Lecture notes, lecture index laws PDF

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Maths Learning Service: Revision

Mathematics IA

Index Laws

Mathematics IMA Intro. to Fin. Maths I

Index laws are the rules for simplifying expressions involving powers of the same base number.

am × an = am+n (am )n = amn am = am−n an 1 a−m = m a

First Index Law Second Index Law Third Index Law

a0 = 1 1

an =

√ n

a

Examples: Simplify the following expressions, leaving only positive indices in the answer. (a)

36 24 34 36 = 4 × 24 3 = 32 24

Notes:

(1)

3

32 × 3−5

(b)

= 3−3 1 33 1 = 27 =

(c)

9 (x2 ) 3xy 2 1 9 x6 × 2 = × y x 3 1 = 3 × x5 × 2 y 3x5 = 2 y

(d)

√ a−1 a 1

= a−1 a 2 1

= a− 2 =

1 1 1 or √ a a2

More involved fractional powers can be dealt with by noting that 1 m a n = (a n )m by the Second Index Law. For example, √3 1 2 2 (27) 3 = (27 3 ) = ( 27)2 = (3)2 = 9.

(2)

Watch out for powers of negative numbers. For example, (−2)3 = −8 and (−2)4 = 16, so (−x)5 = −x5 and (−x)6 = x6 .

Index Laws (3)

2007 Maths IA, IMA & Intro. Fin. Maths I Revision/2 In general (ab)n = an bn . For example, (3x2 y )3 = 33 (x2 )3 y 3 = 27x6 y 3 .

Exercises 1. Simplify the following expressions, leaving only positive indices in the answer. (a)

42 × 4−3

(b)

32 (22 ) 23

(d)

(y 4 )6

(e)

(g)

x2 z −3 × (xz 2 )2

(j)

(a2 × a)5

(m)

x−1 y 4 x−5 y −3

1

2

(p)

(a ×

(s)

3 5

√

−2

(c)

x5 x8

(−3)3

(f)

(4ab2 c)3

(h)

2n × (2−n )3 × 22n

(i)

3m × 27m × 9−m

(k)

(−2ab)2 2b

(l)

(−a4 b)3 (ab)5 −a8 b8

(o)

√ x3x

(r)

(3a)−1 × 3a−1

(u)



10a3 b−2 5a−1 b2

(n)

!−1

1

a)

2

2x2 x x2

(q) (t)

32



4 25

3 2

1

43



1

23



Terms involving the “√ ” symbol are known as a radicals or surds. √ √ √ √ √ Notes: (1) a + b 6= a + b . For example 144 + 25 = 169 = 13 √ √ but 144 + 25 = 12 + 5 = 17. √ √ √ (2) Similarly, a − b 6= a − b. √ √ √ √ √ √ √ (3) ab = a × b . For example 4 × 9 = 36 = 6 and 4× 9 = 2 × 3 = 6. s √ √ r 16 √ 4 a a 16 (4) = √ . For example = 4 = 2 and √ = = 2. b 2 4 4 b These techniques can be used to simplify radicals. For example √

√ √ 9× 2 = 3 2. √ √ √ 75 = 25 × 3 = 5 3 . 18 =

√

9×2 =

√

When asked to simplify radical expressions involving fractions, you are required to produce a single fraction (as in ordinary algebra) with no radicals in the denominator. For example

Index Laws

2007 Maths IA, IMA & Intro. Fin. Maths I Revision/3 √ √ 3 2 3 √ +√ = √ × 2 6 3 3 = √ + 6 5 = √ 6

√ 2 √3 + √ 6 2 2 √ 6

√ 6 5 = √ ×√ 6 6 √ 5 6 = 6

Exercises (continued) 2. Simplify the following expressions √ √ (a) 50 (b) 72 √ √ 2 3 1 1 (e) √ + √ (d) √ − √ 10 3 5 2

(c) (f)

√

12 +

√

27 √ 2 3 1 √ −√ 3 15

Index Laws

2007 Maths IA, IMA & Intro. Fin. Maths I Revision/4

Answers to Exercises 1. (a)

1 4

(b)

9 9 = 7 2 127

(c) x13

(d) y 24

(e) −27

(f) 64a3 b6 c3

(g) x4 z

(h) 1

(i) 32m

(j) a15/2

(k) 2a2 b

(l) a9

(m) x4 y 7

(n) 21 a−4 b4

(o) x4/3

(p) a5

(q) 2x−1/2

(r) a−2

(s) 8

(t)

√ (b) 6 2 √ √ 5 3−6 5 (f) 15

√ (c) 5 3

(u) 2 √ 2. (a) 5 2 √ 5 6 (e) 6

√ √ 2 5 − 10 (d) 10

8 125...