1. MAT 163
Surds, Indices, and Logarithms
Radical
Definition of the Radical
For all real x, y > 0 , and all integers a > 0 ,
a
x = y if and only if x = y a
where a is the index
is the radical
x is the radicand.
Surds
A number which can be expressed as a fraction of integers (assuming the denominator is never 0)
5 4
is called a rational number. Examples of rational numbers are , − and 2.
2 5
A number which cannot be expressed as a fraction of two integers is called an irrational number.
Examples of irrational numbers are 2 , 3 7 and π .
An irrational number involving a root is called a surd. Surds occur frequently in trigonometry,
calculus and coordinate geometry. Usually, the exact value of a surd cannot be determined but an
approximate value of it can be found by using calculators or mathematical tables. In this chapter,
a means the positive square root of a while − a means the negative square root of a.
General Rules of Surds
Multiplication of surds
a × b = a×b
For example (i) 3 × 12 = 3 ×12 = 36 = 6
(ii) 32 × 2 = 32 × 2 = 64 = 8
(iii) 5 × 5 = 5 × 5 = 25 = 5
1

2. MAT 163
Division of surds
a a
a÷ b= =
b b
72
For example (i) 72 ÷ 2 = = 36 = 6
2
45
(ii) 45 ÷ 5 = = 9 =3
5
These rules are useful for simplifying two or more surds of for combining them into one single
surd.
Note, however, that 3 + 6 ≠ 3 + 6 and 6 − 2 ≠ 6 − 2 which can be easily checked by a
calculator; and, therefore, in general a + b ≠ a + b and a − b ≠ a − b .
Example 1
(i) 3 5 + 5 = (3 + 1) 5 (ii) 40 = 4 ×10 = 4 × 10
=4 5 = 2 10
Example 2
Simplify (i) 243 − 12 + 2 75
(ii) 50 + 8 + 32
Solution:
(i) 243 − 12 + 2 75 = 81× 3 − 4 × 3 + 2 25 × 3
= 81 × 3 − 4 × 3 + 2 25 × 3
= 9 3 − 2 × 3 + 10 × 3
= (9 − 2 + 10) 3
= 17 3
(ii) 50 + 8 + 32 = 25 × 2 + 4 × 2 + 16 × 2
= 25 × 2 + 4 × 2 + 16 × 2
= 5× 2 + 2× 2 + 4× 2
= (5 + 2 + 4) 2
= 11 2
Try This 1
Simplify (i) 27 (ii) 28 − 175 + 112 (iii) 5 × 125 × 8
2

3. MAT 163
Rationalization of the Denominator
3
When a fraction has a surd in its denominator, e.g. , it is usual to eliminate the surd in the
2
denominator. In fact, the writing of surds in the denominators of fractions should be avoided.
The process of removing this surd is called rationalizing of the denominator.
m + n and m − n are specially related surds known as conjugate surds. The product of
conjugate surds is always a rational number.
( m + n )( m − n ) = ( m )2 − ( n ) 2 = m − n
For example ( 9 + 5)( 9 − 5) = ( 9)2 − ( 5) 2 = 9 − 5 = 4
( 7 + 3)( 7 − 3) = ( 7)2 − ( 3)2 = 7 − 3 = 4
Example 3 Example 4
5 4
Simplify . Simplify .
3 7+ 3
Solution: Solution:
5 5 3 4 4 7− 3
= × = ×
3 3 3 7+ 3 7+ 3 7− 3
5 3 4( 7 − 3)
= =
3 7−3
= 7− 3
Example 5
1 1
Simplify, without using tables or calculators, the value of + .
3− 2 3+ 2
Solution:
1 1 (3 + 2) (3 − 2)
+ = +
3 − 2 3 + 2 (3 − 2)(3 + 2) (3 + 2)(3 − 2)
(3 + 2) + (3 − 2)
=
9−2
6
=
7
3

4. MAT 163
Try This 2
3 1 7 +2
Simplify (i) (ii) (iii)
12 3+ 7 7 −2
Answers to Try This
3
1. (i) 3 3 2. (i)
2
7− 3
(ii) 7 (ii)
4
11 + 4 7
(iii) 50 2 (iii)
3
4

5. MAT 163
Indices
If a positive integer a is multiplied by itself three times, we get a3 , i.e. a × a × a = a 3 . Here a is
called the base and 3, the index or power. Thus a 4 means the 4th power of a.
In general, a n means the nth power of a, where n is any positive index of the positive integer a.
Rules of Indices
There are several important rules to remember when dealing with indices.
If a, b, m and n are positive integers, then
(1) am × an = am + n e.g. 35 × 38 = 313
(2) am ÷ an = am − n e.g. 514 ÷ 53 = 511
(3) (a m )n = a m n e.g. (52 )6 = 512
(4) a m × b m = ( a × b) m e.g. 35 × 25 = 65
m 4
a 5
(5) a ÷b =  
m m
e.g. 5 ÷3 =  
4 4
b 3
(6) a0 = 1 e.g. 50 = 1
1 1
(7) a−n = e.g. 5 −3 =
an 53
1 1
(8) an = a
n
e.g. 83 =38
m 2
(9) an = a = ( a)
n m n m
e.g. 8 3 = 3 82 = ( 3 8)2
5

6. MAT 163
Example 1
1 3 3
−
−3
Evaluate (i) 2 (ii) 83 (iii) 16 4 (iv) 25 2
Solution:
1
−3 1 1
(i) 2 = 3= (ii) 83 =38=2
2 8
3 3
− 1
(iii) 16 4 = ( 16)
4 3
(iv) 25 2 = 3
25 2
1
= 23 =
( 25)3
1 1
=8 = 3=
5 125
Try This 1
Evaluate each of the following without using a calculator
3 2
−
−1 0 3
(i) 7 (ii) 17 (iii) 49 2 (iv) 8
4
3 1 − −2
5  1  3 1
(v) 243 (vi) 814 (vii)   (viii)  
 27  4
Example 2
1 2 1 1
Simplify (i) a3 × a5 ÷ a2 (ii) (a 3b 2 )4 (iii) 3
a ÷ 5 a 2 × ( a −1 ) 2
Solution:
1 2 1 1
−1 2
(i) a3 ×a 5 ÷ a2 (ii) 3 2 4
(a b ) (iii) 3
a÷ a 5 2
× (a )
1 2 1 1 2 1
+ − −
3×4 2×4
= a3 5 2 =a b = a3 ÷ a5 ×a 2
7 1 2 1
−
=a 30
=a b 12 8
= a3 ÷ a5 ×a 2
1 2  1
− + − 
=a 3 5  2
17
−
=a 30
Try This 2
Simplify each of the following, giving your answer in index form:
5 3 1 2
− −
(i) a 3 ÷ a −4 × a 2 (ii) 16a 2 ÷ 4a 2 (iii) (a 3 × b 5 )15
6

7. MAT 163
Solving Exponential Equations
Example 3
Solve the following exponential equations
(i) 2 x = 32 (ii) 4 x + 1 = 0.25
Solution:
(i) 2 x = 32 (ii) 4 x + 1 = 0.25
1
2 x = 25 4x + 1 =
4
x +1
∴x = 5 4 = 4− 1
x + 1 = −1
∴ x = −2
Try This 3
Solve the following equations:
1
(i) 3x = 81 (ii) 32 x = 8 (iii) 7x =
49
(iv) 5x = 1 (v) 34 x = 27 x + 3 (vi) 4 × 32 x = 6
x
Example 4
Solve the equation 22 x + 3 + 2 x + 3 = 1 + 2 x .
Solution:
22 x + 3 + 2 x + 3 = 1 + 2 x
2 x × 2 x × 23 + 2 x × 23 = 1 + 2 x
Let y = 2 x
8 y2 + 8 y = 1 + y
8 y2 + 7 y −1 = 0
(8 y − 1)( y + 1) = 0
1
y = or −1
8
1
When y= when y = −1
8
1
2x = 2 x = −1 (inadmissible)
8
2 = 2− 3
x
∴ x = −3
7

8. MAT 163
Try This 4
Solve the equation 32 x + 1 + 9 = 3x + 3 + 3x .
Example 5
1
If 3x × 92 y = 27 and 2 x × 4− y = , calculate the values of x and y.
8
Solution:
3x × 92 y = 27 ⋯⋯⋯ (1)
1
2 x × 4− y = ⋯⋯⋯ (2)
8
From (1): 3x × (32 ) 2 y = 33
3x × 34 y = 33
3x + 4 y = 33
x + 4y = 3 ⋯⋯ (3)
1
From (2): 2 x × (22 )− y =
23
2 x × 2 −2 y = 2−3
2 x − 2 y = 2 −3
x − 2 y = −3 ⋯⋯ (4)
(3) − (4) : 6y = 6
y =1
Substitute y = 1 into (3): x + 4(1) = 3
x = −1
∴ x = −1 and y = 1 .
Try This 5
Solve the simultaneous equations 3x + y = 243 , 22 x − 5 y = 8
8

9. MAT 163
Answers to Try This
1 1
1. (i) (ii) 1 (iii) 343 (iv)
7 4
(v) 27 (vi) 3 (vii) 81 (viii) 16
4
2. (i) a9 (ii) (iii) a 5b 6
a
3
3. (i) x=4 (ii) x= (iii) x = −2
5
1
(iv) x=0 (v) x=9 (vi) x=
2
4. x = −1 or 2
5. x = 4, y = 1
9

10. MAT 163
Logarithms
Definition:
For any number y such that y = a x ( a > 0 and a ≠ 1 ), the logarithm of y to the base a is defined
to be x and is denoted by log a y .
Thus if y = a x , then log a y = x
For example, 81 = 34 ∴ log 3 81 = 4
100 = 10 2
∴ log10 100 = 2
Note: The logarithm of 1 to any base is 0, i.e. log a 1 = 0 .
The logarithm of a number to a base of the same number is 1, i.e. log a a = 1 .
The logarithm of a negative number is undefined.
Example 1
Find the value of (i) log 2 64 (ii) log9 3
1
(iii) log 3 (iv) log8 0.25
9
Solution:
(i) Let log 2 64 = x (ii) Let log 9 3 = x
64 = 2 x
3 = 9x
26 = 2 x 3 = 32 x
∴x = 6 1 = 2x
1
∴x =
2
1
(iii) Let log 3 =x (iv) Let log8 0.25 = x
9
1
= 3x 0.25 = 8 x
9
1
3−2 = 3x = 23 x
4
−2
∴ x = −2 2 = 23 x
3x = −2
2
∴x = −
3
10

11. MAT 163
Laws of Logarithms
(1) log a mn = log a m + log a n e.g. log3 5 + log 3 2 = log 3 10
m 5
(2) log a = log a m − log a n e.g. log3 5 + log 3 4 = log 3  
n 4
(3) log a m p = p log a m e.g. log10 5 = 2 log10 5
2
Example 2
41 41
Without using tables, evaluate log10 + log10 70 − log10 + 2 log10 5 .
35 2
Solution:
41 41
log10 + log10 70 − log10 + 2 log10 5
35 2
 41 41 
= log10  × 70 ÷ × 52 
 35 2 
= log10 100
= log10 102
= 2 log10 10
=2
Try This 1
Simplify 2 log 3 5 − log 3 10 + 3log 3 4 .
Changing the Base of Logarithms
Logarithms to base 10 such as log10 5 and log10 ( x + 1) are called common logarithms. An
alternative form of writing log10 5 is lg 5 .Common logarithms can be evaluated using a
scientific calculator.
Logarithms to base e such as log e 3 and log e x are called Natural logarithms or Napierian
logarithms. Natural logarithms are usually written in an alternative form, for example, log e 3 is
written as ln 3 . (Note: e = 2.718... )
log c b
If a, b, and c are positive numbers and a ≠ 1 , then log a b = .
log c a
11

12. MAT 163
Example 3
Find the value of log5 16 .
Solution:
log10 16 1.204
log5 16 = = = 1.722
log10 5 0.699
Try This 2
Find the value of log 4 54 .
Solving Logarithmic Equations
Example 4
Solve the equation 3x = 18 .
Solution:
3x = 18
Taking logarithms to base 10 on both sides,
log10 3x = log10 18
x log10 3 = log10 18
log10 18 1.2553
x= =
log10 3 0.4771
= 2.631
Try This 3
Solve the equation 5 x + 1 = 30 .
Example 5
Given that log10 4 + 2 log10 p = 2 , calculate the value of p without using tables or calculators.
Solution:
log10 4 + 2 log10 p = 2
log10 (4 × p 2 ) = 2
4 p 2 = 102
100
p2 =
4
p = 25
2
p = ±5
Since p cannot be −5 because log10 (−5) is not defined, p = 5 .
12

13. MAT 163
Try This 4
3x + 1
Solve the equation log 2 = 3.
2x − 7
Example 6
Solve the equation log10 (3x + 2) − 2 log10 x = 1 − log10 (5 x − 3) .
Solution:
log10 (3x + 2) − 2 log10 x = 1 − log10 (5 x − 3)
log10 (3 x + 2) − log10 x 2 + log10 (5 x − 3) = 1
(3x + 2)(5 x − 3)
log10 =1
x2
(3 x + 2)(5 x − 3)
2
= 101
x
15 x − 9 x + 10 x − 6 = 10 x 2
2
5x2 + x − 6 = 0
(5 x + 6)( x − 1) = 0
6
∴ x = − or x = 1
5
Since x cannot be negative, x = 1 .
Try This 5
Solve the equation log 2 x 2 = 4 + log 2 ( x − 3) .
Answers to Try This
1. log3 160
2. 2.877
3. x = 1.113
57
4. x=
13
5. x = 4 or 12.
13