082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

1.2 Laws of indices 5
(c)
(d)
x 9
x 5 =x9−5 =x 4
y 6
y =y6−1 =y 5
Consider the expression 23
23. Usingthe second law of indices we may write
2 3
2 = 3 23−3 = 2 0
But, clearly, 23
2 3 = 1,and so 20 = 1.This illustratesthe general rule:
a 0 =1
Any expression raised tothe power 0 is1.
1.2.3 Negativeindices
Consider the expression 43
45. We can write thisas
4 3
4 = 4.4.4
5 4.4.4.4.4 = 1
4.4 = 1 4 2
Alternatively, using the second law of indices we have
4 3
4 5 = 43−5 = 4 −2
So wesee that
4 −2 = 1 4 2
Thus we are able to interpret negative indices. The sign of an index changes when the
expression is inverted. Ingeneral wecan state
a −m = 1
a m a m = 1
a −m
Example1.6 Evaluate the following:
(a) 3 −2 2
(b) (c) 3 −1 (d) (−3) −2 (e) 6−3
4 −3 6 −2
Solution (a) 3 −2 = 1 3 = 1 2 9
2
(b)
4 = −3 2(43 ) = 2(64) = 128
(c) 3 −1 = 1 3 = 1 1 3
(d) (−3) −2 = 1
(−3) = 1 2 9
6 −3
(e)
6 = −2 6−3−(−2) = 6 −1 = 1 6 = 1 1 6