INDICES Concept of Indices Repeated Multiplication
Example:
1) 5 3 =
2) 25 =
3) 8 6 =
23 = 2 × 2 × 2 b4 = b × b × b × b
4) a 2 =
5) k 5 =
6) m 4 =
Example : Write the following integers in index notation. 8 = 2 × 2 × 2 = 23
1) 125=
2) 32=
a0 = 1
1) 30 =
2) (−5)0 =
2 3) ÷ = 5
4) a −3 =
5) 4− m =
2 6) ÷ = 3
7) 22 × 20 =
8) (−4)3 × (−4) 2 =
10) −4a 3 × 2a 5 =
11) 4a 3 × a 5 =
13) 7 8 ÷ 7 5 =
14) k 8 ÷ k −2 =
16) 3m6 ÷ 27 m3 =
17) −22n3 ÷ 4n3 =
18) 18m2 ÷ 27 m3 =
19) (9 3 ) =
20) ( r 4 ) =
21) ( m −4 ) =
Index notation
3)
25 = 49
Laws of Indices
Example 40=1 a−m =
1 am
−3
Example 1 23 a m × a n = a m+ n
0
2−3 =
Example 2 3 × 2 4 = 23 + 4 = 2 7
Example 4
5
4 +5
2a × 3a = 6a = 6a 9
a m ÷ a n = a m −n
Example 23 ÷ 2 2 = 23− 2 = 21 = 2 Example 8a 5 8a 5 ÷ 2 a3 = 3 2a = 4 a 5 − 3 = 4a 2
(a ) m
n
Indices
= a mn
1 2
5
4
1
2
3
2
2
9) ÷ × ÷ = 3 3
12)
1 4 2 5 a × a = 2 3
2
4
2
2
15) ÷ ÷ ÷ = 5 5
3
Example
(2 )
2 3
= 2 2×3 = 2 6
22) ( 3m 4 ) =
Example
( 2r ) 4
3
3
(m n )
= 23 r12
4 2
= 8r12 3
25) ( r 4 n 2 )
= m12 n6
−2
2
=
n
am = (
n
5
1
1
27) ( a −4b 2 ) 4 =
2
28) 8 3 =
a )m
24) ( 2m 4 n3 ) =
2
26) ( m 4 n 2 ) 2 =
=
5
m an
23) m 2 ÷ = 3
2
29) 125 3 =
30) 64 3 =
Example : 3
42 = (
2
4)3
= 23 = 8 Example: 1 (4m −1n 2 ) 2
2
1
2 −2 1 = (4 m n )
31) (3m −2 n 4 )2 =
32) (2m 3 n −2 )3 =
1
33) (27 a 3b −6 ) 3 =
= 16m −2 n
A.
Questions Based on Examination Format
1. m
3 4
C. -
can also be written as
A. 4 m 3 C. ( m 4 )3
B. a6 D. a27
3. 76 ÷ 72 = A. 72 C. 74
B. 73 D. 75
1 9
D. -9
7. 3 x 32 x 23 = A. 24 B. 144 C. 216 D. 648 8. Simplify (m3)2 ÷ m4. A. m B. m2 9 C. m D. m10 9. 52 x 5 x 54 = A. 56 B. 57 C. 58 D. 59
4. Simplify (r4)2 ÷ r3 = A. r B. r3 5 C. r D. r13 3
5. 81 2 x 9-2 =
Indices
B. 9
6. Given that 3125 = 5n, find the value of n. A. 4 B. 6 C. 5 D. 7
B. 3 m 4 D. ( m 3 ) 4
2. a9 ÷ a3 = A. a3 C. a12
1 9
10. Simplify a3 x a4
2
÷
a5 =
A. a2 C. a7
B. a4 D. a 12
C. 48
11. Given that y4 = 81, find the value of y
1 1 1 1 1 x x x x = 2 2 2 2 2 1 1 A. B. 32 16 5 5 C. D. 16 32
19.
1 3 1 D. 9
A. 3
B.
C. 9 c8 = c2 A. c2 C. c6
D. 415
12.
20. The value of 5-3 = A. 125 B. -125
B. c4 D c7
C.
1 125
D. -
1 125
13. 2 x 23 = p, find the value of p A. 4 B. 8 C. 12 D. 16 14. (a4b2)2 = A. a6b2 C. a8b2
Past Year SPM questions November 2003
B. a6b4 D. a8b4
1
15. The value of ( -0.3)3 is = A. 0.27 B. - 0.27 C. 0. 027 D. – 0.027 1
1
A B
1 p
3 C. ( )
17. 5 a2 x a4 x 4a2 = A. 5a8 C. 20a6
B. p
2
C
4mn p3
D
16mn 2 p6
3
Find the value of 1
1
D. ( p )6
B. 4a6 D. 20a8
3 5
Indices
m2 p3 p6
(8 × 3 −6 ) 3 ÷ (2 3 × 3 −4 ) 1 A C 4 4 9 1 B D 9 9 4 July 2004 3 Simplify (mp 2 )3 × p −4 . A mp C m3p 4 B mp D m3p2
3
4 18. (4 ) = A. 43
4n p3 16n
p
1
16. p x p x p , expressed in index notation is 3 A. p
1
Simplify (4m −1 n 2 ) 2 ×
B. 45 3
4 3
2 4 Evaluate 27 × (3 ) . 812
A B
3 27
C D
9 81
November 2004
5
Simplify
m 6 × (9e 2 ) 3 6
A B
C
B
1
C
(m e ) C
9m 2 e 9m 3 e
1 1
D
.
1 2
3m 5 e
3m 5 e4
1 12 4 −5 mn ÷ ÷ m n 2 = A mn C m-1n4 4 B mn D m-3n4
(
)
November 2005 9 Simplify (m 2 ) 4 ÷ m 5
A B
m m3
C D
m11 m13
9 3 x = 2 x , find the value Given that 10 3 of x.
Indices
3
3p 3p2
C D
4p 4p2
November 2006
Which of the following is equivalent to 2-3 ? A C 1 1 3 2 32 B D 1 1 − 3 − 2 2 3 3
Simplify p × ( 2 p −1 ) ÷ ( 2 p −4 ) . A B
1 2
1 3
2 3 1 2
July 2006
July 2005
8
2
3
r 4 can also be written as A 4 r3 C ( r 4 )3 B 3 r4 D ( r 3 )4
6
7
A
4
2 13 8 ×7 Simplify 2 14 A 24 × 7 −3 B 26 × 7 −2
3
÷. ÷ ÷ C 212 × 7 −5 D 216 × 7 −1