Indices and standard form

# 10 2 10 3 10 6 10 2 f 5 4 52 a without using a

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Unformatted text preview: 0 6 ÷ 10 2 (f) 5 4 ÷ 52 (a) Without using a calculator, write down the values of k and m. 64 = 8 2 = 4 k = 2 m (b) Complete the following: 2 15 = 32 768 2 14 = (KS3/99Ma/Tier 5-7/P1) 3.2 Laws of Indices There are three rules that should be used when working with indices: When m and n are positive integers, 1. am × an = am + n 2. a ÷a = a 3. (a ) m m n n m−n am or n = a m − n a ( m ≥ n) = am × n These three results are logical consequences of the definition of a n , but really need a formal proof. You can 'verify' them with particular examples as below, but this is not a proof: 2 7 × 2 3 = (2 × 2 × 2 × 2 × 2 × 2 × 2 ) × (2 × 2 × 2) = 2×2×2×2×2×2×2×2×2×2 = 2 10 (here m = 7, n = 3 and m + n = 10 ) 40 MEP Y9 Practice Book A or, 27 ÷ 23 = 2×2×2×2×2×2×2 2×2×2 = 2×2×2×2 = 24 (2 ) 7 Also, 3 (again m = 7, n = 3 and m − n = 4 ) = 27× 27× 27 = 2 21 (using rule 1) (again m = 7, n = 3 and m × n = 21) The proof of the first rule is given below: Proof am× an = a × a × ... × a × a × a × ... × a 14 244 4 3 14 244 4 3 m of these n of these = a × a × ... × a × a × a ... × a 14444 244444 4 3 (m + n) of these = am+n The second and third rules can be shown to be true for all positive integers m and n in a similar way. We can see an important result using rule 2: xn = xn − n = x0 n x but xn = 1, so xn x0 = 1 This is true for any non-zero value of x, so, for example, 30 = 1, 270 = 1 and 10010 = 1 . 41 MEP Y9 Practice Book A 3.2 Example 1 Fill in the missing numbers in each of the following expressions: (a) 24× 26= 2 (b) 37× 39 = 3 (c) 36 ÷ 32 = 3 (d) (10 ) (b) 37× 39 = 37 + 9 4 3 = 10 Solution (a) 24× 26 = 24+6 = 2 10 (c) = 3 16 36 ÷ 32 = 36 − 2 (d) = 34 (10 ) 4 3 = 10 4 × 3 = 10 12 Example 2 Simplify each of the following expressions so that it is in the form a n , where n is a number: a4× a2 3 6 7 (a) a × a (b) (c) ( a 4 ) 3 a Solution (a) a6× a7 = a6+7 = a 13 (b) a4× a2 a4+2 = a3 a3 = a6 a3 = a6−3 = a3 (c) (a ) 4 3...
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## This document was uploaded on 03/03/2014 for the course O-LEVEL Mathematic at Beaconhouse School System.

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