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• NCERT Solutions for Class 8 Maths

# Class 8 NCERT Solutions- Chapter 12 Exponents and Powers – Exercise 12.1

### Question 1. Evaluate:

Solution:

(i) 3–2

3-2                (Property used: a-n)

(ii) (– 4)– 2

(-4)-2        (Property used: a-n)

(iii) () -5

()-5  = (2)5 = 32           (Property used: )

### Question 2. Simplify and express the result in power notation with a positive exponent.

Solution:

(i) (-4)5 ÷ (-4)8

= (-4)5-8 = (-4) -3                      (Property used: am ÷ an= am-n)

(ii)

(Property used: (am)n = am×n)

(iii) (-3)4 × ()4

= ((3)4 ×                                (Property used: (a/b)n = an/ bn  & (-a)n = an if a is positive number and n is even)

= 54

(iv) (3-7 ÷ 3-10) × 3-5

= 3 (-7-(-10)) × 3-5                                     (Property used: am ÷ an= am-n)

= 3 (-7+10) × 3-5

= 33 × 3-5

= 3 (3+(-5))                                                   (Property used: am × an = a m + n)

= 3-2                                                             (Property used: a-m  =)

(v) 2-3 × (-7)-3

= (2 × (-7))-3                                           (Property used: am × bm = (a×b)m)

= (-14)-3                                                     (Property used: a-m  =)

=

### Question 3. Find the value of

Solution:

(i) (30 + 4-1) × 22

= (1 + ()) × 4                     (a0 = 1  a 0)

= () × 4

= 5

(ii) (2-1 × 4-1) ÷ 2-2

= (2 × 4)-1 ÷                                (Property used: am × bm = (a×b)m)

= (8)-1 ÷

= () ÷

=() × 4

= ()

(iii) (1/2)-2 + (1/3)-2 + (1/4)-2

= 22 + 32 + 42                                        (Property used:   =am)

= 4 + 9 + 16

= 29

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

= ()0                             (a0 = 1 (a 0)

= 0

(v) {()-2}2

= () -2×2                                      (Property used: (am)n = am×n

= ()-4    = ()4                     (Property used: (b/a)-n = an/bn)

### Question 4. Evaluate

Solution:

(i) (8-1 × 53) / 2-4

= ( × 125) / (2-4)                      (Property used: (b/a)-n = an/bn)

= () × 125 × 24

= 250

(ii) (5-1 × 2-1) × 6-1

= (5 × 2)-1 × 6-1                           (Property used: am × bm = (a×b)m)

= 10-1 × 6-1

= (10 × 6)-1                                     (Property used: am × bm = (a×b)m)

= 60-1

### Question 5. Find the value of m for which 5m ÷ 5– 3 = 55

Solution:

5m-(– 3) = 55                       (Property used: am ÷ an= am-n)

5m+3 = 5

m+3 = 5

m = 5-3

m = 2

### Question 6. Evaluate

Solution:

(i) {()-1 – ()-1}-1

= (31 – 41) -1                          (Property used: (1/a)-m  = am)

= (-1)-1

= (1/(-1))1

= (-1)

(ii) ()-7 × ()-4

= ()7 × ()-4                       (Property used: (b/a)-n = (a/b)n)

= () 7+ (-4)                               (Property used: am × an = a m + n)

= ()3  = 83/53

### Question 7. Simplify

Solution:

(i)  (t ≠ 0)

(Property used: am ÷ an= am-n ) (25 = 52)

=

(ii)

(Property used: (a×b)m = am × bm)

= (3-5-(-5) × 2-5-(-5) × 5 (-5)+3+7)                                      (Property used: am ÷ an= am-n )

= (30 × 20 × 55)                                                                (a0 = 1 (a ≠ 0)

= 55

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