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# Exponents and Powers

## Exponent and Powers :

Let 'a' be any real number and 'n' be any positive integer .

an = a x a x a x a x a x ..........n times. In an , 'a' is called base and ' n ' is called exponent or power.

In an , a is multiplied 'n' times by itself . an is read as " a is raised to the power n " (or) "nth power of a " (or) " a to the power n " .

Multiplying an integer several times by itself and writing the product in this form is known as exponential notation .

Examples :

1) 3 x 3 x 3 x 3 = 34 2) (-5)5 = (-5) x(-5) x (-5) x (-5) x (-5)

Q1 : Evaluate : (i) $(\frac{3}{4})^2$ (ii) $(\frac{-2}{3})^3$

Solution : (i) $(\frac{3}{4})^2 =\frac{3}{4}.\frac{3}{4} = \frac{9}{16}$

(ii) $(\frac{-2}{3})^3 =\frac{-2}{3}.\frac{-2}{3}.\frac{-2}{3} =\frac{-8}{27}$

## Exponential Notation

Q1 : Express each of the following rational numbers in exponential notation .

(i) 81/256 (ii) -32/243 (iii) -1/343 (iv) 256/625 (v) 8/27

Solution : (i) We can write , 81 = 3 x 3 x 3 x 3 = 3 4 and

256 = 4 x 4 x 4 x 4 = 44

Therefore $\frac{81}{256}=\frac{3^4 }{4^4 } =(\frac{3}{4})^4$

(ii) -32/243

We can write , -32 = (-2) x (-2) x (-2) x (-2) x (-2) = (-2)5 and

243 = 3 x 3 x 3 x 3 x 3 = 35

Therefore $\frac{-32}{243}=\frac{(-2)^5 }{3^5 } =(\frac{-2}{3})^5$

(iii) -1/343

We can write , -1 = (-1) x (-1) x (-1) = (-1)3

343 = 7 x 7 x 7 = 73

Therefore $\frac{-1}{343}=\frac{(-1)^3 }{7^3 } =(\frac{-1}{7})^3$

(iv) 256/625

256 = 4 x 4 x 4 x 4 ; 625 = 5 x 5 x 5 x 5

Therefore $\frac{256}{625}=\frac{4^4 }{5^4 }=(\frac{4}{5})^4$

(v) 8/27

8 = 2 x 2 x 2 ; 27 = 3 x 3 x 3

Therefore $\frac{8}{27}=\frac{2^3 }{3^3 }=(\frac{2}{3})^3$

## Substitution

Q1 : Find the value of (i) x3 (ii) 3 x when x = 5 .

Solution : x = 5

(i) x3 = x . x . x

53 = 5 . 5 . 5 = 125

(ii) 3 x = 3 . x

3 . 5 = 15 .

Q2 : Find the values of the following :

(i) a7 when a = 1 (ii) (2a)5 when a = 5 (iii) 2 x3 when x = 3 (iv) 3(2x -1) when x = 9

(v) 3(x-1)2 when x = 6 (vi) 4 (x2 - 4 ) when x = 4. (vii) 4( x2 - 9) when x = -3.

Solution : (i) a = 1

a7 = 17 = 1 . 1 . 1 . 1 . 1 . 1 . 1 = 1

(ii) (2a)5 when a= 5

(2a)5 = [(2)(5)]5 = (10)5 = 10 . 10 . 10 . 10 . 10 = 100,000

(iii) 2 x3 when x = 3

2 x3 = 2 (3)3 = 2 . 3 . 3 . 3 = 54

(iv) 3(2x - 1) when x = 9

3(2x - 1) = 3 [ 2(9) -1 ] = 3 ( 18 - 1) = 3 ( 17 ) = 51

(v) 3 ( x - 1 )2 when x = 6

3 (x - 1)2 = 3( 6 - 1 )2 = 3 (5)2 = 3 . 5 . 5 =75

(vi) 4( x2 - 4 ) when x = 4

4( x2 - 4 ) = 4 ( 42 - 4 ) = 4 ( 16 - 4 ) = 4 ( 12 ) = 48

(vii) 4 ( x2 - 9 ) when x = -3

4 ( x2 - 9 ) = 4 [ (-3)2 - 9 ] = 4 (9 - 9 ) = 4 ( 0 ) = 0

Q3 : Write the following in exponential notation .

(i) a . a . a . (ii) 4 . x. . x . y . y (iii) a . a . b .b . b . c . c . c (iv) 3 .4 .5 . x . y . y . z . z . z

(v) 4 . ( x + y ) . ( x + y ) . ( x + y ) . ( x + y ) (vi) ( a + 1) raised to the power 3 .

(vii) product of 20 and ( x - y ) raised to the power 4 (viii) product of 25 and ( a + b ) raised to the power 5 .

Solution : (i) a3 (ii) 4 x2 y2 (iii) a2 b3 c3 (iv) 60x y2 z3 (v) 4 ( x + y )4 (vi) (a + 1 )3

(vii) 20 (x - y )4 (viii) 25 ( a + b )5