Page 46 - SM inner class 7.cdr

+ 4 % 1 ¼ 3 ÷ ¾
×
Chapter5 6 ½ –

Exponents

Introduction So lu tion :
´
´
=
We know how to calculate the expression 5 5´ . (i) (-3 ) (-3 ) (-3 ) (-3 ) 3
This expression can be written in a shorter (ii) 2 2 2 2 2 2 2´ ´ ´ ´ ´ ´ = ( 2) 7
way using something called exponents. æ 1 ö æ 1 ö æ 1 ö æ 1 ö æ 1 ö 4
5 ´ 5 = 5 2 (ii) ç ÷ ´ ç ÷ ´ ç ÷ ´ ç ÷ = ç ÷
è
è
4 ø
4 ø
4 ø
è
è
4 ø
è
4 ø
The number 5 is called the base, and the æ 7 æ 7 æ 7 2
- ö
- ö
- ö
number 2 is called the exponent. The exponent (iv) ç ÷ ´ ç ÷ = ç ÷
è 12 ø è 12 ø è 12 ø
corresponds to the number of times the base is
used as a factor. Ex am ple 2 : Iden tify the base and ex po nent of
An expression that represents repeated each num ber.
multiplication of the same factor is called a 25 æ 7 9
- ö
power or exponents. (i) 13 (ii) ç è 11 ÷ ø
3 1 3 to the first power 3 (iii) a m (iv) (-426 ) 11
n t
4 to the second power or
æ x
4 2 4.4 æ a ö (vi) - ö
4 squared (v) ç ÷ ç ÷
è b ø è y ø
5 to the third power or 5
3
5 5.5.5 So lu tion :
cubed
9
- ö
2 6 2 to the power of six 2.2. 2. 2.2.2 25 æ 7
(i) 13 (ii) ç ÷
è 11 ø
Here the meaning of a dot is to multiply or ' 'X -7
Similarly, base = 13 base =
11
2
11 11 can be written as 11 . We read it as 11 exponent = 25 exponent = 9
´
to the power of 2 where 11 is the base and 2 is m 11
(iii) a (iv) (-426 )
the exponent.
base = a base = -426
From the above examples we can conclude that
exponent = m exponent = 11
if a number “a” is multiplied with itself n – 1 n t
n
æ x
a ö
æ
times, then the product will be a , i e. . (v) ç ÷ (vi) - ö
ç
÷
n
´
a = a a a´................... ´ a (n -1 times è b ø è y ø
´
multiplications of “a” with itself) base = a base = -x
We read it as “a to the power of n” or “n th b y
power of a ”where “a” is the base and “n” is the exponent = n exponent = t
exponent.
Ex am ple 3 (a). Write the following in the
Ex am ple 1. Express each of the following in sim plest form.
ex po nen tial form. æ 2ö 2 æ - ö 1 4
´
´
(i) (-3 ) (-3 ) (-3 ) (i) (-5 ) 3 (ii) ç ÷ (iii) ç è 4 ÷ ø
è
3 ø
(ii) 2 2 2 2 2 2 2´ ´ ´ ´ ´ ´
So lu tion :
æ 1 ö æ 1 ö æ 1 ö æ 1 ö
3
(iii) ç ÷ ´ ç ÷ ´ ç ÷ ´ ç ÷ (i) (-5 ) = (-5 ) (-5 ) (-5 )
´
´
è 4 ø è 4 ø è 4 ø è 4 ø
(
= +25 ) ´ - )
5
(
æ 7 æ 7
- ö
- ö
(iv) ç ÷ ´ ç ÷ = -125
è 12 ø è 12 ø
3
Thus, (-5 ) = -125
46
Mathematics In Focus - 7