6 Exponential Functions
6 Exponential Functions
4 2  ( 4 )
b
6
2b 6

3
8.
4b 2  6b  4 
9.
x 4
x 4
 3 13
1 3
(2 x )
2 x
Review Exercise 6 (p. 6.5)
1.
(2) 4  (1) 4 (2) 4
 1  16
 16
x  4  ( 3 )
8
1
x

8
1

8x

3
2.
3.
4.
 1
1
    (1)3  
 3
3
1
 1 3
3
1

27
4
 
5
2
3
3 1 5
5 3 5 1 5
10. (2 x y )  (2) x y
 32 x15 y  5
1

4
 
5
1
 2
4
52
52
 2
4
25

16
2

2 0
3
11. ( 2 pq )  8 p q  1 

4
47  44  47  ( 4)
4
1
43
1

64
(23  60 ) 2  (23  1) 2
13. (a)
 ( 23 )  2
5n  5n 1  5n  ( n 1)
 52 n 1
 2 6
1
26
1

64

(b)
3n  3n 1
 3n  ( n 1)  ( 2 n 1)
32 n 1
 3 2

 5   3   5  3 
           
 3   5   3  5 
 (1) 2
1
2
7.
2
4
 3b  4 
  1 
 a 
3 4 b  4( 4 )

a 1( 4 )
b16

81a 4

6.
8q
p3
p3
8q
 3b 3 
 3b 31 
12.  1    1 
a b
 a 
 43
5.
32 x15
y5
2
14. (a)
6n  6 
 
3n  3 
 2n
1
9
n
n2
n 1
n2
2 n 1
(b) 2  4  2  (2 )
a 5  a 3  a 53
 2n  2  22 n  2
 a 2
1
 2
a
 2( n  2 )  ( 2 n  2 )
 23 n
1
NSS Mathematics in Action (2nd Edition) 4B
15. (a)
Full Solutions
Classwork (p. 6.10)
1. (a) 
(c) 
27 n
(33 ) n
 n
n 1
3 9
3  (32 ) n1
n
33n
3  32 n  2
 33nn( 2 n2)

n
1
1
2.
(a)
1 1

3
x2  x3  x2
1
 3 2
1

9
 x6
2
3
(b) ( x y ) 3  x
3
2
3
2
y3
2
4 n3  16 n (2 2 ) n3  (2 4 ) n

8 2 n1
(23 ) 2 n1
(b)
(b) 
(d) 
 x2 y 3
2 2 n6  2 4 n
2 6 n 3
( 2 n  6 )  4 n ( 6 n 3 )
2

(c)
x
 
 y

4
3
x

y
 23

4
3

4
3
4
8
y3

4
x3
Activity
1
Warm-Up Activity (p. 6.6)
1. (a) 5, –5
(b) 3, –3
(c) 4
2.
3.
x5  ( x5 ) 2
(a)
5
 x2
no
1
Activity 6.1 (p. 6.21)
1. y = 2x
–2
–1
x
0.3
0.5
y
(b) ( 4 x ) 2  ( x 4 ) 2
1
0
1
1
2
2
4
3
8
 x2
4
16
1
y=
x
y
(c)
3x
–2
0.1
–1
0.3
0
1
1
3
2
9
3
1
 ( x2) 3
x2
3
27
x

2
3
Classwork (p. 6.12)
1. 2.62
3. 0.432
2.
4.
Classwork (p. 6.24)
1. 
2.
4. 
5.


Quick Practice
Quick Practice 6.1 (p. 6.7)
(a) ∵ 10 4  10 000
∴
2.
Yes, (0, 1)
3.
(a) (i)
(b) (i)
No
increases
(b) ∵
(ii) Yes
(ii) 0
∴
Classwork
10 000  10
(3)3  27
3
 27   3
3
(c) ∵
Classwork (p. 6.8)
1. 2.88
3. 1.04
4
2.
4.
3.22
0.904
∴
2
1
1
  
4
64
 
3
1
1

64 4
0.0922
2.95
3.

6 Exponential Functions
Quick Practice 6.2 (p. 6.11)
(a)
1
3
(b)
64  64
x

3
2

2
3
27
8

3
4

3
(x 2 )
 27 
 
 8 

2
3
4
 3 3 
x    
 2  
(b) 8 3  (3 8 ) 4
 24
 16
(c)
 1 
 
 36 

3
2
 36
3
2
 ( 36 )3
 63
3
 
2
2
2
 
3
4

9
2

2
3
 216
Quick Practice 6.5 (p. 6.17)
Quick Practice 6.3 (p. 6.13)
a
(a)
1
4
a
3

a
a
a
(32 ) x  3 3
3
4

32 x  3
1 3

2 4

1
27
9x 
1
2
3
2
3
2
3
x
4
2x  
1
4
a
1
 1
∴
a4
Quick Practice 6.6 (p. 6.17)
2 x  2 x1  48
1 1
(b)
3
a  a  (a  a 2 ) 3
1
 (a
a
a
2(2 x 1 )  2 x1  48
1 1
2 3
)
2 x1 (2  1)  48
3 1

2 3
3(2 x1 )  48
1
2
2 x1  16
1
3
2 x1  2 4
x 1  4
x5
2 1
(c)
2
3
16ab  (a b)

3
2

(42 ab 3 ) 2
1
3
( a b)
∴
3
2
Quick Practice 6.7 (p. 6.20)
1
1


1
4a 2 b 3
1

2
a b
(a)
3

2
1  1 1  3
  
  
 2 3  2
 4a 2
 4ab
8
b
11
6
1.8
(b)
Quick Practice 6.4 (p. 6.13)
4
(a)
1
f (1)  p(1)     5(0.61 )
5
53
1
f (1.8)  p(1.8)     [5(0.61.8 )]
5
 0.0277 (cor. to 3 sig. fig.)
x 3  16
4 3
Quick Practice 6.8 (p. 6.21)
(a) The required number of words that Jason can remember
 50(1  0.95 )
3
( x 3 ) 4  16 4
3
x  (24 ) 4
 20 (cor. to the nearest integer)
 23
8
3
NSS Mathematics in Action (2nd Edition) 4B
Full Solutions
(b) Number of words that Jason can remember after studying
for 6 minutes
 50(1  0.96 )
The required percentage increase
50(1  0.96 )  50(1  0.95 )

 100%
50(1  0.95 )
 14.419%
(b) 4 x  3 8 x 
(2 2 ) x
x
( 23 ) 3
22 x
2x
 22 x  x

 2x
 12%
∴ The percentage increase in the number of words
that Jason can remember in the 6th minute exceeds
12%.
3.
(a)
Quick Practice 6.9 (p. 6.26)
1

x
 x
 1
x  2
x
2




1
1 1
2 2
 (x
)
1 1
2 2
 (x )
1
 x4
3
1
1
1 3

2
1
 x2
 x2 y
3
(a) (i)
( 0.5 )3  0.5 2

From the graph, when x 
∴
3
, y  0 .4 .
2
(c)
1

2
From the graph, when x  
∴




x

y 
1
1
 3 64 x 3 y 

y
 (4 xy 3 )
x

1 11 1 3
x y
4

y3
1
2  1 .4
2
5  (5 2 ) x  1
165  ( 4 16 )5
5 2 x 1  50
 25
 32
2x  1  0
∴
3
x
(216) 2  (3  216 ) 2
 (6) 2
 36
1
2
16 x  2  4 x 1
(b)
( 2 4 ) x  2  ( 2 2 ) x1
2 4 x  21 2 ( x1)
2.
(a)
4x 2
Further Practice (p. 6.17)
1. (a)
5  25 x  1
Further Practice (p. 6.14)
(b)
y
 (4 3 x 3 y ) 3
x
1
1
, y  1.4 .
2
Further Practice
4
5
2
5
∴ The solution of 0.5 x  6 is x  2.6 .
(a)
x
6
x
3 6
1
2 x
2 4 x  2 2 x 3
( 3 ) x  27  (3 )  (3 )
x
4x  2x  3
x
 32  32
3
3
3
x2
(b) From the graph, when y  6 , x  2.6 .
1.
y2
y2
( 0.5 )3  0.4
2  0.51  0.5
(ii)
3
xy  x 2 y 3  x 2 y 2  x 2 y 3
(b)
2x  3
x x

2 2
∴
x
4
x
3
2
6 Exponential Functions
2.
(a)
Exercise
2 x  y  23  (1)
 x 1
y
 (2)
3  3
Exercise 6A (p. 6.14)
Level 1
From (1),
2 x y  23
1
1.
∴ x  y  3(3)
From (2),
3x1  3 y
7
x3  (x3 ) 7
3
 x7
∴ x  1  y (4)
(3) – (4): y  1  3  y
2y  4
1
2.
(3 x )  8  ( x 3 )  8
x
y2
By substituting y  2 into (3), we have
x23
1
3.
x
x 1

2 x  y  1 (1)
(b)  x  y
3  9 (2)
From (1),
2 x y  1
1
( x3 ) 2
1
3
x2
x
2 x y  20
∴ x y 0
From (2),
3x y  9
8
3
1

3


3
2
4.
3
10  2.15 (cor. to 3 sig. fig.)
5.
5

6.
23 4  2.19 (cor. to 3 sig. fig.)
7.
1.75
8.
∵
23  8
∴
3
∵
27
3
  
5
125
 
∴
3
(3)
4
  0.894 (cor. to 3 sig. fig.)
7
3x y  32
1
(4)
∴ x y 2
(3) + (4): 2 x  2
x 1
By substituting x  1 into (3), we have
1 y  0
y 1
Further Practice (p. 6.27)
1. (a) G1 : y  0.5 x , G2 : y  0.2 x , G3 : y  2 x , G4 : y  1.2 x
(b) (0, 1)
(c) G1 : y  0.5 x and G3 : y  2 x
2.
(a) (i)
(ii)

2
3
 0.689 (cor. to 3 sig. fig.)
8 2
3
9.
From the graph, when t  2 , H  2.0 .
27
3

125 5
2  Q2
Q 2
(4)3  64
10. ∵
(3  64 ) 2  (4) 2
∴
 16
(b) Number of households with a computer at the
beginning of 2017
11.
 ( 2 )7 thousand
93  ( 9 ) 3
 33
 27
 11.31 thousand
 10 thousand
∴
There will be more than 10 thousand
1
12. 32 5  5 32
2
households with a computer at the beginning
of 2017.
5
NSS Mathematics in Action (2nd Edition) 4B
13. 216

1
3
Full Solutions
1
 (3 216 ) 1
21.
8
3
a  4 a3  a 8  a 4
 61
1

6
1 3

4
 a8
7
 a8
3
1
14. 16 4  ( 4 16 )3
2
8
 9 
15.  
 49 
3
2

3
22.
3
3
a
a3
 a3 
a2
27
3
1 3

a3 2

3
3
 49  2
 
 9 

 49 


 9 


7
 
3
343

27
3
4
16. (a ) 2  a
4
3
3a 6
3
1 1
23.
4
a  3 a  (a  a 3 ) 4
1
 (a

2
3
1
 a3

2
3
24.
a2
1
3
3 4 2
a
 2
 6   
 3
a2

(a b )
a
1 3

3 2
1
2
6 4   2 
a
2
 3a
19. a 0  a

3
4
1
5
 a 6  1 a
a

3

3
4
5
 a6
1
 3
 a 4  a2
25. 
6
 b

3 5
 
4 6

19
12
3
a

a
a
a
2
3  31

a 4 2
  6

 b


 1
a 4
 6
 b

1
19
a 12
a
1
a2
 6
b
7
4
a
20.
3
2
a 2
 6
b
5  1

4
a 2 b6
2

b
a2

 a4
5
)
 (a )
3
2
 (a  6 )
18. 6a 4  2a
1 1
3 4
4 1
3 4
 a6
 1 
17.  6 
a 
7
a 6

3
1
 7

1
2
a
1 2
 
4 3
b
1
3
6

2
3
1
a 6
 4
b
1
 1
1 1

2 3
1
 a6
a 6 b4
6
2
3




2
3




6 Exponential Functions
26.
3
a 2b 6  a

4
3
1
 ( a 2b 6 ) 3  a
a
2
1
1
6
3
3
b
2 4

3 3
a
b

4
3
a

9
 3 36
32. 
 36

4
3
  36 
  3

  6 
 

1
 (6 3 ) 9
2
 63
 216
2

3
 a b2
b2

2
33.
a3
3
q 1
q4 
q
4
 q3 
3
4
q 1
3
q4
4
1
 ( 1) 
 q3
x4  3
27.
9
1
q
( x 4 ) 4  34
x  81

3
4
5
12
1

5
q 12
2
x5 
28.
1
9
4
1
(x )   
9
2 5
5 2
5
2
34.
4
p5
 p 
2
3
(5 p ) 2
2
5
29.
x
5

3
5
3


3
5
(x )
4 2 2
 
5 3
 p5

4
 p 15
1
 4
5
p 15

1
4 5
 a
35.  1
 2
a
 32
 32
 p3
p5
 1  2  2
x    
 3  
1
 
3
1

243
2
p5
3
5
5 1

2
1
a4 2
 4

a

a4

1 1
 

a 2 2

5

3

5 5
x  (2 )
a
 23
1
a8
1

4
a4
5  1 1
   
 4 4
 a8
1

8
5
 a8
3
Level 2
3
30. 3 2  27

1
6
3
 3 2  (33 )
3

 32  3

1
6
50 s 
5
36.
s5
3
5
2
 50 s 
2s
1
2
1
2s 2

3 1

2
s5
50
5 3 1
 
5 2
s2
2
 32
13

3
50 5
s
2
13
31. 8
5

3
5

3 3
 43  (2 )
 5s 5
 ( 4 )3
5
 2  23
2
37.
 5  ( 3)
4
1
1
3
1
27 a 2  9a 3  (27 a 2  3a 2 ) 4
1 3 1

2 4
 2 2
 (27  3  a 2
1
2 4
1

4
 (34 a )
1
 3a 2
7
)
NSS Mathematics in Action (2nd Edition) 4B
2
3
38. (a
2
3
Full Solutions
1
2
43.
ab  2 ) 4  (a a b 1 ) 4
2
3
1
4
2
3
m n  ( 2m n)
2
7
 27 

 

 m
 27  3

 m n  (2m n)  

 m
14
 a 3 b4
1
4
2
3
1
4
2
3
2
3
2
1
14
4
3
2
2
p
4
 ( pq )  p 3 q
2
3
1

4
p q
 4  3 27  m 4
2
3
17
3
7
3
2
3
p q

p
q
q
2
n3
2
8
 12m 12 n 3
q
p
1
m6
1 4 1
 
3 6
39.
27 3
 m n 2 m n 
a3
 4
b
2
3
1
3
1
 (a 6 b 1 ) 4
3

1 2
 
4 3
11

12
44. (a)
7
3
6
 1
 b   b2 

  
3 b   1 
3


b 
6
1 1

3 6
 (b 2
11
12
)
1
6 6
 (b )
b
2
40.
a 4b 1
2
1
1
 3 a2 
  (a 4b 1 ) 2  (a 3 b  2 ) 2

 b 


 a 2b
a
2

4
3
2
3
a b
1
1
(b) (i)
4
1
2
 a 3 b 1
b
1
  ( 1)
2
x  0 and
∵
x x
1
1
1
b
x
3
42.
( m)
3
 (m
n)

8n
m
3
2
2n

m
1
3
1

6
m n
1  1
  
2  2
∵
3
1
∴
n

y 0
y
3
y
6
 y

16  
3 y 


y  1 (By (a))
2
4
3
y is undefined.
y y
3  1
    1  1
2  6
4 3
m3n

2
7
12
∴
1
1
and
.
3
2
(or any other reasonable answers)
4

1
x
y  0 and
∴
1

2
x 0
3
(ii) Unless y  0 , otherwise
 a 2b
1
3
x is undefined.
 x

  16
3 x 


∴
x  1 (By (a))
∴ Two possible values of x are 2 and 3.
(or any other reasonable answers)
1 1
1 1
 1 1 
2
2 2
3
3
6
1
 a2
∴
∴
1
2

a
 ab 2  a 2 b  a 2 b 2  ab 2
b
41. a 2 b 
Unless x  0 , otherwise
∴ Two possible values of y are
m3
7
2n 12
1
45.
3
5 x  25 x  (5 x ) 3  (52 ) x
x
 5 3  52 x
x
 53
2x
7x
53
8
6 Exponential Functions
46.
3
2x ( 4x ) 2x ( 4 )x

81 x
( 23 )1 x
50.
64 x 2  27
3
x2 
2 x (2 x )
 33 x
2
 2 x  x  (33 x )
27
64
3
x2   
4
3
 25 x  3
3
2
 3 3  3
( x )    
 4  
3 2
2 3
2

 (33 x ) 3
  2( x  2)  4 x
3
3

2
x 3
47.
(27 )
9 x  2  81 x
1
2




3
x 
4
9

16
1
 (32 x  2 ( x  2 )  ( 4 x ) ) 2
2
1
 (34 x  4 ) 2
 32 x  2
51. 80( x  1)
( 4)  a
48. (a)
4
3
5  0
( x  1)
k
3

k

43  a
4
[( x  1) 3 ]

4
3

3
4
k
( 4 3 )3  a 3
4 a
k
∴

5
80
 1 
 
 16 
x  1  (2 4 )
3

3
4

3
4
x  1  23
x9
k
43  a
(b)
k
(4 3 ) 2  a 2
52.
k
2 3
(4 )  a 2
2
x2
3
x2 
k
3
∴ 16  a 2
3
49. (a)
2
x3 
72m  b
3
m
1
x 
4
1

64
49 3  b
m
(49 3 )3  b3
49m  b3
7
(7
2m
3
2m
3
b
3
2
) b
7 b
m
a p  aq
3
2
a2
3
2
3
2
a
9
2 p 
2 p
 b4
9
∴
q
2
ap a
3
2
( 7 )  (b )
3m
2
3
a2
53. (a)
3
m 2
7
1
16
 1  2  2
( x )    
 4  
m
(b)
2
32
2 3
3 2
(7 2 ) 3  b
∴
 32
3
73m  b 4
∴
(b) ∵
∴
9
q
2
1
 a0
 a0
q
0
2
q
 2 p
2
q  42p
p and q are non-negative integers.
The possible values for p and q are
‘ p  0, q  4 ’ or ‘ p  1, q  2 ’ or
‘ p  2, q  0 ’. (any two pairs)
NSS Mathematics in Action (2nd Edition) 4B
Full Solutions
Exercise 6B (p. 6.18)
Level 1
3x  6  81x
8.
3x  6  (34 ) x
x
2
3 9
1.
3 x  6  34 x
x  6  4x
x
3 2  32
∴
x 1
1
25 x  2   
5
(52 ) x  2  (51 ) x 1
9.
1
125
1
x
5  3
5
5 x  5 3
5x 
2.
∴
52 x  4  5 x 1
2x  4  x  1
3x  5
3
10.
1
2 3
27 x  9 x  2
(33 x ) 3  (32 ) x  2
2
3x  32 x  4
7 x 1  7 3
2
x 1 
3
1
x
∴
3
x  2x  4
∴
x4
3 x 1  3 x  12
11.
3(3 x )  3 x  12
4 8
3x
( 2 2 ) 3 x  23
3 x (3  1)  12
2 6 x  23
4(3 x )  12
6x  3
x
∴
3x  3
1
2
∴
1
216
1
(6 2 ) x  2  3
6
6 2 x  4  6 3
2 x  4  3
5(52 x 1 )  52 x 1  4
36 x 2 
52 x 1 (5  1)  4
4(52 x 1 )  4
52 x 1  1
1
2
x
∴
x 1
52 x  52 x 1  4
12.
5.
5
3
1
7 x 1  (7 )
4.
x
∴
x  3
7 x 1  3 49
3.
x2
∴
x
2
2
x4
∴
52 x 1  50
2x  1  0
1
x
2
x
16 4  32
6.
x
4 4
2 x 1  3( 2 x 1 )  8
13.
1
5 2
(2 )  (2 )
2x  2
2 2 ( 2 x 1 )  3( 2 x 1 )  8
2 x 1 ( 4  3)  2 3
5
2
2 x 1  2 3
5
x
2
∴
x 1  3
∴
7.
4( 4 )  4
x
2 x 1
4 x 1  4 2 x 1
x 1  2x 1
x2
∴
10
x4
6 Exponential Functions
3(4 x 1 )  2(4 x 1 )  35
14.
7 2 x  49 x 1  50
19.
3(4 x 1 )  2  4 2 (4 x 1 )  35
4
x 1
7 2 x  (7 2 ) x 1  50
(3  32)  35
7 2 x  7 2 x  2  50
35(4 x 1 )  35
7 2 (7 2 x  2 )  7 2 x  2  50
7 2 x  2 (7 2  1)  50
4 x 1  1
50(7 2 x  2 )  50
4 x 1  40
x 1  0
72 x2  1
x 1
∴
72 x  2  70
Level 2
53 x  125(25 x )
15.
∴
53 x  53 (52 x )
x
4( 2 x )  4 2 
20.
2 x 3
5 5
3x  2 x  3
3x
x
3
4
3
4( 2 x )  2 x 
4
3
x
2 ( 4  1) 
4
3
x
3( 2 ) 
4
1
x
2 
4
2 x  2 2
82 x
4
3 2x
(
2
)
2 x 1 
22
2 x 1  26 x  2
2 x 1 
x  1  6x  2
5x  1
1
x
∴
5
∴
2(4 2 x )  4 1 (4 2 x )  4 2 x  11
32 x (33 x 1 )  (33 ) 2 x 1
1 

4 2 x  2   1  11
4 

11 2 x
(4 )  11
4
42 x  4
2x  1
32 x  (3 x 1)  36 x 3
35 x 1  36 x 3
5x  1  6 x  3
x4
∴
18.
162 x
8 x 1
(24 ) 2 x
 3 x 1
(2 )
4x 3 
(22 ) x  3
x2
2(4 2 x )  4 2 x 1  16 x  11
21.
9 x (33 x 1 )  27 2 x 1
17.
3
4
4( 2 x )  ( 2 2 ) 2 
∴ x3
16.
2x  2  0
x 1
∴
1
2
1
8
1
(6 2 ) x 1  6 2 x  8(6 2 x ) 
8
1
2x2
2x
2x
6
 6  8(6 ) 
8
1
2
2x
2x
2x
6 (6 )  6  8(6 ) 
8
1
2x
2
6 (6  1  8) 
8
36 x 1  6 2 x  8(36 x ) 
22.
22 x  6  28 x  (3 x  3)
x
2 2 x  6  25 x  3
2 x  6  5x  3
3x  3
∴
x 1
1
216
62 x  63
2 x  3
62 x 
∴
11
x
3
2
NSS Mathematics in Action (2nd Edition) 4B
Full Solutions
 (1)
23. 23 x  4 y 1

y 2
 x2  1 
 (2)
5   5 
 

From (1),
23 x  4 y 1
3x  9 y
26.
3 x  32 y
∴
23 x  ( 2 2 ) y 1
4 x  8z
2 y 2
2 2
3x  2 y  2
∴
From (2),
3x
x  2y
x 2

y 1
2 2 x  23 z
 (3)
∴
y 2
1
5 x 2   
5
x 2
5  5 ( y  2 )
∴
∴
x  2  y  2
∴
x  4  y  (4)
∴
By substituting (4) into (3), we have
3( 4  y )  2 y  2
2 x  3z
x 3

z 2
x : y  2 :1
x :z 3 :2
x : y : z  6:3: 4
( x  y) : ( y  z )  (6  3) : (3  4)
 9:7
12  3 y  2 y  2
27. ∵
∴
∴
10  5 y
y2
By substituting y  2 into (4), we have
x  42
4 2  2 4 and 4 4  4 4
x  2 and x  4 satisfy the equation.
Peter’s claim is agreed.
Exercise 6C (p. 6.27)
Level 1
2
1. (a) f (2)  2(4 )
2
 32
24. 3  9  0  (1)
 x
2 y 1
 (2)
36  6
x
y
(b)
From (1),
3x  9 y  0
f ( 1)  2(4 1 )
1

2
3x  9 y
1
3 x  32 y
∴
x  2y
From (2),
36 x  6 2 y 1
(c)
 (3)
6 2 x  6 2 y 1
∴ 2 x  2 y  1  (4)
(4) – (3):
2 x  x  (2 y  1)  2 y
x 1
(d)
2.
(a)
1
4 f    4  2(4 2 )
2
 16
f (0) 2(4 0 )

2
2
1
f (1.2)  g (0)  51.2  0.9 0
 6.90 (cor. to 3 sig. fig.)
By substituting x = 1 into (3), we have
1  2y
2
1
y
2
(b)
25. 10 a  100 b  1000 c
10 a  (10 2 ) b  (10 3 ) c
(c)
10 a  10 2b  10 3c
2
2 2
f    g    5 3  0.9 3
3 3
 3.86 (cor. to 3 sig. fig.)
f (2)
5 2

3g (2) 3(0.9 2 )
 0.0165 (cor. to 3 sig. fig.)
a  2b  3c
∴
a 2
b 3


and
b 1
c 2
a : b  2 :1
∴
b : c  3: 2
a : b : c  6 : 3: 2
∴
(d)
2 f (3)  3g (2)  2(5 3 )  3(0.9 2 )
  2.41 (cor. to 3 sig. fig.)
12
6 Exponential Functions
f (2)  5
3.
9.
2
a(1.4 )  5
5
1.4  2
 5 1.42
 9.8
a
4.
(a)
f (3)  10
1.25(b 3 )  10
10
1.25
b3  8
b3 
10.
b3 8
2
(b)
5.
6.
f (5)  2 f (4)  1.25(25 )  2  1.25(2 4 )
 40  40
0
500
Price of each hamburger  $150(0.995 )
 $12.2 (cor. to the nearest $0.1)
11. (a) (0, 1)
(a) Number of students absent on the 3rd day
 50(1  0.73 )
(b)
 33 (cor. to the nearest integer)
(b) Increase in the number of students absent
 50(1  0.7 4 )  50(1  0.73 )
 5 (cor. to the nearest integer)
7.
(a) Original number of rabbits in the pet shop
 50(1.120 )
 50
12.
(b) Number of rabbits in the shop after 8 weeks
 50(1.128 )
 123.798
 100
∴ There will be more than 100 rabbits in the
shop after 8 weeks.
8.
(a) Weight of the substance after 200 days of decay
200
 400(0.5 25 ) mg
 400(0.58 ) mg
 1.5625 mg
(a) (i)
t
25
(b) Since W  400(0.5 )  0 for all real values of t,
the weight of the radioactive substance will not
reduce to 0 mg.
From the graph, when x  1.4, y  7.0.
∴
1
 
4
1.4
 7.0
1.8
1
(ii) 0.251.8   
4
From the graph, when x  1.8, y  12.0.
∴ 0.251.8  12.0
13
NSS Mathematics in Action (2nd Edition) 4B
Full Solutions
2.2
Level 2
1
(iii) 42.2   
4
From the graph, when x  .2, y  21.0.
∴ 42.2  21.0
f (3)  g (3) 
14. (a) (i)
33
 3
2
(b) From the graph, when y  2 , x  0.5 .
30.5
f (0.5)
 8
g (0.5)  4  0.5
 
9
30.5 40.5

 0.5
8
9
(ii)

3
9

8
4

3 3

8 2

∴
2.5
1
1.5
2
5
      2.51.5
5
2
From the graph, when x  1.5 , y  0.2 .
1.5
∴
2
 
5
 0.2
1
(iii)
3
125  5   2
   
8
 2  
5
 
2
∴
15.
2
1
4
 6

f (5)  8 f (2)
a5
8
a2
a3  8
3
, y  4.0 .
2
( a  0)
a3 8
2
125
 4 .0
8
16. (a)
f ( a )  20
5(8 a )  20
(b) From the graph, when y  9 , x  2.4 .
∴
2
a 5  8a 2
3
2
From the graph, when x 
kf ( 1)  g (1)  2
 31   4 
 
k 

 8  9
k
9

3 8 4
k
24
k
 1.6
1.5
(ii)
(b)
From the graph, when x  0.5 , y  1.6 .
3 3
16
8a  4
The solution of 2.5x  9 is x  2.4 .
( 23 ) a  2 2
23a  2 2
3a  2
a
14
2
3
3
 2  2 
   
 3  
33 2 6

2 3 36
8

27
13.
0.5
4
 
9

x
1
∴ The solution of    2 is x  0.5 .
4
(a) (i)
33
8
3
6 Exponential Functions
(b)
f (k  1)  f (k )  70
5(8
k 1
(b) Number of bacteria in the test tube after 1 day
 1000(1.2224 )
)  5(8 )  70
k
 6 320 000 (cor. to 3 sig. fig.)
5  8(8 )  5(8k )  70
k
8k (40  5)  70
21. (a) Number of cakes sold per day with $5500 spent on
advertising
 500(1.03)5.5
35(8k )  70
8k  2
 588 (cor. to the nearest integer)
23k  2
3k  1
k
(b) Number of cakes sold per day with $10 000 spent on
advertising
 500(1.03)10
Total profit  $50 × 500(1.03)10
 $33 597.9
> $10 000
∴ The spending is wise since the profit is greater
than the advertising expense.
1
3
17. (a) When t  0, L = 4.
4  9  A(0.37) 0
4  9  A(1)
A5
22. (a)
(b) Blood glucose level 1.5 hours after the injection
 [9  5(0.37)1.5 ]units
 7.87 units
 7 units
∴ The patient’s blood glucose will be restored to
the normal level 1.5 hours after the injection.
18. (a) ∵ Jennifer receives $1081.6 after 2 years.
∴
1081.6  1000(1  r %) 2
(1  r %) 2  1.0816
1  r %  1.04
2.1
2.1
r4
(i)
(b) Interest she will receive in the 3rd year
 $[1000(1  4%)3  1081.6]
7
3
   
3
7
From the graph, when x  2.1 , y  6.0 .
∴
 $43 (cor. to the nearest dollar)
7
 
3
2.1
 6.0
1
19. (a) V  400 000(1  10%)
t
(ii)
 400 000(0.9)t
3
 
7
(b) Value of the car after 5 years  $400 000(0.9) 5
The required percentage decrease
400 000  400 000(0.9) 5

 100%
400 000
∴
(b)
20. (a) When t = 0, N = 1000.
1000  ab2( 0)
a  1000
When t = 1, N = 1440.
1440  ab 2 (1)
1440  1000b 2
b 2  1.44
∴

3
2
From the graph, when x  
 40.951%
 40%
∴ The percentage decrease in the value of the
car after 5 years will be more than 40%.
∴
3
343  7   2
   
27  3  
b  1 .2
15
343
 3.5
27
3
, y  3 .5 .
2
NSS Mathematics in Action (2nd Edition) 4B
Full Solutions
(ii) From the graph, when N  1600 , t  3.5.
∴ Time required  3.5 hours
x
(i)
3
8   36  0
7
x
3
8   36
7
∴ N  2000  1800(0.65) t  2000
∴ The number of bacteria in the beaker will not
exceed 2000.
∴ Peter’s claim is disagreed.
x
9
3
  
2
7
From the graph, when y =
9
, x = –1.8.
2
25. (a) (i)
From the graph,
when t = 0, V  4000 ,
when t = 5, V  3100 .
(ii) When t = 0, V  4000 .
x
3
∴ The solution of 8   36  0 is x = –1.8.
7
x
7
   0 .1
3
(ii)
1800(0.65) t  0 for all real values of t.
(b) ∵
4000  a(0.95)0
1
a  4000
 7  x 
    0.11
 3  
(b) The value of the mobile phone after 1 year
 $4000(0.95)12
x
3
   10
7
From the graph, when y  10 , x  2.7 .
 $2160 (cor. to 3 sig. fig.)
x
From the graph, when n  0 , V  10 .
∴ The value of the savings fund at the
beginning of 2014 is 10 thousand dollars.
(ii) From the graph, when n  1 , V  10.5 .
∴ The value of the savings fund at the
beginning of 2015 is 10.5 thousand dollars.
7
∴ The solution of    0.1 is x  2.7 .
3
23. (a) (i)
x
–4
–3
–2
–1
0
1
2
y
7.7
4.6
2.8
1.7
1
0.6
0.4
26. (a) (i)
(b) When n  0 , V  10 .
10  Pr 0
(ii)
P  10
When n  1 , V  10.5 .
10.5  10r1
r  1.05
(b)(i)
(c) The value of the savings fund at the beginning of
2024
 10(1.0510 )
∴ Percentage change in the value of the savings
fund
10(1.0510 )  10

 100%
10
  62.9% (cor. to 3 sig. fig.)
x
(b) (i)
3
  4
5
0 .6 x  4
From the graph, when y  4, x  2.7 .
Check Yourself (p. 6.35)
1. (a) 
(b) 
(c) 
(d) 
(e) 
x
∴
3
The solution of    4 is x = –2.7.
5
(ii) 3x  2(5 x )  0
3x  2(5 x )
2.
(a) (0, 1)
(c) 0  b  1
3.
(a) ∵
x
3
   2
5
0.6 x  2
24. (a) (i)
(2)3  8
3
∴
∵ The graph of y  0.6 x  0 for all real
values of x.
∴ 3x  2(5x )  0 has no real solutions.
8   2
4
(b)
27 3  (3 27 ) 4
 34
 81
From the graph, when t = 0, N = 200.
∴ Initial number of bacteria  200
16
(b) x
(d) y
6 Exponential Functions
(c)
9
 
4

1
2
 9


 4


3
 
2
2

3
4
4.
(a)
3
1
2.
∵
(2)5  32
∴
5
 32   2
1
3.
1
1
 4 4 3
813
(3 )
4
1
 1 4
  12 
3 
1
 3
3
1

27
2 4

3
b2  b 3  b 3
 b2
2
(b)
3
8a 2  3 8 a 3
 2a
(c)
 a3 


 b



1
3
3
2
3
4.
3
36 2  (62 ) 2
 63
 216
a

 1
3   
 3
5.
b
a 1

b

2
2
1  1
  
2  3
3 3
 1  3  1  
     
 27 
 3  
1
 
3
1

9
1
6
1
b6

a
5.
3
4
6.
x  64
(a)
3 4
4 3

0.0001
1
4
2
 (10 4 )

1
4
 101
4
6 3
 10
( x )  (2 )
x2
 256
8
1 1
7.
1
(a  a 2 ) 4  (a
 (a )
1
5
2 x
(5 )  5 1
3
 a8
5 2 x  5 1
2 x  1
∴
x
)
3 1
2 4
25 x 
(b)
1 1
2 4
3
8.
1
2
a3
3

a4
a2
4
a3
3 4

3
 a2
7 x 1  7 x  42
(c)
1
 a6
7(7 x )  7 x  42
7 x (7  1)  42
9.
6(7 x )  42
7x  7
∴
3
3
5
2
 1

3
 a5


5
  (3a )   3




2
x 1

3
5
a5
 (3a )  2
3
Revision Exercise 6 (p. 6.36)
Level 1
1. ∵ 103  1000
∴

 3
(3a )   1
 5
a

3 2

3 55
a
32

1 5
a
3
1
1
1000  10

1
3a 5
17





2
NSS Mathematics in Action (2nd Edition) 4B
a  b 1
10.
b  a 2
Full Solutions
1
3
32 x  31
1

a 2 b 1
9x 
16.
1
b 2 a 2
1
 a2
( 2 )
5
 a 2b

1
b
2 x  1
1
∴ x
2
1
2
3
2
5

a2
b
16 x  32
17.
3
2
2 4 x  25
4x  5
1
3

1 6
4
11.
(a b )
a
3
4
1

8
3
1 
a b

a
 a4 b
1

4
∴
1
8
x
5 2  (52 ) x1
x
5 2  5 2 x 2
x
 2x  2
2
3x
2
2
4
∴ x
3
a 4b8
3

(8a) 6 b  3  (8a ) 6 b  3

1
3
 (8a) 2 b 1
 64a 2b 1
64a 2
b

2
x 3  49
2
22 x
3
2 x 1
8
2 x 1
 3
2
 2( x 1)  3
4x 
19.
13.
5
4
5 x  25 x1
18.
1

8
a b
1
 1 1
12.
x
3
22 x
( x 3 ) 2  (7 2 ) 2
22 x  2 x  4
x  73
 343
∴
2x  x  4
x4
3
4
x
2
4
14.
3
5
3
5(5 x )  2(5 x ) 
5
3
5 x (5  2) 
5
3
x
3(5 ) 
5
1
x
5 
5
5 x  51
5 x 1  2(5 x ) 
20.
3
x4  8
4
3
4
( x 4 ) 3  ( 23 ) 3
x  24
 16
15. 2 x

3
2
 54  0
x

3
2
2
3


3
2
(x )
 27
 (3 )
3

∴
2
3
x  3 2

1
9
18
x  1
6 Exponential Functions
(2 x 1 ) 2  2 2 x  5
21.
1
(b)
22 x  2  22 x  5
2
2x2
 5(3 1000 )  5(3  1000 )
 5(10)  5(10)
 2 2 (2 2 x  2 )  5
2 2 x  2 (1  2 2 )  5
 100
5(2 2 x  2 )  5
22 x  2  1
25. (a)
2 2 x  2  20
2x  2  0
h( x ) 

8 x 1  23 x 1  15
22.
 3[2
( 23 ) x 1  23 x 1  15
h(b)  192
(b)
15( 23 x  3 )  15
3( 4b )  192
23 x  3  1
4b  64
23 x  3  2 0
3x  3  0
∴
x 1
27. (a) The amount of drug intake  25(0.85)0 mg




 25(1) mg




 25 mg
0
(b) Percentage of drug left after 4 hours
25(0.85) 4 mg
 100%
25 mg
 52.2% (cor. to 3 sig. fig.)

  4  1 
3 f (1)
 3(6 1 )  4  
4 g (1)
  3  

28. (a)
3   3 
 4 
6   4 
1
3
2
1

6

24. (a)
4b  43
b3
26. The body surface area of the person
 0.096 (660.7) m2
 1.80 m 2 (cor. to 3 sig. fig.)
2
4
(b) 5 f (0)  g (0)  5(6 0 )   
3
 5(1)  1
6
(c)
]
 3( 4 x )
23 x  3 ( 2 4  1)  15
23. (a)

 3( 2 )
2 4 ( 23 x  3 )  23 x  3  15
4
f ( 2)  g ( 2 )  6 2   
3
 16 
 36 
 9 
 64
x (  x )
2x
23 x 1  23 x  3  15
∴
f ( x)
g ( x)
  1 x 
 9(2 x )  3  
  2  
x
 9(2 )  3(2  x )
x 1
∴
1
f (1000)  f (1000)  5(1000 3 )  5(1000) 3
 4% 
A  25 0001 

2 

 25 000(1.02) 2 n
2n
(b) Amount received by Mr Chan after 3 years
 $25 000(1.02)2×3
 $28 154 (cor. to the nearest $1)
f (125)  25
29.
1
3
k (125 )  25
k (3 125 )  25
5k  25
k5
19
NSS Mathematics in Action (2nd Edition) 4B
Full Solutions
30. (a)
31. (a) (i)
x
y
–3 –2 –1
0.1 0.2 0.5
0
1
1
2
3
2.2 4.8 10.6
(ii)
(a)(ii)
(a)(i)
(i)
From the graph, when x  0.5, y  1.2.
∴
2
 
3
0 .5
 1.2
8
2
  
27
3
(ii)
1
3
(b)
5  5 2
 
11  11 
3
 11 
 
5
 2 2
 
3
From the graph, when x 
∴
3
, y = 0.5.
2

1
2
From the graph, when x  
8
 0.5
27
∴
5
 0.6
11
(i)
 11 
4   24
5
1
, y  0 .6 .
2
(c)
(b)
(i)
From the graph, when y  2 , x  –1.7.
x
 2
∴ The solution of    2 is x  –1.7.
 3
x
x
(ii)
1
3
  
2
3
 
x
 11 
  6
5
From the graph, when y  6 , x  2.3 .
1
1
 3  x 
1
     
3
 2  
x
 11 
∴ The solution of 4   24 is x  2.3.
5
x
2
  3
3
From the graph, when y  3 , x 2.7.
x
(ii)
1
5
  
11
3
 
1
1
 5  x 
1
     
3
 11  
x
1
3
∴ The solution of   
is x 2.7.
3
2
x
 11 
  3
5
From the graph, when y  3 , x  1.4 .
x
1
5
∴ The solution of   
is x  1.4 .
11
3
 
20
6 Exponential Functions
32. Consider a > 1.
1
37.
x
 1 
When x > 0, ax > 1 and a 2 x   2   1
a 
2 x
x
When x = 0, a = 1 and a  1
4
3 1
3
33. 25
 125  (5 )
5
5
5

4
3
1
1
1
6
1
2
 (5 )
x
4
 53

3
1
2
1
6

(5 x )  k  8
0.5 k  8
1
2
( 21 )  k  23
1 1
 
2 2
∴
 31
1
3
2 k  23
k 3
5 x 1  5 x  5 x 1  29
40.
52 (5 x 1 )  5(5 x 1 )  5 x 1  29
1
2 3
35.
1
x
5 kx  8
1
3

y
x y
 (32 ) 4
3
3
2
(3a )
(3) a
 (3 a ) 4 
a
9a 3
9a 3
3

 27
a
9

 3a
3
 1
5 x 1 (52  5  1)  29
4
3
29(5 x 1 )  29
5 x 1  1
3 4
 3
2 3
5 x 1  50
x 1  0
1
6
3 x 3  2(3 x  2 )  30  3 x
41.
3 3 (3 x )  2  3 2 (3 x )  3 x  30
3 x (33  2  3 2  1)  30
2
5 2
 6a  5 
  1  (6 a )
36. (5ab  6 ) 0  
3
 5 b3 


(b 5 ) 2
10(3 x )  30
3x  3
x 1
∴
3
(b 5 ) 2

(6 a  5 ) 2

x 1
∴
a6

1
2
1  1
1 1
  ( 1) 1    
2  2
2 2
1 0
0

1
2

1
2
 x y  ( x y )  ( x 1 y
4
3 9
4 4
 
3 3
 4 9  (33 )
3
1
38. x 2 y  (6 xy )3  ( xy 2 ) 1  x 2 y  ( xy ) 6  ( x 1 y
39.

3
9
1
34. 27
3 1
 y8
Level 2
2

2
3
1
 y8  y4
 1 
When x < 0, ax < 1 and a 2 x   2   1
a 
∴ It is obvious that the graphs of y = ax and y = a–2x do
not intersect except at the point P(0, 1).
∴ The two graphs intersect at only one point P(0, 1).
4
9
1
 (y 2 )4 (y 2 )2
x
2

3
1
y y  y y  (y  y2 )4 (y y 2 )2
16 x ( 4 x )  8 x 1  0
42.
6
5
16 x ( 4 x )  8 x 1
b
36a 10
( 2 4 ) x ( 2 2 ) x  ( 23 ) x 1
6
10 5
2 4 x  2 x  23 x  3
a b
36
2 6 x  23 x  3
∴
21
6 x  3x  3
x  1
)

1
2
)
NSS Mathematics in Action (2nd Edition) 4B
Full Solutions
27(5 x )  125(3 x )  0
43.
3 x 1  2 y 1  1
 (1)
46.  x
y
4(3 )  3(2 )  24  (2)
27(5 x )  125(3 x )
27 3 x

125 5 x
3
3 3
   
5 5
x3
∴
Let a  3x and b  2 y .
From (1),
3 x1  2 y 1  1
x
3(3 x )  2(2 y )  1
3a  2b  1(3)
From (2),
4(3 x )  3(2 y )  24
3 p  27 q  81r
44.
3 p  (33 ) q  (34 ) r
4a  3b  24......(4)
(3) × 3 + (4) × 2:
3(3a  2b)  2(4a  3b)  3  2(24)
3 p  33q  34 r
∴ p  3q  4r
Let p  12s , where s is a non-zero real number,
then q  4s and r  3s.
1
1
(a)
( p  q)  (12s  4s )
r
3s
8s

3s
8

3
(b)
17a  51
a3
By substituting a = 3 into (3), we have
3(3)  2b  1
b4
a  3 and
∵
∴
b4
3x  3
and
2y  4
x 1
and
2 y  22
y2
pq
12 s ( 4 s )

pr  qr 12 s (3s )  4 s (3s )
2 f ( k )  g (2k )  80
47.
48s 2
36 s 2  12 s 2
48s 2

24 s 2
2



2(4 k )   3(2 2 k )  80
2(4 )  3( 4 )  80
k
k
4 k (2  3)  80
5( 4 k )  80
4 k  16
4 x  2 y  64  (1)
45.  2 x  y
 1  (2)
16
From (1),
4 x2 y  64
x2 y
∴
48.
f ( 2)  8
ka2  8
4
4
∴ x  2 y  3 (3)
From (2),
162 x  y  1
2x y
4k  42
k2
3
(1)
1
f   1
2
1
ka 2  1
16
 16
∴ 2 x  y  0(4)
(3) + (4) × 2:
( x  2 y )  2(2 x  y )  3
0
ka2
(1)  ( 2) :
ka
a
5x  3
x
(2)
2
1
2
1
2

8
1
8
3
2
3
5
a 8
3
2
2
2
( a ) 3  ( 23 ) 3
3
By substituting x  into (4), we have
5
a  22
4
3
2   y  0
5
6
y
5
By substituting a  4 into (1), we have
k (4 2 )  8
16k  8
k
22
1
2
6 Exponential Functions
52.
2
49. (a) By substituting b  1.36 and m  64 into b  Am 3 ,
we have
2
1.36  A(64) 3
A
1.36
2
( 43 ) 3
 0.085
2
(b) Weight of Mary’s brain  0.085  75 3 kg
 1.51 kg (cor. to 3 sig. fig.)
50. (a) When t  2, V  4418.
4418  A(0.94) 2
x
(a) (i)
4418
0.94 2
 5000
A
9
  7
5
1 .8 x  7
From the graph, when y  7 , x  3.3.
x
9
∴ The solution of    7 is x  3.3.
5
(b) Volume of water running out of the tank in the
5th minute
 [5000(0.94)4  5000(0.94)5 ] cm3

x
5
5   1
9
(ii)
  234.22 cm3
< 250 cm3
∴ Peter’s claim is disagreed.
x
1
5
  
9
5
 
1
1
 5  x 
1
     
5
 9  
51. (a) When t  1 , N  3300 .
3300  ab1  ab (1)
When t  3 , N  3993 .
x
( 2)  (1) : b 2  1.21
b  1 .1
9
  5
5
1 .8 x  5
From the graph, when y  5 , x  2.7 .
By substituting b  1.1 into (1), we have
1.1a  3300
5
∴ The solution of 5   1 is x  2.7 .
9
3993  ab
3
(2)
x
a  3000
(b) (i)
5(9 x )  9(5 x )  0
5(9 x )  9(5 x )
(b) The required increase in the number of people
 3000(1.1)10  3000(1.1)9
9x  9

5
5x
 707 (cor. to the nearest integer)
x
(c) The required percentage increase
3000(1.1)10  3000(1.1) 0
 100%
3000(1.1) 0
 159% (cor. to the nearest integer)

∵
∴
(ii)
9
9
  
5
5
1.8 x  1.8
The graph of y  1.8 x  0 for all real
values of x.
5(9 x )  9(5x )  0 has no real solutions.
2 (3 x )  2( 5 ) x  0
2 (3 x )  2( 5 ) x
x
 3 
2


 5  2


2
 3  x   2  2

  

 5    2 


x
9
  2
5
1.8 x  2
23
NSS Mathematics in Action (2nd Edition) 4B
Full Solutions
From the graph, when y  2 , x  1.2 .
∴ The solution of
x  1.2 .
(a) (i)
2 (3 )  2( 5 )  0 is
x
x
From the graph, when x  28, y  2.0.
∴ The required percentage of sunlight intensity
 2.0%
(ii) From the graph, when y  5, x  17.
∴ The corresponding depth  17 m 
53. (a) ∵ The graph passes through (0, 2).
∴ By substituting (0, 2) into y  ka x , we have
(b) (i)
The rate of change in percentage of sunlight
intensity decreases as the depth below sea level
increases.
(ii) When x = 100,
y  20(0.975) 3.28(100)
2  ka0
k2
∵ The graph passes through (1, 6).
∴ By substituting (1, 6) and k = 2 into y  ka x ,
we have
6  2  a1
 4.95 10 3
1
∴ The percentage of sunlight intensity will
be less than 1% at a depth of 100 m below
sea level.
Alternative Solution
From the graph, when x  50, y  0.3 < 1.
When x increases from 50 to 100, y will
decrease. As a result, y will remain less than 1.
Therefore, the percentage of sunlight intensity
will be less than 1% at a depth of 100 m below
sea level.
a3
f (m)  108
(b)
2(3m )  108
108
4
3m 
3m  27
3
3m  3 2
∴
Multiple Choice Questions (p. 6.41)
1. Answer : B
f (3)  f (2)  (43  33 )  (4 2  32 )
3
m
2
 37  7
 30
54. (a) When x  0 , S  5 and T  20 .
∴
5  ab 0 and 20  mn0
a5
m  20
2.
Answer : B
1
3
4 x  4 x3  2 x 2  x 4
(b) (i)
From the graph, when x  1 ,
S  6.5 and T  16.0
1 3

4
 2x 2
5
(ii) When x = 1, S  6.5 and T  16.0 .
∴
 2x 4
6.5  5b1 and 16.0  20n1
b  1.3
n  0.8
3.
Answer : A
1
ab
1 ( ab) 2

 2 
a
a
b2
b4
1
a
 1  1 1
1
(c) From the graph, the population of the salmon will
exceed that of tuna after 3 years, i.e. starting from
the beginning of 2017.
4
b2
55.
a
1
a 2b 2
1
2
1
b
 a 2 b 1

24
a
b
1 1
 
2 2
6 Exponential Functions
4.
Answer : C
9.
3
2
125 x  27  0
3
2
x 
681
 1  4510
N  N0  
2
N
 0.901
N0
27
125
2
 3 3  3
( x )    
 5  
3
2
Answer : D
When t  681,
2
3
3
x 
5
9

25
 90% (cor. to the nearest integer)
2
10. Answer : A
When x = 0,
y  5 ( 0 )
∴
5.
Answer : A
3 x  27 x  9
3 3  3
x
3x
11. Answer : C
The graph of I may be the graph of y  a x , where
0  a  1.
2
3 x 3 x  3 2
The graph of II may be the graph of y  a x , where
a  1.
34 x  32
∴
6.
4x  2
1
x
2
Answer : B
4 x 1  22 x 1  9
22 x  2  22 x 1  9
The graph of y  a x always lies above the x-axis.
∴
∴
The graph of III cannot be the graph of y  a x .
The answer is C.
1
For II, as x increases, the rate of decrease of  
7
becomes smaller.
∴ II is false.
23 (22 x 1 )  22 x 1  9
2 2 x 1 (23  1)  9
9(22 x 1 )  9
x
22 x 1  20
2x  1  0
1
2
HKMO (p. 6.43)
7.
Answer : D
25 x  (52 ) x


2
 (2)

1.
 (5 )
x 2
  
2 ...
2
 k2
2
2


  (4) 4
 
2 m1
4
2 m1
 22( 4
  
4 ...
4 n1
n1 )
2m 1  2(4n 1 )
8.
x
1
For III, the graphs of y  7 x and y    show
7
reflectional symmetry with each other about the y-axis.
∴ III is false.
∴ The answer is A.
22 x 1  1
x
∵
12. Answer : A
For I, the two graphs intersect at one point (0, 1) only.
∴ I is true.
(22 ) x 1  22 x 1  9
∴
1
The coordinates of A is (0, 1) .
Answer : D
4 x  32 y
2m 1  2(22 ) n 1
2m 1  21 2 ( n 1)
( 2 2 ) x  ( 25 ) y
2m 1  22 n 1
2 2 x  25 y
m  1  2n  1
m  2n
2x  5 y
x 5

y 2
∴ x : y  5:2
∴
25
m
2
n
k 2

4
NSS Mathematics in Action (2nd Edition) 4B
Full Solutions
Exam Focus
x y  x y  2 2
2.
1
2 2
xy
xy 
Exam-type Questions (p. 6.45)
1. Answer : D
For I:
From the graph, we have
a  1, 0  b  1 and 0  c  1 .
2
 y 1 
 x  y   (2 2 ) 2
x 

1
x2 y  2 y  2  8
x
1
2y
x  2y  2  4
x
∴ a  1  bc
i.e. a  bc
∴ I must be true.
For II:
Consider the values of b x and c x when x = 1.
From the graph, we have
b1  c1
2
 y 1 
x  y   4
x 

( x y  x y )2  4
k2  4
bc
∴ II must be true.
For III:
From the graph, we can see that all the three graphs
always lie above the x-axis, i.e. they do not intersect the
x-axis.
∴ III must be true.
k  2 or  2 (rejected)
3.
48 x  2 and
48 y  3
6 x8 x  2 and 6 y8 y  3
6 x8 x  6 y 8 y  2  3
6 x  y8 x  y  6
8
x y

2.
6x  y
1
Answer :A
m 1
m 1


3
9m  9
m 1
m 1

3
3 m 1
m 1
m 1


3 m 1 3 m 1
0
6
1
(8 x  y ) 1 x  y  (61 x  y ) 1 x  y
x y
81 x  y  6
b6
∴
3.
1
2
x x
4.
1


1
2
Answer :C
 23 n  4 n

 8

3
1
( x 2  x 2 ) 2  32
x  x 1  2  9
2
  23n 2 2 n 
 

  23 
 

3n  2 n  3 2
 (2
)
 ( 25 n  3 ) 2
x  x 1  7
1 2
(x  x )  7
 (2 2 )5 n  3
2
 45 n  3
x 2  x  2  2  49
x 2  x  2  47
1
2

4.
Answer :B
x n 1  x n 1 x 2 ( x n 1 )  x n 1

x n 1
x n 1
n 1
x ( x 2  1)

x n 1
2
 x 1
5.
Answer :B
8 x 1  216
1
2
( x  x )( x  x 1 )  3  7
3
1
x2  x2  x

1
2
x
3
2
x 3 x
3
2
x x
b



3
2

2

3
2
3

2

3
2
 21
 21
 18
3
2
x  x 3
x 2  x 2  2
18  3
47  2
15
45
1
3
8(8 x )  216
8 x  27
23 x  33
2
2
(23 x ) 3  (33 ) 3
22 x  32
4x  9
26
6 Exponential Functions
6.
Answer :B
3.
1
When x = 0, y  0  1 .
a
∴ The graph passes through (0, 1).
∴ C and D must not be the required graph.
∵ a 1
1
∴ 0  1
a
1
∵ When 0   1 , the value of y decreases as x
a
increases and the graph always lies above the x-axis.
∴ The answer is B.
6 4  1296
6
∴ 3 4  36
3 4  81
∴
34  4 3
∴
63  6 4
4
6 3  216
∴
36  63
∴
43  4 6
6
4
3
46  64
∴
b
1
a
bk  a k
7
8
7
6
6 7 , 68 , 7 6 , 78 , 8 6 , 8 7
6
(b) 7 8  5 764 801
87  2 097 152
78  87
8
7
8
6
7
6
From 1(a), 6 7  6 8
68  1 679 616
8 6  262 144
68  8 6
From 1(a), 7 6  7 8
7 6  117 649
6 7  279 936
∴
67  7 6
From 1(a), 8 6  8 7
8
8
7
Consider 6 7 , 7 6 and 8 6 .
67  7 6
8
8
From 1(b), 6 7  7 6
68  86
7
From 1(b), 68  8 6
8
7
4
 64
3
∴ 6 4 is the smallest.
k
∴
36  63
3
b
  1
a
bk
1
ak
∴
3
36  63
kb  ka
8
3
46  6 4
(b) For b  a  1 and k  0,
(a)
3
3
Consider 3 6 , 6 4 and 4 6 .
4
2.
3
36  729
kb
1
ka
∴
4
4 3  64
(a) For k > 1 and b > a,
k ba  1
∴
46  6 4
∴
Investigation Corner (p. 6.47)
1.
4 6  4096
7
7
∵
6 7  68  8 6
∴
67 is the greatest.
8
27