Page 14 - Spotlight A+ SPM Additional Mathematics Form 4 & 5

Form
4
Chapter 4 Indices, Surds and Logarithms Additional Mathematics

3 2
4 2
(3x y ) 3 (x ) (y ) (a b ) = a b np Solution:
3 4 2
2
mp
n p
m
(d) =
5
12x y 12x y (a) 7 x + 2 − 7 x + 1 = 6
5
9x y 7 ·7 − 7 ·7 = 6
1
x
x
2
6 8
=
12x y 7 (7 − 7 ) = 6
2
x
1
5
x
3 7 (49 − 7) = 6
y
= x 6 − 5 8 − 1 x
4 7 (42) = 6
3 x 6
= xy 7 7 =
4 42
1 Express the equation
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x
7 = of both sided in the
7 same base
Try question 2 in Formative Zone 4.1
7 = 7 −1
x
x = −1 CHAP
4
Example 3
x
–1
Simplify 2 + 2 n + 2 − 2 n + 1 in the form of k(2 ) • If 7 = 7 , then x = –1.
n
n
7
7
where k is a constant. Hence, determine the • If x = 4 , then x = 4.
value of k.
Solution: (b) 81·(3 x + 2 ) = 1
2 + 2 n + 2 − 2 n + 1 = 2 + 2 ·(2 − 2 ) 3 ·(3 x + 2 ) = 1
n
1
2
4
n
n
n
0
1
= 2 (1 + 2 − 2 ) 3 4 + x + 2 = 1 3 = 1
2
= 2 (1 + 4 − 2) 3 6 + x = 3°
n
= 2 (3) 6 + x = 0
n
= 3(2 ) x = −6
n
Thus, k = 3.
Al te rnative Me thod
Alternative Method
Let u = 2 , then
n
n
2 + 2 n + 2 − 2 n + 1 = 2 + 2 ·2 − 2 ·2 1 Change 1, 3 and 81 to a common base, that
n
2
n
n
= u + 4u − 2u is 3. Hence, equate the indices to solve for x.
= 3u
= 3(2 )
n
Try question 5 in Formative Zone 4.1
Try question 3 and 4 in Formative Zone 4.1
Example 5
Show that 5 n + 3 − 5 − 5 n + 2 is divisible by 11
n
Solving problems involving indices for all positive integers of n.
1. When solving an equation involving indices,
we should consider the following statement: Solution:
n
n
3
n
If a = a , then m = n or if a = b , then 5 n + 3 − 5 − 5 n + 2 = 5 ·5 − 5 − 5 ·5 2
n
n
m
m
m
a = b when a > 0 and a ≠ 1. = 5 (5 − 1 − 5 )
n
2
3
= 5 (125 − 1 − 25)
n
= 5 (99)
n
Example 4
Since 5n(99) is divisible by 11. Thus,
Solve the following equations. 5 n + 3 − 5 − 5 n + 2 is divisible by 11.
n
(a) 7x + 2 – 7x + 1 = 6
(b) 81·(3x + 2) = 1 Try question 6 in Formative Zone 4.1
4.1.1 4.1.2 61