A and
B respectively.
By the definition of sound intensity level, we have
I 
60 = 10 log  1 
 I0 

I
  log 1

I

 0








1
I2
= 10 2
I1
= 3.16 (cor. to 3 sig. fig.)




∴ The sound intensity after adjustment is 3.16 times to
that before adjustment.
I
∴ 75 – 60 = 10 log  2
 I0
 I
15 = 10 log 2
  I 0
I
3
= log  2
2
 I1

I
 – 10 log  2

I

 0
 I
–32 = 10 log 1
  I 0
Exercise 5E (p.236)
I
and 75 = 10 log  2
 I0
Let I1 and I2 be the sound intensities after the party and
during the party respectively.
By the definition of sound intensity level, we have
I 
35 = 10 log  1 
 I0 
I
and 67 = 10 log  2
 I0
 (1)
20.  x  y  1

 ( 2)
log
y

log
x

1

From (2),
log y = log x – log 10
log x = log y + log 10
log x = log 10y
x = 10y
……(3)
By substituting (3) into (1), we have
10y – y = 1
9y = 1
1
y=
9
1
By substituting y =
into (3), we have
9
1
10
x = 10( ) =
9
9
10
1
∴ The solution is x =
,y= .
9
9
1.
Exponential and Logarithmic Functions

I
 – 10 log  1

I

 0

I
  log 1

I

 0








4.
Let E1 and E2 be the relative energies released by the
earthquakes measured 8 and 2 respectively.
By the definition of the Richter scale, we have
8 = log E1 and 2 = log E2
i.e. E1 = 108
and E2 = 102




14
Certificate Mathematics in Action Full Solutions 4A
E1 10 8

E 2 10 2
= 106
∴ The strength of an earthquake measured 8 is 106
times to that measured 2.
∴
5.

= 87.8 (cor. to 3 sig. fig.)
Let E1 and E2 be the relative energies released by the
earthquakes in the Philippines and Indonesia
respectively.
By the definition of the Richter scale, we have:
For Philippines,
For Indonesia,
7.3 = log E1
6.0 = log E2
E1 = 107.3
E2 = 106.0
7. 3
E1 10

∴
E 2 10 6.0
= 101.3
= 20.0 (cor. to 3 sig. fig.)
∴ The sound intensity level of the Hi-Fi is 87.8 dB
after adjustment.
8.
9.
 I 
∴ 30 = 10 log  
 I0 
Let  dB be the sound intensity level when the master of
ceremony speaks.
 2I 
∴  = 10 log  
 I0 





∴ The sound intensity level is 33.0 dB when the master
of ceremony speaks.
Let I be the original sound intensity of the Hi-Fi, then the
sound intensity is 0.6I after adjustment.
∵ The original sound intensity level is 90 dB.
 I 
∴ 90 = 10 log  
 I0 
Let  dB be the sound intensity level of the Hi-Fi after
adjustment.
 0.6 I 

∴  = 10 log 

 I0 
 0.6 I 
 I
 – 10 log 

I
I
 0 
 0
– 90 = 10 log 
Let E be the relative energy released by an earthquake
measured 5.8.
By the definition of the Richter scale, we have
5.8 = log E
E = 105.8
∴ The relative energy released by an earthquake that is
20 times to that measured 5.8
= 20  105.8
∴ The magnitude of the earthquake on the Richter scale
= log (20  105.8)
= 7.10 (cor. to 3 sig. fig.)
10. Let E be the relative energy released by an earthquake
measured 7.2.
By the definition of the Richter scale, we have
7.2 = log E
E = 107.2
∴ The relative energy released by an earthquake that is
1
times to that measured 7.2
8
1
=
 107.2
8
∴ The magnitude of the earthquake on the Richter scale
1
= log (
 107.2)
8
= 6.30 (cor. to 3 sig. fig.)
  2I 
 I 
– 30 = 10 log   log 
 I 0 
  I 0 
– 30 = 10 log 2
= 30 + 10 log 2
= 33.0 (cor. to 3 sig. fig.)
7.




∴ The increase in the sound intensity level is 0.414 dB.
Let I be the sound intensity of the gathering when the
master of ceremony does not speak, then the sound
intensity is 2I when the master of ceremony speaks.
∵ The sound intensity level in the gathering is 30 dB.
 2I 
 I
 – 10 log 

I
I
 0 
 0

 I
 – 10 log 

I

 0
  1.1I 
 I 
  log 
= 10 log

I 
I
 0 
  0 
= 10 log 1.1
= 0.414 (cor. to 3 sig. fig.)
Level 2
– 30 = 10 log 
Let I be the original sound intensity of the noise, then the
sound intensity is 1.1I after adjustment.
The sound intensity levels before and after adjustment are
 I 
 1.1I 
 dB respectively.
10 log   dB and 10 log 

I
 0
 I0 
 1.1I
∴ 10 log 
 I0
∴ The strength of the earthquake in the Philippines is
20.0 times to that in Indonesia.
6.
  0.6I 
 I 
  log 
– 90 = 10 log

I 
  I 0 
 0 
– 90 = 10 log 0.6
= 90 + 10 log 0.6




15
5
Revision Exercise 5 (p.244)
Exponential and Logarithmic Functions
1
x2
6
(b)
x3
Level 1
(x2 ) 6

1
(x3 ) 2
1
1.
1
5
(a) ( x )  ( x )
7
5
x3

7
3
7
5
x
x2
1 3

2
 x3
(b)
1
3
x4
1


(x4 )
1
x
x
2.
(a)  216 
 125 

4
3
7
2


 216  3 

 

 125  


 36 
 
 25 
25

36
1
3
( x 5 ) 3 3 4 x 1 5
 x  3  (x4 ) 3
x3
x
1
(c)
 6  2 
   
 5  

7
6
x6
4
1
 x 1 8  x 3
x
2

3 3
 6   
    
 5   


 27 
(b)  

 343 

1

4
3

2
3
x
1
3
1
x

4
3
50
3
1

50
x3
1
1
1
1
1
2
1 2
4
2
2 1
2
(d) a b  (ab )  ( ab )  a b  a b
1
1
1 1
 a2 b
a
1


 27  3 

 

 343  



3
 3  
    
 7  

 3
  
 7
7

3
1 8
1
3
a
1

1 1
2b 2
1
2
 1
1
 [(ab) 4 ]2
1
 (ab) 2
1 1
 a 2b 2
1 1 1 1
 

2 2b 2 2
 a 1






1
4.
1
a
2 x  3 16
(a)
1
1
 (2 4 ) 3
4
 23
4
∴ x=
3
1
3.
(a)
3
a 2 b 3 c 1 2  ( a 2 b 3 c 1 2 ) 3
2
3
 a b 1c 4
4 x  64
(b)
1
[(2 2 ) x ] 2  2 6
2
a 3 c4

b
1
(2 2 x ) 2  2 6
∴
2 x  26
x=6
3
5
3
5(5x) – 2(5x) =
5
(c) 5x + 1 – 2(5x) =
16
Certificate Mathematics in Action Full Solutions 4A
1
1

(d) log9 9 + log9   = log9  9  
81
81
 


1
= log9  
9
= log9 9–1
= 1
3
5
1
5x =
5
5x = 5–1
x = 1
3(5x) =
(d)
5.
2x + 2 – 2x = 24
22(2x) – 2x = 24
(22 – 1)2x = 24
3  2x = 24
2x = 8
2x = 23
x =3
(a) ∵
(e)
1
= 10–4
10 000
(f)
 1 
 = 4
∴ log 
 10 000 
(b) ∵
0.000 000 1 = 10–7
∴ log 0.000 000 1 = 7
(c) ∵
64 = 82
∴ log8 64 = 2
(d) ∵
1
5
8.

6.
7.
(b) ∵
∴
102x = 20
2x = log 20
x = 0.651 (cor. to 3 sig. fig.)
400
+ log 2
8
= log 50 + log 2
= log (50  2)
= log 100
=2
(a) log 400 – log 8 + log 2 = log
(b) log 125 + log 80 = log (125  80)
= log 10 000
= log 104
=4
log 5
log 5 2
4 log 5

1
log 5
2
4

1
2
8
1
(b)
2x – 1 = 6
log 2x – 1 = log 6
(x – 1) log 2 = log 6
log 6
x–1=
log 2
log 6
x=
+1
log 2
= 3.58 (cor. to 3 sig. fig.)
(c)
32x + 3 = 33
log 32x + 3 = log 33
(2x + 3) log 3 = log 33
log 33
2x + 3 =
log 3
log 33
2x =
–3
log 3
x = 0.0913 (cor. to 3 sig. fig.)
 1 
1

log 5 

2
 5
10x = 45
x = log 45
= 1.65 (cor. to 3 sig. fig.)

5x = 10
log 5x = log 10
x log 5 = 1
1
x=
log 5
= 1.43 (cor. to 3 sig. fig.)
1
2
(a) ∵
∴
log 5 4
log 625
(a)
5
∴
log 256 log 4 4

log 16
log 4 2
4 log 4

2 log 4
2
(d) log (3x + 1) = 2
log (3x + 1) = log 100
3x + 1 = 100
3x = 99
x = 33
(c) log6 108 + log6 2 = log6 (108  2)
= log6 216
= log6 63
=3
(e)
17
x 
log  1 = –1
2 
5
Exponential and Logarithmic Functions
 I
10 log 2
  I 0
x 
1
log  1 = log  
2


 10 
1
x
–1=
10
2
x
= 1.1
2
x = 2.2

I
  log 1

I

 0

 = 8


I
log  2
 I1
 4
=
 5

4
I2
= 10 5
I1
4
I2 = 10 5 I1
(f) log2 (6x – 4) = 5
log2 (6x – 4) = log2 25
6x – 4 = 32
6x = 36
x=6
∴ The percentage increase of the sound intensity:
I I
= 2 1 100%
I1
4
9.
10 5 I 1  I 1
 100%
=
I1
(a) log 144 = log (24  32)
= log 24 + log 32
= 4 log 2 + 2 log 3
= 4x  2 y
(b) log 288 = log 288
= 531% (cor. to 3 sig. fig.)
12. Let E1 and E2 be the relative energies released by the
earthquakes measured 7.5 and 6.5 respectively.
By the definition of the Richter scale, we have
7.5 = log E1 and 6.5 = log E2
i.e. E1 = 107.5
and E2 = 106.5
7.5
E1 10

∴
E 2 10 6.5
= 10
∴ The strength of an earthquake measured 7.5 is 10
times to that measured 6.5.
1
2
1
log (25  32)
2
1
= (log 25 + log 32)
2
1
= [5 log 2 + 2 log 3]
2
1
= (5 x  2 y )
2
=
10. Let I1 and I2 be the sound intensities of groups A and B
respectively.
By the definition of sound intensity level, we have
I 
45 = 10 log  1 
 I0 
I
and 53 = 10 log  2
 I0




I
∴ 53 – 45 = 10 log  2
 I0
 I
8 = 10 log 2
  I 0
I
4
= log  2
5
 I1

I
 – 10 log  1

I

 0

I
  log 1

I

 0












4
I2
= 10 5
I1
= 6.31 (cor. to 3 sig. fig.)
∴ The sound intensity of group B is 6.31 times to that
of group A.
11. Let I1 and I2 be the sound intensities of the cassette player
before and after adjustment.
∵ The sound intensity level increases by 8 dB.
I
∴ 10 log  2
 I0

I
 – 10 log  1

I

 0

=8


13. Let E1 and E2 be the relative energies released by the
earthquakes in Indonesia and Pakistan respectively.
By the definition of the Richter scale, we have:
For Indonesia,
For Pakistan,
7.5 = log E1
4.6 = log E2
E1 = 107.5
E2 = 104.6
7.5
E1 10

∴
E2 10 4.6
= 102.9
= 794 (cor. to 3 sig. fig.)
∴ The strength of the earthquake in Indonesia is 794
times to that in Pakistan.
14. (a) log (4x – 2) – log 2x = y
 4x  2 
log 
=y
 2x 
2x 1
= 10y
x
2x – 1 = x10y
x (2 – 10y) = 1
1
x=
2  10 y
1
1
=
=1
2  10 0 2  1
1
1
When y = 1, x =
=
8
2  101
(b) When y = 0, x =
∴ A possible solution is x = 1, y = 0 or x = 
1
,
8
y = 1. (or any other reasonable answers)
18
Certificate Mathematics in Action Full Solutions 4A
Level 2
(b)
15. (a)
(4 a ) 2
3
a

1
3

1
4
(a ) 2
a
a

a
1
3
a
1
3
3
4
 a a

1
3

a
(b)
3
5
6
p3
5
p2


5
p
3
5

( p3 )
1
17. (a)
3
5

p2
2

3
5

1
2
1
 p 10
1

4
2
y
xy 3 
x
 x3 y
3

4
1
4
1

1
3
2
3
1
2
3
 x4 y4  y2  x4  x3 y
1 3

3 1

2
 x4 4 y4
2
5
2
 x3 y
 xy 4  x 3 y
x
1
1
3
2
3
x y
5
y4

1
1
3
b4

1
3
1
(  )
3

1

1
1
 a 2b 2

1
1
 a 2b 2
1
13 1

2
19
5x + 1 + 5x – 5x – 1 = 29
(5)5x + 5x – (5– 1)5x = 29
5x(5 + 1 – 5– 1) = 29
1
5x(6 – ) = 29
5
29
5x(
) = 29
5
5x = 5
∴
x =1
(b) 3x + 3 – 2(3x + 2) + 3x = 30
(33)3x – 2(32)3x + 3x = 30
3x(27 – 18 + 1) = 30
3x(10) = 30
3x = 3
∴
x =1
1
2
 ( xy )  y 2  x 4  x 3 y
3
1
2
1
p2
16. (a)

 a 6 b12
1
6
p5
p
2
13
5 1

1

1
2
 a 3 2 b12
( p2 )5
p


 a 3 b12  a 2 b 2
1
p

2
1 (  )
3
7

1
 ab 4  a 3 b
1
3
a6
p
2
3
1
2
 ab 4  ( a 3 b 3 ) 1  a 2 b 2
1

a
 ab  [(a b) ] 
1
6

1
3 1
2
3
1
6
a


1
2
3
b
1
2
a
a
1
3
a
ab  ( a b )   
b
3
4



1
2
(c)
1
2
1
2
1
2
1
(  )
2
7
4
22x – 5(2x) + 4 = 0
(2x)2 – 5(2x) + 4 = 0
Let y = 2x, the equation becomes
y2 – 5y + 4 = 0
(y – 1)(y – 4) = 0
y = 1 or y = 4
∴ 2x = 1
or
2x = 4
2x
or
or
2x = 22
x=2
∴
(d)
20
=
x =0
42x – 10(4x) + 16 = 0
(4x)2 – 10(4x) + 16 = 0
Let y = 4x, the equation becomes
y2 – 10y + 16 = 0
(y – 2)(y – 8) = 0
y = 2 or y = 8
∴ 4x = 2
or
4x = 8
22x = 21 or 22x = 23
2x = 1 or 2x = 3
1
3
∴ x=
or x =
2
2
19
5
18. (a) 4 x  2 y  64  (1)
 2 x y
 1  ( 2)
16
From (1),
4x + 2y = 43
∴ x + 2y = 3
From (2),
162x – y = 160
∴ 2x – y = 0
y = 2x
Consider the simultaneous equations:
 x  2 y  3  (3)

 ( 4)
 y  2x
By substituting (4) into (3), we have
x + 2(2x) = 3
5x = 3
3
x=
5
3
By substituting x =
into (4), we have
5
3
6
y = 2( ) =
5
5
3
6
∴ The solution is x = , y = .
5
5
32
log
64
(b) log 32  log 64 
log 128
log 128
1
log
2

log 128
log 2 1
log 2 7
 log 2

7 log 2
1

7

(c)
log 9 1
log 9
 log 9

log 9
 1
(d)
20. (a)
1
19. (a) 

1
log 625 – log 4 = log 625 2 – log 4
2
1
= log
– log 4
625
1
= log 25
4
1
= log
100
= log 10–2
= 2
3 log 3  log 243 log 33  log 243

log 9
log 9
27
log
243

log 9
1
log
9

log 9

(b) 3 x y  9 2 x y  (1)
 3 x y
 9  ( 2)
9
From (1),
3x + y = (32)2x – y
3x + y = 34x – 2y
∴ x + y = 4x – 2y
y=x
From (2),
93x + y = 91
∴ 3x + y = 1
Consider the simultaneous equations:
 (3)
y  x

 ( 4)
3
x

y

1

By substituting (3) into (4), we have
3x + x = 1
4x = 1
1
x=
4
1
1
By substituting x =
into (3), we have y =
4
4
1
1
∴ The solution is x = , y = .
4
4
Exponential and Logarithmic Functions
1.5 log 16  2 log 2 1.5 log 2 4  2 log 2

log 36  2 log 3
log 36  log 32
1.5(4 log 2)  2 log 2

log 36  log 9
(6  2) log 2

36
log
9
4 log 2

log 4
4 log 2

log 2 2
4 log 2

2 log 2
2
log a 4  log b 8 log a 4 b 8

log ab 2
log ab 2

log( ab 2 ) 4
log ab 2
4 log ab 2
log ab 2
4

20
Certificate Mathematics in Action Full Solutions 4A
(b)
(c)
2x  26 = 96
128x = 96
3
x=
4
4 log x 2  2 log x 4 8 log x  8 log x

log x 3
3 log x
(8  8) log x

3 log x
16 log x

3 log x
16

3
log xy  log x y
log 3 xy  log( x  3 y )
1

(c) log3 (4x – 2) – log3 (x + 1) = 1
 4x  2 
log3 
 = log3 3
 x 1 
4x  2
=3
x 1
4x – 2 = 3(x + 1)
4x – 2 = 3x + 3
x =5
1
log( xy ) 2  log xy 2
1
1
log( xy ) 3  log xy 3
1

1
1
log x 2 y 2  log xy 2
1
1
(d) log5 (3x + 3) – log5 (x – 1) = 2 log5 5
1
log x 3 y 3  log xy 3

x y
log 
1
 xy 2


 13 13
x y
log 
1
 xy 3

1
2

log x

1
2

2
1
2
 3x  3 
2
log5 
 = log5 (5 2 )
 x 1 
3x  3
=5
x 1
3x + 3 = 5(x – 1)
3x + 3 = 5x – 5
2x = 8
x=4
1










(e) Let y = log3 (2x + 3), then the equation
[log3 (2x + 3)]2 – 4 log3 (2x + 3) + 3 = 0 becomes
y2 – 4y + 3 = 0
(y – 1)(y – 3) = 0
y = 1 or y = 3
For y = 1, log3 (2x + 3) = 1
2x + 3 = 3
2x = 0
x=0
For y = 3, log3 (2x + 3) = 3
2x + 3 = 27
2x = 24
x = 12
∴ x = 0 or 12
log x 3
1
 log x
 2
2
 log x
3
3

4
 a 2b 

log 
log a b  log ab
 ab 

(d)
log ab 2  log a 2 b 2
 ab 2 
log  2 2 
a b 
2
 (1)
22. (a) log x  2 log y  2

 ( 2)
3x  2 y  2
From (1),
log x + log y2 = log 100
log xy2 = log 100
xy2 = 100
……(3)
From (2),
3x = 2 + 2y
2  2y
x=
……(4)
3
By substituting (4) into (3), we have
 2  2y  2

 y = 100
 3 
2y2 + 2y3 = 300
3
y + y2 – 150 = 0
(y – 5)(y2 + 6y + 30) = 0
log a
1
log
a
log a

log a 1
log a

 log a
 1

21. (a) log (3x – 2) + log 2 = 1
log [(3x – 2)  2] = log 10
(3x – 2)  2 = 10
3x – 2 = 5
3x = 7
7
x=
3
y=5
(b) log 2x + 6 log 2 = log 96
log 2x + log 26 = log 96
log (2x  26) = log 96
or
y=
=
21
 6  6 2  4(1)(30)
2(1)
 6   84
(rejected)
2
5
By substituting y = 5 into (4), we have
2  2(5)
x=
=4
3
∴ The solution is x = 4, y = 5.
(b) Let E be the relative energy released by an
earthquake measured 8.
By the definition of the Richter scale, we have
8 = log E
E = 108
∴ The relative energy released by the earthquake
in City B is half of that measured 8
= 0.5  108
∴ The magnitude of the earthquake in City B on
the Richter scale
= log (0.5  108)
= 7.70 (cor. to 3 sig. fig.)
 (1)
(b) log( 6 x  5 y)  1

 ( 2)
log xy  log 2  1
From (1),
6x – 5y = 10
6x = 10 + 5y
10  5 y
x=
……(3)
6
From (2),
xy
log
= log 10
2
xy
= 10
2
xy = 20
……(4)
By substituting (3) into (4), we have
 10  5 y 

 y = 20
 6 
10y + 5y2 = 120
y2 + 2y – 24 = 0
(y – 4)(y + 6) = 0
y = 4 or y = –6
By substituting y = 4 into (3), we have
10  5(4)
x=
=5
6
By substituting y = –6 into (3), we have
10  5( 6)
10
x=
=
6
3
∴ The solution is x = 5, y = 4 or x = 
Exponential and Logarithmic Functions
Multiple Choice Questions (p.246)
1.
Answer: B
1
( x )3
4
x3

( x 2 )3
1
(x3 ) 4
3

x2
3
x4
3 3

4
 x2
3
 x4
2.
 18 
log 0.18 = log 

 100 
= log 18 – log 100
= log (2  32) – log 100
= log 2 + log 32 – log 100
= log 2 + 2 log 3 – log 100
= a  2b  2
10
, y = –6.
3
23. Let I be the original sound intensity, then the increased
sound intensity is 1.8I.
The change of the corresponding sound intensity level

 1.8I 
 I 
  10 log  dB
= 10 log

I 
I
 0 
 0 

3.
  1.8I 
 I 
  log  dB
= 10 log

I 
I
 0 
  0 
= (10 log 1.8) dB
= 2.55 dB (cor. to 3 sig. fig.)
Answer: C
9(32x) – 10(3x) + 1 = 0
9(3x)2 – 10(3x) + 1 = 0
Let y = 3x, the equation becomes
9y2 – 10y + 1 = 0
(9y – 1)(y – 1) = 0
1
y = or y = 1
9
∴ 3x =
∴ The corresponding sound intensity level is increased
by 2.55 dB.
24. (a) Let E be the relative energy released by an
earthquake measured 2.
By the definition of the Richter scale, we have
2 = log E
E = 102
∴ The relative energy released by the earthquake
in City A is 1.5 times to that measured 2
= 1.5  102
∴ The magnitude of the earthquake in City A on
the Richter scale
= log (1.5  102)
= 2.18 (cor. to 3 sig. fig.)
Answer: C
1
9
or
3x = 3–2 or
∴ x = 2 or
4.
3x = 1
3x = 30
x=0
Answer: A
3 x  y  27  (1)
 2 x y
 1  ( 2)
2
From (1),
3x + y = 33
∴ x+y=3
From (2),
22x – y = 20
∴ 2x – y = 0
22
Certificate Mathematics in Action Full Solutions 4A
Consider the simultaneous equations:
 x  y  3  (3)

2 x  y  0  ( 4)
(3) + (4),
(x + y) + (2x – y) = 3 + 0
3x = 3
x=1
By substituting x =1 into (3), we have
1+y=3
y=2
∴ The solution is x = 1, y = 2.
5.
6.
7.
8.
9.
10. Answer: D
Since the graph cuts the y-axis, it cannot be a logarithmic
function.
∴ A and C cannot be the answer.
Read from the graph, the function passes through the
point (1, 3).
The point (1, 3) satisfies the function y = 3x but does not
satisfy the function y = 10x.
∴ The answer is D.
Answer: C
32x = 23y
log 32x = log 23y
log (32)x = log (23)y
x log 32 = y log 23
x log 9 = y log 8
x log 8

y log 9
log 8
x:y=
log 9
Answer: C
log3 (3x + 12) – log3 (2x + 1) = 1
 3x  12 
log3 
 = log3 3
 2x 1 
3 x  12
=3
2x 1
3x + 12 = 3(2x + 1)
3x + 12 = 6x + 3
3x = 9
x =3
11.
Answer: B
f(5 +2x) f(5 – 2x) = 52(5 +2x) + 152(5 – 2x) + 1
= 510 +4x + 1+ 10 – 4x + 1
= 522
12.
Answer: B
x4 log x = 10
log x4 log x = 1
4 log x log x = 1
1
(log x)2 =
4
1
log x =
or
2
log x = 
1
(rejected)
2
1
x = 10 2
= 10
HKMO (p. 248)
Answer: D
(log x)2 – 2 log x2 + 3 = 0
(log x)2 – 4 log x + 3 = 0
Let y = log x, then the equation (log x)2 – 4 log x + 3 = 0
becomes
y2 – 4y + 3 = 0
(y – 1)(y – 3) = 0
y = 1 or y = 3
For y = 1, log x = 1
x = 10
For y = 3, log x = 3
x = 1000
∴ x = 10 or 1000
Answer: A
Since log 1 x is undefined for x  0, the graph of
2
y  log 1 x does not cut the y-axis.
2
∴ C and D cannot be the answer.
1
1
When x = , y  log 1 = 1.
2
2 2
∴ The graph passes through the point (
Answer: B
When x = 0, y = a0 = 1.
∴ The graph of y = ax passes through the point (0, 1).
∴ III cannot be the answer.
1
, 1).
2
∴ The answer is A.
23
1.
4a = 10
and
25b = 10
a
log 4 = 1
and
log 25b = 1
a log 4 = 1
and b log 25 = 1
1
1
= log 4
and
= log 25
a
b
1 1
∴
+ = log 4 + log 25
a b
= log (4  25)
= log 100
=2
2.
∵ logx t = 6
∴
x6 = t
∵ logy t = 10
∴
y10 = t
∵ logz t = 15
∴
z15 = t
(xyz)60 = x60y60z60
= (x6)10(y10) 6(z15)4
= t10t 6t4
= t20
∴
logxyz t20 = 60
20 logxyz t = 60
logxyz t = 3
∴
d =3
5
3.
Exponential and Logarithmic Functions
S = log144 3 2 + log144 6 3
1
1
= log144 2 3 + log144 3 6
2
1
1
= log144 2 6 + log144 3 6
1
1
= log144 4 +
log144 3
6
6
1
= (log144 4 + log144 3)
6
1
= log144 12
6
1
1
log144 144 2
6
1 1
= 
6 2
1
=
12
=
24