NCERT Class 7 Mathematics

(Ron) #1

252 MATHEMATICS


(ii) 432 = 2 × 216 = 2 × 2 × 108 = 2 × 2 × 2 × 54
= 2 × 2 × 2 × 2 × 27 = 2 × 2 × 2 × 2 × 3 × 9
= 2 × 2 × 2 × 2 × 3 × 3 × 3
or 432 = 2^4 × 3^3 (required form)
(iii) 1000 = 2 × 500 = 2 × 2 × 250 = 2 × 2 × 2 × 125
= 2 × 2 × 2 × 5 × 25 = 2 × 2 × 2 × 5 × 5 × 5
or 1000 = 2^3 × 5^3
Atul wants to solve this example in another way:
1000 = 10 × 100 = 10 × 10 × 10
= (2 × 5) × (2 × 5) × (2 × 5) (Since10 = 2 × 5)
= 2 × 5 × 2 × 5 × 2 × 5 = 2 × 2 × 2 × 5 × 5 × 5
or 1000 = 2^3 × 5^3
Is Atul’s method correct?
(iv) 16,000 = 16 × 1000 = (2 × 2 × 2 × 2) ×1000 = 2^4 ×10^3 (as 16 = 2 × 2 × 2 × 2)
= (2 × 2 × 2 × 2) × (2 × 2 × 2 × 5 × 5 × 5) = 2^4 × 2^3 × 5^3
(Since1000 = 2 × 2 × 2 × 5 × 5 × 5)
= (2 × 2 × 2 × 2 × 2 × 2 × 2 ) × (5 × 5 × 5)
or, 16,000 = 2^7 × 5^3
EXAMPLE 6 Work out (1)^5 , (–1)^3 , (–1)^4 , (–10)^3 , (–5)^4.
SOLUTION
(i) We have (1)^5 = 1 × 1 × 1 × 1 × 1 = 1
In fact, you will realise that 1 raised to any power is 1.
(ii) (–1)^3 = (–1) × (–1) × (–1) = 1 × (–1) = –1
(iii) (–1)^4 = (–1) × (–1) × (–1) × (–1) = 1 ×1 = 1
You may check that (–1) raised to any odd power is (–1),
and (–1) raised to any even power is (+1).
(iv) (–10)^3 = (–10) × (–10) × (–10) = 100 × (–10) = – 1000
(v) (–5)^4 = (–5) × (–5) × (–5) × (–5) = 25 × 25 = 625

EXERCISE 13.1



  1. Find the value of:
    (i) 2^6 (ii) 9^3 (iii) 11^2 (iv) 5^4

  2. Express the following in exponential form:
    (i) 6 × 6 × 6 × 6 (ii) t × t (iii) b×b×b×b
    (iv) 5 × 5× 7 × 7 × 7 (v) 2 × 2 × a×a (vi) a× a ×a×c×c×c×c× d


(–1)odd number = –1
(–1)even number = + 1
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