Surds, Indices, and Logarithms Radical
MAT 163
Surds, Indices, and Logarithms Radical Definition of the Radical
For all real , , and all integers a ,
x y > >
a a
x y if and only if x y
= =
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.
3 Examples of irrational numbers are 2 , 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 a means the positive square root of a while − 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
× = × = =
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
5 (3 1) 5 (i) = + (ii) 40 = 4 10 × = 4 ×
- 3 5
10
= 4 5 = 2 10 Example 2
12 2 75
- Simplify (i) 243 −
50
8
- (ii)
32 Solution : 12 2 75 = 81 3 × − 4 3 × + 2 25 3 ×
- (i) 243 − =
81 × + 3 − 4 × 3 2 25 ×
3
= 9 3 − ×
2 3 10 + ×
3
= − + (9 2 10) 3
17 3 =
50
- (ii)
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 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.
2
2
( m n )( m n ) = ( m ) − ( n ) = − m n
− +
2 For example
( 9 5)( 9 − 5) = ( 9) − ( 5) = − = 9 5
- 2
4
2
( 7 3)( 7 − 3) = ( 7 ) − ( 3) = − = 7 3
- 2
4 Example 3 Example 4
5
4 Simplify . Simplify .
7
- 3
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 .
2
3
2
− Solution:
- 3
1 (3 2) (3 − 2)
- 1
- =
3 −
2
3 + 2 (3 − + + 2)(3 2) (3 2)(3 − 2)
- (3 2) (3 + − 2) =
9 2 −
6
=
7
- −
3
(iii) 50 2 (iii) 11 4 7
−
4
3
7
2 (ii) 7 (ii)
3
1. (i) 3 3 2. (i)
Answers to Try This
2
7
2
7
7
3
1
12 (ii)
3
Simplify (i)
Try This 2
MAT 163
+
(iii)
MAT 163
Indices
3
3 If a positive integer a is multiplied by itself three times, we get a , i.e. a a a a . Here a is × × =
4 th
called the base and 3, the index or power. Thus a means the 4 power of a.
n In general, a 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
5
8
- m n m n
13 (1) a × a = a e.g.
3 × =
3
3
m n m n
14
3
11 − (2) a ÷ a = a e.g.
5 ÷ =
5
5
m n m n 2 6
12
(3) ( a ) = a e.g. (5 ) =
5
m m m
5
5
5 (4) a × b = × ( a b ) e.g.
3 × =
2
6
m
4 a
5
m m
4
4 a b
5
3 (5) e.g.
÷ = ÷ = b
3 (6) a = 1 e.g. 5 =
1
1
1
n
3 −
− a e.g.
5 (7) =
= n
3 a
5
1
1 n
3 n
8
3 (8) a a e.g.
8
= = m
2 n
3 m n m
2
3
2 n
3
(9) a = a = ( a ) e.g. 8 = 8 = ( 8)
MAT 163
Example 1
1
3
3 −
3 −
3
4
2 Evaluate (i)
2 (ii) 8 (iii) 16 (iv)
25 Solution :
1
1
1
3 −
3
3
2 (ii)
8
8
2 (i) = = = =
3
8
2
3
3
−
1
3
4
4
2
(iii) 16 ( 16) (iv)
25
= =
3
2
25
1
3
2
= =
3
( 25)
1
1
8
= = =
3
5 125
Try This 1
Evaluate each of the following without using a calculator
3
2 − −
1
3
2
(i) (ii) (iii) (iv)
7
17
49
8
4
3 1 − −
2
3
1
1
5
4
(v) (vi) (vii) 243 81 (viii)
27
4 Example 2
1
2
1 1 3 2 4
5
2
1 3 −
3
5
2
2
( a b ) Simplify (i) a × a ÷ a (ii) (iii) a ÷ a × ( a )
Solution :
1
2
1 1 3 2 4
5
2
1 3 −
3
5
2
2
(i) a × a ÷ a (ii) ( a b ) (iii) a ÷ a × ( a )
1
2
1
1
2
1 − +
− 3 4 2 4 × ×
3
5
2
3
5
2 = a = a b = a ÷ a × a
7
1
2
1 −
30 12 8
3
5
2 a a b a a a
= = = ÷ ×
1
2
1 − + −
3 5 2 = a
17 −
30 = a
Try This 2
Simplify each of the following, giving your answer in index form:
5
3
1
2 − −
3 −
4
2
15
2
2
3
5
(i) a ÷ a × a (ii) 16 a ÷ 4 a (iii) ( a × b )
MAT 163
Solving Exponential Equations Example 3
Solve the following exponential equations
1
- x x
(i) 2 = 32 (ii) 4 =
0.25 Solution:
1 + x x
(i)
2 32 (ii)
4
0.25
= =
1
5 x
1
- x
2
2
4
= =
4
1
1
− x- x
5
4
4
∴ = = x
1
1
- = −
x
2
∴ = − Try This 3
Solve the following equations:
1
x x x
(i)
3 81 (ii)
32 8 (iii)
7
= = =
49
4 x x 3 x 2 x
- x
(iv)
5 1 (v)
3 27 (vi)
4
3
6
= = × =
Example 4Solve the equation
2 2 1 2 .
- 2 3 x x
- 3 x
= + Solution:
2 3 x x 3 x
2 1 2
- 2
= + x x 3 x 3 x
2 × × + × = +
2
2
2 2 1 2
x
Let y =
2
8 y 8 y 1 y
- 2
= +
8 y 7 y − =
- 2
1 (8 y − 1)( y + = 1)
1
y = or 1 −
8
1
y y
1 When = when = −
8
x
1
x
1 2 = 2 = − (inadmissible)
8
x −
3
2
2
= x
3
∴ = −
MAT 163
Try This 4
3 x Solve the equation .
3 + =
9
3
- 2 1 x x
3 Example 5
1
x 2 y x − y
2 4 , calculate the values of If 3 × 9 = 27 and × = x and y.
8 Solution:
x 2 y
3 × 9 = 27 (1)
⋯⋯⋯
1
x − y
2 × 4 = (2)
⋯⋯⋯
8
x 2 2 y
3 From (1):
3 × (3 ) =
3
x 4 y
3
3
3
3
× =
4 y
3
- x
3 =
3
- x
4 y = 3 (3)
⋯⋯
1
x 2 − y
From (2): 2 (2 )
× =
3
2
x 2 y
3 − −
2 × 2 =
2
x 2 − y −
3
2
2
= x −
2 y = − 3 (4)
⋯⋯
(3) (4) : y
6
−
6 =
y
1
=
1 into (3): x 4(1)
3
= = x
- Substitute y
1
= − x
1 y
1 and .
∴ = − = Try This 5
x y 2 5 x y
−
Solve the simultaneous equations , 3 = 243 2 =
8
MAT 163
x =
3
5
x =
(iii) 2 x
= −
(iv)
(v) 9 x
=
=
(vi)
1
2
x = 4.
1 x
= −
(ii)
3. (i) 4 x
Answers to Try This
4 (v)
1. (i)
1
7 (ii)
1 (iii)
343 (iv)
1
27 (vi)
5 6 a b
3 (vii)
81 (viii) 16
2. (i)
9 a (ii)
4
a
(iii)
or 2 5. 4, = 1 x y =
MAT 163
Logarithms Definition: x a a
1 For any number y such that y = a ( > and ≠ ), the logarithm of y to the base a is defined to be x and is denoted by log y .
a x y a
Thus if , then log y = x
= a
4 For example, 81 3 = ∴ log 81 =
4
3
2
100 10 = ∴ log 100 =
2
10 Note: The logarithm of 1 to any base is 0, i.e. log 1 = .
a
The logarithm of a number to a base of the same number is 1, i.e. log a = 1 .
a The logarithm of a negative number is undefined.
Example 1
Find the value of (i) log 64 (ii) log 3
2
9
1 log log 0.25 (iii) (iv)
3
8
9 Solution: (i) Let log 64 (ii) Let log 3
= x = x
2
9 x x
2
9 64 = 3 =
6 x 2 x
2 =
2 3 =
3
x
6 2x
∴ =
1 =
1
x ∴ =
2
1
x x
(iii) Let log = (iv) Let log 0.25 =
3
8
9
1
x x
3
0.25
8
= =
9
1
− 2 x 3 x
3
3
2
= =
4
2 3 x − x
2
2
2
∴ = − = x
2
3
= −
2
x ∴ = −
3
MAT 163
Laws of Logarithms
(1) log mn log m log n e.g. log 5 log 2 log 10
= + + = a a a
3
3
3 m
5
(2) = − log 5 log 4 = log
- log log m log n e.g.
a a a
3
3
3 n
4
p
2
log m p log m e.g. log 5 2 log 5 (3) =
= a a
10
10 Example 2
41
41 Without using tables, evaluate log log 70 log 2 log 5 .
− + +
10
10
10
10
35
2 Solution:
41
41
- log log 70 log − 2 log 5
10
10
10
10
35
2
41 41
2
log
70
5
= × ÷ × 10
35 2 log 100
=
10
2 = log 10
10 = 2 log 10
10 =
2 Try This 1 + Simplify 2 log 5 log 10 3log 4 − .
3
3
3 Changing the Base of Logarithms
Logarithms to base 10 such as log 5 and log ( 1) are called common logarithms. An
x
10
10
alternative form of writing log 5 is lg 5 .Common logarithms can be evaluated using a
10 scientific calculator. x
Logarithms to base e such as log 3 and log are called Natural logarithms or Napierian
e e
logarithms . Natural logarithms are usually written in an alternative form, for example, log 3 is
e e 2.718...written as ln 3 . (Note: = ) log b
c a
1 If a, b, and c are positive numbers and ≠ , then log b = .
a
log a
c
MAT 163
Example 3 Find the value of log 16 .
5 Solution:
log 16 1.204
10
log 16 = = = 1.722
5
log 5 0.699
10 Try This 2 Find the value of log 54 .
4 Solving Logarithmic Equations Example 4 x
Solve the equation 3 18 .
= Solution : x
3 =
18 Taking logarithms to base 10 on both sides,
x
log 3 log 18
=
10
10
log 3 log 18
x =
10
10
log 18 1.2553
10 x
= =
log 3 0.4771
10
2.631
= Try This 3
1 + x
Solve the equation
5 30 .
= Example 5
Given that log 4 2 log 2 , calculate the value of + p = p without using tables or calculators.
10
10 Solution:
2
- log 4 2 log
p =
10
10
2
log (4 p )
2
× =
10
2
2
4 p =
10 100
2 p
=
4
2 p
25
= p
5
= ±
because log ( 5) is not defined, p 5 . Since p cannot be 5 − − =
10
MAT 163
Try This 4
3 x
1 log 3 . Solve the equation
- =
2
2 x
7
− Example 6 Solve the equation log (3 2) 2 log 1 log (5 3) . x + − x = − x −
10
10
10 Solution:
log (3 2) 2 log 1 log (5 3)
x x x
- − = − −
10
10
10
2
log (3 x + − + 2) log x log (5 x − = 3) 1
10
10
10
- (3 x 2)(5 x 3) log
1
− =
10
2 x
- (3 x 2)(5 x 3)
1
10
− =
2 x
2
15 x − 9 x 10 x − = 6 10 x
- 2
2
5 x + − = x
6
(5 − =
- x 6)( x 1)
6
x
1
∴ = − x or =
5
x
1 Since x cannot be negative, = .
Try This 5
2 Solve the equation log x = + 4 log ( x − 3) .
2
2 Answers to Try This
1. log 160
3
2. 2.877
x 1.113
3. =
57 4. x =
13
x
4 5. = or 12.