This document discusses surds, indices, and logarithms. It begins by defining radicals, surds, and irrational numbers. Some general rules for operations with surds like multiplication, division, and simplification are provided. The document then covers rules and operations for indices like exponentiation, roots, and properties like distributing exponents. Examples are given to demonstrate applying these index rules. The document concludes by defining logarithms as the inverse of exponentiation and provides an example equation.

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