Page 13 - Spotlight A+ SPM Additional Mathematics Form 4 & 5

Form
4 Additional Mathematics Chapter 4 Indices, Surds and Logarithms

4.1 Laws of Indices 1
2
(c) × 32 = 3 × 3
−4
Simplifying algebraic expressions 81 = 3 −4 + 2

involving indices using the laws of indices = 3 −2
1
1. A number can be written in index notation or = 3 2
index form as a where a is a base and n is an 1
n
index or power. = 9
2. An algebraic expression can be simplified using 1 1
the laws of indices. (d) 2 × 32 ÷ 16 = 2 × (2 ) ÷ 2
5
5
5
5 5
4
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1
4
5
3. The following indices should be known first = 2 × 2 ÷ 2
5 + 1 − 4
before we simplified the algebraic expressions = 2 2
= 2
CHAP using the laws of indices. = 4
4 • a = 1 Try question 1 in Formative Zone 4.1
–1
a (Negative index)
• a = 1 (Zero index) Calculator
0
1
n
• a = a (Fractional index) Based on Example 1(a), the algebraic expression
n
5
2 × 16 can be evaluated using calculator as follows:
−2
m
• a = ( a) m (Fractional index)
n
n
1. Press 2 ^ 5 × 16 ^ (–) 2 =
4. The following shows the laws of indices: 2. The answer will be displayed as 0.125 .
b
• a × a = a n + m 3. Press SHIFT a  to convert the answer to
c
n
m
• a ÷ a = a n – m fraction 1 8 .
m
n
• (a ) = a nm
n m
• (ab) = a b Example 2
n n
n
a n
•   = a b n n Simplify the following algebraic expressions.
b
3
(a) 3 × 27 3 − n (b) 2 n + 1 ÷ 4 × 16 4
2
2
(2p) × p −2 (3x y )
3
3 4 2
Example 1 (c) pq 3 (d) 12x y
5
By using the laws of indices, evaluate the
following algebraic expressions. Solution:
2
2
3 3 − n
(a) 2 × 16 (b) (5 ÷ 5 ) (a) 3 × 27 3 − n = 3 × (3 )
−3
−4 2
−2
5
2
1 1 = 3 × 3 9 − 3n
(c) × 32 (d) 2 × 32 ÷ 16 = 3 2 + 9 − 3n
5
5
81
= 3 11 − 3n
Solution: 3 3
(b) 2 n + 1 ÷ 4 × 16 = 2 n + 1 ÷ (2 ) × (2 )
4
2
2 2
4 4
4 −2
(a) 2 × 16 = 2 × (2 ) = 2 n + 1 − 4 + 3
−2
5
5
= 2 × 2 = 2 n
5
−8
= 2 5 + (−8) a × a = a m+n (2p) × p −2 2 p × p −2
m
n
3
3
3
−3
= 2 (c) pq 3 = pq 3
1
= 8p × p −2
3
2 3 =
1 pq 3
8 = 8p 3−2
pq 3
(b) (5 ÷ 5 ) = (5 ) ÷ (5 )
−3
−3 2
−4 2
−4 2
= 5 ÷ 5 −8 = 8p 3
−6
= 5 −6 − (−8) a ÷ a = a m−n pq
m
n
= 5 2 = 8
= 25 q 3
60 4.1.1