SURDS AND INDICES
IMPORTANT FORMULAE
1. LAWS OF INDICES:
(i) am x an = am + n
(ii) am / an = am - n
(iii) ( am )n = amn
(iv) ( a b )n = anbn
(v) ( a / b )n = ( an / bn )
(vi) a0 = 1
2. SURDS: Let a be a rational number and n be a positive integer such that a1/n = nsqrt(a)
is irrational. Then nsqrt(a) is called a surd of order n.
3. LAWS OF SURDS:
(i) n√a = a1/2
(ii) n √ab = n √a * n √b
(iii) n √a/b = n √a / n √b
(iv) (n √a)n = a
(v) m√(n√(a)) = mn√(a)
(vi) (n√a)m = n√am
Examples:
Ex. 1. Simplify : (i) (27)2/3 (ii) (1024)-4/5 (iii)( 8 / 125 )-4/3
Sol . (i) (27)2/3 = (33)2/3 = 3( 3 * ( 2/ 3)) = 32 = 9
(ii) (1024)-4/5 = (45)-4/5 = 4 { 5 * ( (-4) / 5 )} = 4-4 = 1 / 44 = 1 / 256
(iii) ( 8 / 125 )-4/3 = {(2/5)3}-4/3 = (2/5){ 3 * ( -4/3)} = ( 2 / 5 )-4 = ( 5 / 2 )4 = 54 / 24
Ex. 2. Evaluate: (i) (.00032)3/5 (ii)l (256)0.16 x (16)0.18.
(ii) (256)0. 16 * (16)0. 18 = {(16)2}0. 16 * (16)0. 18 = (16)(2 * 0. 16) * (16)0. 18
=(16)0.32 * (16)0.18 = (16)(0.32+0.18) = (16)0.5 = (16)1/2 = 4.
Ex. 3. Find the largest from among 4√6, √2 and 3√4.
Sol. Given surds are of order 4, 2 and 3 respectively. Their L.C,M, is 12, Changing each to a surd of order 12, we get:
4√6 = 61/4 = 6((1/4)*(3/3)) = 63/12 = (63)1/12 = (216)1/12.
√2 = 21/2 = 2((1/2)*(6/6)) = 26/12 = (26)1/12 = (64)1/12.
3√4 = 41/3 = 4((1/3)*(4/4)) = 44/12 = (44)1/12 = (256)1/12.
Clearly, (256)1/12 > (216)1/12 > (64)1/12
Largest one is (256)1/12. i.e. 3√4 .
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