Class 8 | NCERT Solution Maths Chapter 12 | Exponents and Powers | Exercise 12.1

Exercise 12.1 Page No: 197

1. Evaluate:
(i) 3-2 (ii) (-4)-2 (iii) (1/2)-5  

Solution:

(i) 3-2 = (1/3)2

= 1/9

(ii) (-4)-2 = (1/-4)2

= 1/16

(iii) (1/2)-5 = (2/1)5

= 25

= 32

2.  Simplify and express the result in power notation with positive exponent:

(i) (-4)4 ÷(-4)8   
(ii) (1/23)2  
(iii) -(3)4×(5/3)4 
(iv) (3-7÷3-10)×3-5  
(v) 2-3×(-7)-3

Solution:

(i)

= (-4)5/(-4)8

= (-4)5-8

= 1/(-4)3

(ii) (1/23)2  

= 12/(23)2

= 1/23×2 = 1/26

(iii) -(3)4×(5/3)4 

= (-1)4×34×(54/34 )

= 3(4-4)×54

= 30×54 = 54

(iv) 

=   (3-7/3-10)× 3-5

= 3-7 – (-10) × 3-5

= 3(-7+10)×3-5

= 33×3-5

= 3(3+-5)

= 3-2

=1/32

(v) 2-3×(-7) – 3

= (2×-7)-3

(Because am×bm = (ab)m)

= 1/(2×-7)3

= 1/(-14)3

3. Find the value of :

(i) (30+4-1)×22
(ii) (2-1×4-1)÷2 – 2
(iii) (1/2)-2+(1/3)-2+(1/4)-2
(iv) (3-1+4-1+5-1)0
(v) {(-2/3)-2}2

Solution:

(i)(30+4– 1)×22 = (1+(1/4))×22

= ((4+1)/4 )×22

= (5/4)×22

= (5/22)×22

= 5×2(2-2)

= 5×20

= 5×1 = 5

(ii)(2-1×4-1)÷2-2

= [(1/2)×(1/4)] ÷(1/4)

= (1/2×1/22 )÷ 1/4

= 1/23÷1/4

= (1/8)×(4)

= 1/2

(iii) (1/2)-2+(1/3)-2+(1/4)-2

= (2-1)-2+(3-1)-2+(4-1)-2

= 2(-1×-2)+3(-1×-2)+4(-1×-2)

= 22+32+42

= 4+9+16

=29

(iv) (3-1+4-1+5-1)0

= 1

(v) {(-2/3)-2}2 = (-2/3)-2×2

= (-2/3)-4

= (-3/2)4

= 81/16

4. Evaluate

(i) (8-1×53)/2-4
(ii) (5-1×2-2)×6-1 

Solution:

(i) (8-1×53)/2-4

=

= 2×125 = 250

(ii) (5-1×2-2)×6-1 

= (1/10)×1/6

= 1/60

5. Find the value of m for which 5m ÷ 5-3 = 55

Solution:

5m ÷ 5-3 = 55

5(m-(-3) ) = 55

5m+3 =55

Comparing exponents both sides, we get

m+3 = 5

m = 5-3

m = 2

6. Evaluate

(i) 

(ii) 

Solution:

(i)

= 3-4

= -1

(ii)

=

=

 =

=  512/125

7. Simplify.

(i)     

(ii) 

Solution:

(i)

=
 =

(ii)

=

=

=
 =

= 1×1×3125

= 3125