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5
142
NON-FOUNDATION
Name :
Date :
Mark :
Exponential and LogarithmicFunctions
5A
○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○
5.1 Integral Indices
Key Concepts and Formulae
Laws of indices
1. a0 = 1 (a ≠ 0) 2. a an
a n
− = ≠10 ( )
3. am × an = am + n 4. am ÷ an = am − n (a ≠ 0)
5. (am) n = amn 6. (ab) n = anbn
7. a
b
n n
n
ab
b
= ≠ ( )0 8. a an n
1
= (n is a positive integer)
9. a a am
n n m mn= =( ) (m and n are integers where a > 0, n > 0)
Express each of the following in the form x p,where p is a rational number. (1 – 3)
1. x46
Solution
x x
x
x
46 4=
=
=
( )(
(
(
)
)
)
2. 13 x
Solution
1 13 1
3
1
3
xx
x
=
=−
3. 158
x
Solution
1 1
1
585
1
8
5
8
5
8
xx
x
x
=
=
=−
( )
1
6
4
6
2
3
143
5 Exponential and Logarithmic Functions
Find the values of the following withoutusing a calculator. (4 – 6)
4. 8
27
2
3
Solution
8
27
2
3
2
3
=
=
=
)
)
( )( )
(
(
5. 9
4
3
2
−
Solution
9
4
9
4
3
2
3
2
27
8
8
27
3
2
23
2
1
3 1
1
32
1
=
=
=
=
−
−
−
−
−
=
6. −
−125
64
4
3
Solution
− −
−
−
=
=
=
=
=
−−
−
−
−
125
64
125
64
5
4
5
4
625
256
256
625
4
3
4
3
1
4
3
1
4 1
1
3
2
3
3
23
2
4
9
144
Number and AlgebraNumber and Algebra
Simplify and express each of the followingwith positive indices. (7 – 14)
7. a2(−a)3 ÷ a−5
Solution
a2(−a)3 ÷ a−5 = a2 ⋅ (−a( )) ÷ a−5
= −a2 ⋅ a( ) ÷ a−5
= −a ( ) + ( ) − ( )
= −a ( )
9. m n
m n
5
4
1
3
3
3
4
1
−
Solution
m n
m n
m n
m n
m n
m n
m n
5
4
1
3
3
3
4
1
5
43
1
3
3
4
5
4
12
4
4
12
9
12
7
4
5
12
7
4
5
12
7
4
5
12
1
1
1
1 1
=
=
=
=
=
−
− − −
− − −
− − −
− × − − × −
( )
( )
( )
( ) ( )
10. ( )( )
xx
x
1
2
4
3
3 5
1
2
÷
Solution
( )( )
xx
x
x
x
x
x
x
x
x
x
x
x
1
2
4
33 5
1
2
1
2
4
3
1
35
1
2
2
3
5
3
1
2
2
3
10
6
3
6
2
3
7
6
4
6
7
6
3
6
1
2
1
2
1
÷ = ÷
=
=
=
=
=
××
− −
− −
−
−
−
−
=
=
8. ( )a b a b1
2
3
4 23
4
1
2− −
÷
Solution
( )a b a b a b a b
ab a b
a b
a b
a
b
1
2
3
4 2 2 2
1
1
3
4
1
2
1
2
3
4
3
4
1
2
3
2
3
4
1
2
3
4
3
2
1
2
1
4
1
4
− −
− − −
÷ = ÷
= ÷
=
=
=
× − × −
− −
−
−
3
3
2 3 −5
10
145
5 Exponential and Logarithmic Functions
11. p pp
451
4
7
31⋅ ⋅
−−
Solution
p p p p
p p p
p
p
p
pp
45 11
4
7
34
5
1
4
7
3
4
5
1
4
7
3
4
5
1
4
7
3
48
60
15
60
140
60
173
60
1⋅ ⋅
−−
− −
−
+ − +
− +
= ⋅ ⋅
= ⋅ ⋅
=
=
=
−
( )
12. x y x yy
x
−−
−⋅ ⋅
3
4
2
5
2
1
25 34
Solution
x y x y
x y x y x y
x y
x y
x y
y
x
−−
−
− −
− + + − +
− +
⋅ ⋅
= ⋅ ⋅
=
=
=
3
4
2
5
2
1
2
3
4
5
2
2
5
3
4
1
8
3
42
3
4
5
2
2
5
1
8
100
40
16
40
5
40
89
40
5 34
2
2
2
13. m
n
n
m
m n38
2 1
5
4
4
3
1
2
1
3
2
5( )
( )−
−−
÷ ⋅
Solution
m
n
m
n
n
m nn
m
m n m n
m n m n m
m n
38
2 1
5
4
4
3
1
2
3
8
2
5
4
1
2
4
3
1
22
5
1
3
3
8
5
8
2
3
2
5
1
3
3
8
5
8
2
5
2
3
1
3
2
5
2
2
( )( )
( )
−
−
−
× − × − − −
− − − −
− − − −
÷ ⋅
= ÷ ⋅
= ÷ ⋅
=
−
−
⋅ −
=
1
3
3
5
7
3m n
14. 4 33
4
1
2
1
2 3 4
3
2
4
3
3
21 2p q p qq
p
÷− −
⋅ −( )( )
Solution
4 3
3
32
3
4
1
2
1
2 3 4
3
2
4
3
3
2
3
8
1
4
3
2
4
3
3
2
3
2 3
8
3
2
1
4
4
3
3
2 39
8
43
12
3
2
39
8
43
1 2
3 2
3 2
1
2
3
2
3
2
p q p q
p q p q p q
p q
p q
q
p
p q
÷
= ÷
=
=
=
−−
⋅
⋅
−
− − − − −
− − − − − −
− −
⋅
( )( )
( )
1212
5
146
NON-FOUNDATION
Name :
Date :
Mark :
Exponential and LogarithmicFunctions
5B
○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○
5.2 Exponential Equations
Solve the following exponential equations.(1 – 6)
1. 4 64243
x
=
Solution
4 64
2 2
2 2
243
2 243
x
x
x
=
===
( ) ( )
( )
( )
( ) ( )
∴
3. 2x + 1 − 2x = 16
Solution
2x + 1 − 2x = 16
2(2x) − 2x = 16
2x(2 − 1) = 16
2x = 16
2x = 24
∴ x = 4
4. 2(4x) − 3(4x − 2) = 116
Solution
2(4x) − 3(4x − 2) = 116
2(16)(4x − 2) − 3(4x − 2) = 116
32(4x − 2) − 3(4x − 2) = 116
4x − 2 (32 − 3) = 116
4x − 2 = 4
x − 2 = 1
∴ x = 3
2. 3 812
3 +
=x
Solution
3 81
3 3
23
4
6
2
2 4
3
3
+
+
=
=
+ =
=
x
x
x
x∴
6
8x
8
147
5 Exponential and Logarithmic Functions
5. 33x = 27(92x)
Solution
33x = 33(34x)
33x = 33 + 4x
3x = 3 + 4x
∴ x = −3
6. 125(53x) = 25x + 2
Solution
125(53x) = 25x + 2
53(53x) = (52)x + 2
53 + 3x = 52(x + 2)
3 + 3x = 2x + 4
∴ x = 1
Solve the following equations. (7 – 9)
7. 3x + 3 − 3x + 2 − 3x + 1 = 15
Solution
3x + 3 − 3x + 2 − 3x + 1 = 15
9(3x + 1) − 3(3x + 1) − 3x + 1 = 15
5(3x + 1) = 15
3x + 1 = 3
x + 1 = 1
∴ x = 0
8. 52x − 6(5x) + 5 = 0
Solution
52x − 6(5x) + 5 = 0
(5x)2 − 6(5x) + 5 = 0
Let y = 5x, the equation becomes
( )2 − 6( ) + 5 = 0
( )( ) = 0
y = ( ) or y = ( )
∴ 5x = ( ) or 5x = ( )
5x = ( ) or 5x = ( )
∴ x = ( ) or x =
( )
9. 7(72x) − 56(7x) + 49 = 0
Solution
7(72x) − 56(7x) + 49 = 0
7(7x)2 − 56(7x) + 49 = 0
Let y = 7x, the equation becomes
7y2 − 56y + 49 = 0
7(y2 − 8y + 7) = 0
7(y − 1)(y − 7) = 0
y = 1 or y = 7
∴ 7x = 1 or 7x = 7
7x = 70 or 7x = 71
x = 0 or x =
1
y y
y − 1 y − 5
1 5
1 5
50 51
0 1
148
Number and AlgebraNumber and Algebra
Solve the following simultaneous equations.(10 – 12)
10. 2 8
16 4
2
8
x y
y x
+
−
==
......(1)
......(2)
Solution
From (1),
22x + y = 2( )
∴ 2x + y = ( )
y = ( ) − 2x .......(3)
From (2),
(2( )) y − 8x = 22
2( ) = 22
∴ ( ) = 2 .......(4)
By substituting (3) into (4), we have
( ) = 2
( )x = ( )
x = ( )
By substituting x = ( ) into (3), wehave
y = ( )
= ( )
∴ The solution is x = ( ),
y = ( ).
11. 5 25
81 3
3 7
8
x y
x y
+ =
=
+
......(1)
......(2)
Solution
From (1),
53x + 7y = 52
∴ 3x + 7y = 2 ......(3)
From (2),
( )3 3
3 3
1
2
4 1
1
8
2
2
x y
x y
x y
y x
+
+
=
=
=
= −
+
.......(4)
By substituting (4) into (3), we have
3x + 7(2 − x ) = 2
−4x = −12
x = 3
By substituting x = 3 into (4), we have
y = 2 − 3 = −1
∴ The solution is x = 3, y = −1.
3
3
3
4
4y − 32x
4y − 32x
40 10
12 − 8x − 32x
1
4
1
4
3 2
1
4− ×
5
2
5
2
1
4
149
5 Exponential and Logarithmic Functions
12. 27 93
3 2 1
64
x y
x y
+
− −
=
=
4 ( )
......(1)
......(2)
Solution
From (1),
(33)3x + y = 32
39x + 3y = 32
∴ 9x + 3y = 2 ......(3)
From (2),
4−3(x − 2y) = 4−3
∴ − 3(x − 2y) = −3
x − 2y = 1
∴ x = 2y + 1 ......(4)
By substituting (4) into (3), we have
9(2y + 1) + 3y = 2
21y = −7
y = −
1
3
By substituting y = −
1
3 into (4), we have
x = −
+
=
2 11
3
1
3
∴ The solution is x =
1
3, y = −
1
3.
5
150
NON-FOUNDATION
Name :
Date :
Mark :
Exponential and LogarithmicFunctions
5C
○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○
5.4A Definition of Logarithms5.4B Properties of Logarithms
Key Concepts and Formulae
1. If y = 10x, then x = log y.
2. If y = ax, then x = log a y.
3. Properties of Logarithms
(a) log a 1 = 0 (b) log a a = 1
(c) log a (MN) = log a M + log a N (d) log a (MN
) = log a M − log a N
(e) log a Mn = n log a M
Find the values of the following logarithmswithout using a calculator. (1 – 2)
1. log 0.000 01
Solution
Q 0.000 01 = 10( )
∴ log 0.000 01 = ( )
2. log ( )3
1
81
Solution
Q
1
81
1
81
3
4
4
3
=
= −
−
log ( )∴
In each of the following, find the value of xcorrect to 3 significant figures. (3 – 4)
3. 10x = 14
Solution
Q 10x = 14
∴ x = ( )
= ( ) (cor. to 3 sig. fig.)
4. 102x = 0.7
Solution
Q
10 0 7
2 0 7
0 0775
2
0 7
2
x
x
x
=
=
=
= −
.
log .
.
log .
∴
(cor. to 3 sig. fig.)
log 14
−5
−5
1.15
151
5 Exponential and Logarithmic Functions
Find the values of the following expressionswithout using a calculator. (5 – 8)
5. log 5 + log 20
Solution
log 5 + log 20 = log[( ) × ( )]
= log ( )
= log ( )( )
= ( ) log ( )
= ( )
8.−
−
log ( )
log
33 2
3
12
2
8
Solution
− −
−
−
− −=
=
= ×
=
log ( )
log
log
log
log
log
33 2
3
12
3
23
3
32
3
3
2
8
2
2
23
2
32
2
2
3
2
3
4
9
9. Given that log 2 = a and log 5 = b ,express the following in terms of a and b.
(a) log 20 (b) log 200
Solution
(a) log 20 = log[( ) × ( )]
= log ( ) + log ( )
= ( ) log ( ) +
log ( )
= ( )
(b)
log log( )
log( )
log log
log log
200 2 5
2 5
2 5
2 5
3 2
32
1
2
3
2
3
2
3
2
= ×
= ×
= +
= +
= +a b
6. log2 3 − log2 24
Solution
log log log
log
log
log
2 2 2
2
23
2
3 24
2
3 2
3
3
24
1
8
− =
=
=
= −
= −
−
7. log
log
49
343
Solution
log
log
log
log
log
log
49
343
7
7
2 7
3 7
2
3
2
3=
=
=
2
10
2
100
5 20
2
10
22 5
2a + b
22 5
2
5
2
152
Number and AlgebraNumber and Algebra
Find the values of the following expressionswithout using a calculator. (10 – 12)
10. log log log5 5 516 225
8+ −
Solution
log log log
log ( )
log
log
log
5 5 5
5
5
52
5
16 2
16
25
5
2 5
2
25
8
25
8
1
2
+ −
= × ×
=
=
=
=
11. 7
2
1
812 2 213 48log log log ( )+ − −
Solution
7
2
1
81
1
81
1
81
1
48
1
81
2 2 21
2
72
2 21
2
2
2
22
2
3 48
3 48
27 3
27 3 4 3
4
2
2 2
2
log log log ( )
log log log ( )
log ( )
log ( )
log
log
log
+ −
= + −
= × ÷
= × ×
=
=
=
=
−
−
12.0 5 4
2
3125 20 6
2 52
532
. log log log
log log
+ + −
+
Solution
0 5 42
3125 20 6
2 52
532
4 125 20 6
5 32
2 25 20 6
25 4
2 25 20 6
25 4
1000 6
100
3
2
1
2
2
3
2
52
. log log log
log log
log log log
log log
log log log
log log
log( )
log( )
log
log
+ + −
+
+ + −
+
+ + −+
× × −×
−
−
=
=
=
=
=
153
5 Exponential and Logarithmic Functions
Simplify the following expressions, wherex > 0, y > 0 and x, y ≠ 1. (13 – 15)
13.2
1
23 2
log
log log
x
x x−
Solution
2 2
2
2
2
1
2
3
2
1
2
3 2 23
2
4
log
log log
log
log log
log
log log
log
log
x
x x
x
x x
x
x x
x
x
− −
−
−
=
=
=
= −
14.4
5
2log log
log
x x
x x
−
Solution
4
2
1
23
4
2
3
5
2
5
23
4
12 4
52
252
32
12
1
2
1
2
log log
log
log( ) log
log( )
log log
log( )
log log
log
log
log
x x
x x
x x
x x
x x
x
x x
x
x
x
− −
⋅
−
−
−
−
=
=
=
=
=
15.
1
326 3
25
log log
log
x yy
xx
−
Solution
1
32
4
6 3
25
6 3 1 2
2 2
2 2
4
1
3
1
2
2
5
2
5
2
5
2
5
2
5
10
log log
log
log( ) log( )
log
log log
log
log( )
log
log
log
log
log
x yy
x
x
x y x y
x
x y x y
x
x y x y
x
x
x
x
x
−
−
−
÷
=
=
=
=
=
=
−
−
−
5
154
NON-FOUNDATION
Name :
Date :
Mark :
Exponential and LogarithmicFunctions
5D
○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○
5.4C Logarithmic Equations
Solve the following equations and give youranswers correct to 3 significant figures ifnecessary. (1 – 3)
1. 3x = 32
Solution
3x = 32
log 3x = log 32
( ) log ( ) = log 32
∴ x =
( )( )
= ( ) (cor. to 3 sig. fig.)
2. 5(73x + 2) = 48
Solution
5 7 48
7
3 2
3 2
0 279
3 2
3 2 48
5
( )
( )
.
x
x
x
x
x
+
+
=
=
+ =
+ =
= −
−
log 7 log48
5
log 48 log 5
log 7
(cor. to 3 sig. fig.)
3. 42x − 3 = 66 − x
Solution
4 6
2 3 6
2 3 6
6 3
2 3 6x x
x
x x
x x
x
− −
− −
=
=
− = −
− = −
+ = +
log 4 log 6
log 4 log 6
log 4 log 4 log 6 log 6
(2 log 4 log 6) log 6 log 4
2 3 6 x
( ) ( )
∴
x =
=
++
6 3
3 27
log 6 log 4
2 log 4 log 6
. (cor. to 3 sig. fig.)
Solve the following equations. (4 – 12)
4. log(1 − 2x) = 1
Solution
Q log(1 − 2x) = 1
∴ 1 − 2x = ( )
2x = ( )
x = ( )
x 3
log 32
log 3
3.15
10
−9
− 9
2
155
5 Exponential and Logarithmic Functions
5. log2 (5x − 6) = 6
Solution
Q log2 (5x − 6) = 6
∴ 5x − 6 = 26
5x − 6 = 64
5x = 70
x = 14
7. log (4x + 2) − log 5 = 1
Solution
Q
log log 5
log 4x + 2
5
( )4 2 1
1
10
4 2 50
4 48
12
4 2
5
x
x
x
x
x
+ − =
=
=
+ =
=
=
+
∴
∴
8. log5 (2x − 3) − log5 (x − 1) = 0
Solution
Q
log log
log
5 5
5
( ) ( )2 3 1 0
0
1
2 3 1
2
2 3
1
2 3
1
x x
x x
x
x
x
x
x
− − − =
=
=
− = −
=
−−
−−
∴
6. log5 (3x − 1) = 0
Solution
Q log5 (3x − 1) = 0
∴ 3x − 1 = 50
3x − 1 = 1
3x = 2
x =
2
3
156
Number and AlgebraNumber and Algebra
9. [log(x + 2)]2 + 5 log (x + 2) + 4 = 0
Solution
Let u = log (x + 2), then the equation [log(x + 2)]2 + 5 log (x + 2) + 4 = 0 becomes
u2 + 5u + 4 = 0
(u + 1)(u + 4) = 0
u = −1 or u = −4
∴
log ( )x
x
x
+ = −
+ =
= −
2 1
21
10
19
10
or
log
999
10 000
( )x
x
x
+ = −
+ =
= −
2 4
21
10 000
19
or
or
10. 2 log(x + 2) − log (2x − 6) − 1 = 0
Solution
2 2 2 6 1 0
2 2 6 1
1
10
4 4 20 60
16 64 0
8 0
8
2
2
2
2
2
2
2
2 6
4 4
2 6
log log
log log
log
(double root)
( ) ( )
( ) ( )
( )
( )
( )
x x
x x
x x x
x x
x
x
x
x
x x
x
+ − − − =
+ − − =
=
=
+ + = −
− + =
− =
=
+−
+ +−
157
5 Exponential and Logarithmic Functions
11. y x
x y
= ++ =
2 1
1log log
......(1)
......(2)
Solution
By substituting (1) into (2), we have
log x + log( ) = 1
log [( )( )] = 1
(x)(2x + 1) = 10
2x2 + x − 10 = 0
(2x + 5)(x − 2) = 0
x = ( ) or x = ( )
By substituting x = ( ) into (1),we have
y = 2( ) + 1
= ( )
By substituting x = ( ) into (1),we have
y = 2( ) + 1
= ( )
∴ The solutions of the simultaneous
equations are ( , ) and
( , ).
12. x y
x y
− =− =
2
2 3 log log2 2
......(1)
......(2)
Solution
From (1), we have
y = x − 2 ……(3)
By substituting (3) into (2), we have
2 2 3
3
2
8 16
8 16 0
4 0
4
2
23
2
2
2
2
2
log log
log
(double root)
2 2
2
x x
x x
x x
x
x
x
x
x
x
− − =
=
=
= −
− + =
− =
=
−
−
( )
( )
By substituting x = 4 into (3), we have
y = 4 − 2
= 2
∴ The solution of the simultaneous
equations is (4, 2).
2
−4
− 5
2
5
2
2
− 5
2
− 5
2
−4
5
2
2x + 1
2x + 1x
− 5
2
5
158
NON-FOUNDATION
Name :
Date :
Mark :
Exponential and LogarithmicFunctions
5E
○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○
5.4D Applications of Logarithms in Real-life Situations
(In this exercise, give your answers correct to 3 significant figures if necessary.)
1. The sound intensity level produced by two concerts A and B are 85 dB and 95 dB respectively.How many times is the sound intensity produced by concert B to that by concert A?
Solution
Let I1 and I2 be the sound intensities produced by concerts A and B respectively.By the definition of sound intensity level, we have
( ) 10 log
and ( ) log
=
=
I
I
I
I
1
0
2
0
10
∴ − =
−
=
−
=
=
( ) ( )
( )
( )
( )( )
( )
10 log log
log log
log
2 1
I
I
I
I
I
I
I
I
2
0
1
0
0 0
10
10
∴ The sound intensity produced by concert B is ( ) times as that produced by concert A.
2. How many times is the strength of an earthquake measured 6 to that measured 4 on the Richterscale?
Solution
Let El and E2 be the relative energies released by the earthquakes measured 6 and measured 4
on the Richter scale respectively.
By the definition of the Richter scale, we have
6
10 10
1 2
16
24
= =
= =
log and 4 log
i.e. and
E E
E E
85
95
95 85
10
1
10l2l1
l2
l1
10
159
5 Exponential and Logarithmic Functions
Solution
∴
E
E1
2
6
4
10
10
10
100
2
=
=
=
∴ The strength of the earthquake measured 6 is 100 times to that measured 4 on the Richter
scale.
3. The sound intensity level in Peter’s house is 50 dB at 8:00 pm. When Peter was sleeping, thesound intensity in his house is reduced to 40% of the original. Find the sound intensity level inPeter’s house when he was sleeping.
Solution
Let I be the sound intensity in Peter’s house at 8:00 p.m..
Q The sound intensity level in Peter’s house is 50 dB at 8:00 p.m..
∴ 50 10
0
=
log
I
I
Let β dB be the sound intensity level in Peter’s house when he was sleeping.
∴
β
β
β
β
β
=
− =
−
− =
−
− =
= −
=
10
50 10 10
50 100 4
50
50 10
46 0
0 0
0 4
0 0
log
log log
log log .
10 log
log
0.4
5
2
5
2
0
I
I
I
I
I
I
II
II
.
. (cor. to 3 sig. fig.)
∴ The sound intensity level in Peter’s house when he was sleeping is 46.0 dB.
160
Number and AlgebraNumber and Algebra
4. When Grace is playing piano, the sound intensity is increased by 3 times, what is the increaseof the corresponding sound intensity level?
Solution
Let β dB and I be the original sound intensity level and sound intensity respectively.
∴ β =
10 log
I
I0
Let α dB be the sound intensity level when Grace is playing piano.
∴ α =
10
0
log 3I
I
∴
α β
α β
α β
α β
− =
−
− =
−
− =
− =
10 10
10
10
4 77
0 0
0 0
log log
log log
log 3
3
3
I
I
I
I
I
I
I
I
. (cor. to 3 sig. fig.)
∴ The increase of the corresponding sound intensity level is 4.77 dB.
5. If the strength of an earthquake is 8 times to that measured 6.5 on the Richter scale, what is themagnitude of the earthquake on the Richter scale?
Solution
Let R be the required magnitude of the earthquake, E be the relative energy released by the
earthquake measured 6.5 on the Richter scale.
Q The strength of the earthquake is 8 times to that measured 6.5 on the Richter scale.
By the definition of the Richter scale, we have
6.5 = log E and R = log 8E
R = log 8E
= log 8 + log E
= log 8 + 6.5
= 7.40 (cor. to 3 sig. fig.)
∴ The magnitude of the required earthquake is 7.40 on the Richter scale.
5
Name :
Date :
Mark :
161
NON-FOUNDATION
5F
Exponential and LogarithmicFunctions
Multiple Choice Questions
1. 8
27
4
3
=
A.4
9B.
32
243
C.16
18D.
2
3
2.x
x
−
=3
2
5
A. x−
26
5 B. x−
17
10
C. x34
5 D. x−
13
10
3. Simplify m m
m
1
2 3
4 1( )− with positive indices.
A. m15
4 B.1
11
4m
C. m13
4 D. m15
2
4. Simplify x y x y2 3 53
2 with positive
indices.
A. x
y
25
2
B. x y7
2
11
4
C.1
5
2
4
5x y
D. xy5
4
5. Simplify p q p qp q1
2
2
5
1
2
1
2
3
527 33÷ ⋅
−.
A.1
3B.
31
6
q
p
C.q
p
5
6
3D.
q
p
1
6
1
23
6. Solve 6 6 301x x− − = − .
A. x = −1
B. x = 0
C. x = 1
D. x = 2
7. Solve 4(42x) − 65(4x) + 16 = 0.
A. x = 1
4or 16
B. x = −1 or 2
C. x = 1 or −2
D. x = − 1
2or 1
8. Solve 64 165
4x = .
A. x = 5
6
B. x = 5
C. x = 3
2
D. x = 3
4
B
C
A
B
D
C
A
B
162
Number and AlgebraNumber and Algebra
9. Solve 3 1
3 9
6
2
x y
x y
+
−
==
.
A. x y= = −1
4
2
3,
B. x = 2, y = 0
C. x y= =01
6,
D. x y= = −3
2
1
4,
10. log3 729 ⋅ log4 16 =
A. 3 B. 9
C. 12 D. 18
11. Solve log x2 + 3 log x + 2 = 0.
A. x = 10−2 or 10−1
B. x = −2 or −1
C. x = 102
5−
D. x = 101
2−
12. Solve log2 (x + 3) = 2 + log2 (2x − 1).
A. x = 7 B. x = 5
3
C. x = 1
2D. x = 1
13. If log 3 = a and log 4 = b, expresslog 0.75 in terms of a and b.
A. a + b B.a
b
C. ab D. a − b
14. If log , log a a2310= =
A.3
2. B. 10
3
2 .
C. 15. D. 102
3−
.
15. If log x − log y = 2 + log y, x =
A. 100y2.
B. 2 + y2.
C. 2 + 2y.
D. 200y2.
16. Solve 2 2
2 1
2
log
log
log
.x
x
x
= +
−
A. x = 10
B. x = 1
C. x = −1 or 1
D. x = − 1
2
17. Simplify log 9 3 − log
log.
27
3 3
A. −1
2log 3 B. − 1
3
C. log 3 D. 3
18. Given that x < 1 and y < 1, which ofthe following MUST be true?
I. log (xy) < 0
II. log1
0y
<
III. log x 3 < 0
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
D
D
C
C
C
A
A
B
B
D
163
5 Exponential and Logarithmic Functions
19. Given that log2 x = y and log3 y = z,express x in terms of z.
A. x = 6 3
B. x = log 6 z
C. x = 2 3 + z
D. x = 2 3z
20. Given that log x = 3.41, find the value
of log logxx
2 1+ .
A. 3.41
B. 10.15
C. 6.82
D. 0.29
21. Which of the following functions can berepresented by the graph shown below?
−4 −3 −2 −1
5
0
10
15
y
x
A. y = 2x
B. y = −2x
C. y = 1
2
x
D. y = −
1
2
x
B.
0
y
x
1
y
0x
C. y
10x
1
y
0x
D.
23. The inverse function of y x= log5 is
A. yx
= 1
5log.
B. y x= 5 .
C. y x= 5.
D. y x= − log .5
22. Which of the following may be a graph
of y a x= ?
A.
D
C
D
B
A
