NCERT Solutions Class 8 Maths Chapter 12 Exponents and Powers Ex 12.1

NCERT Solutions Class 8 Mathematics 
Chapter – 12 (Exponents and Powers) 

The NCERT Solutions in English Language for Class 8 Mathematics Chapter – 12 Exponents and Powers Exercise 12.1 has been provided here to help the students in solving the questions from this exercise. 

Chapter 12: Exponents and Powers

Exercise – 12.1 

1. Evaluate:
(i) 3-2
(ii) (-4)-2
(iii) \mathbf{\left (\frac{1}{2} \right )^{-5}}

Solution –
(i) 3-2
= \left ( \frac{1}{3^2} \right )              ∵ a-m = \frac{1}{a^m}

= \frac{1}{9}

(ii) (-4)-2 

= \left (\frac{1}{-4^2} \right )          ∵ a-m = \frac{1}{a^m}

= \frac{1}{16}

(iii) \mathbf{\left (\frac{1}{2} \right )^{-5}}

= \left ( \frac{2}{1} \right )^5     ∵ a-m = \frac{1}{a^m}

= 25
= 32

2.  Simplify and express the result in power notation with a positive exponent:
(i) (-4)4 ÷(-4)8
(ii) \left ( \frac{1}{2^3} \right )^2
(iii) -(3)4×\left (\frac{5}{3} \right )^4
(iv) (3-7÷3-10)×3-5
(v) 2-3×(-7)-3

Solution –

(i) (-4)4 ÷(-4)8   

= (-4) 5 – 8 ∵ (a)m ÷(a)n = a m – n 

= (-4)–3

= \frac{1}{-4^3}

= \frac{1}{-64}

(ii) \left ( \frac{1}{2^3} \right )^2  

= \frac{1}{(2^3)^2}    ∵ \left ( \frac{a}{b} \right )^m = \frac{a^m}{b^m} 

= \frac{1}{2^{3\times 2}}                   ∵ (a^{m})^n = a^{m\times n}

= \frac{1}{2^6}

(iii) – (3)4 × \mathbf{\left ( \frac{5}{3} \right )^4}

= (-3) 4 × \frac{5^4}{3^4}                   ∵ \left ( \frac{a}{b} \right )^m = \frac{a^m}{b^m}

= (-1)4 × 34 × \frac{5^4}{3^4}             ∵ (ab)m = ambm

= 1 × 3(4 – 4) × 54  

= 3(4-4)×54                              ∵ (a)m ÷(a)n = am – n 

= 30×54 = 54                        ∵ a0 = 1

(iv) (3-7÷3-10)×3-5

= 3 -7 – (-10) × 3-5                      ∵ (a)m ÷(a)n = am – n

= 3(-7+10)×3-5
= 33×3-5
= 3(3-5)
= 3-2
= \frac{1}{3^2}                                          ∵ a-m = \frac{1}{a^m}

(v) 2-3×(-7) – 3 

= (2×-7)-3                                         ∵ am×bm = (ab)m

= \frac{1}{(2\times -7)^3}                          ∵ a-m = \frac{1}{a^m}

= \frac{1}{-14^3}

3. Find the value of:

(i) (30+4-1)×22
(ii) (2-1×4-1)÷2 – 2
(iii) \left ( \frac{1}{2} \right )^{-2} + \left ( \frac{1}{3} \right )^{-2} + \left ( \frac{1}{4} \right )^{-2}

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

(v) \left \{ \left ( \frac{-2}{3} \right )^{-2} \right \}^2

Solution –

(i)(30+4– 1)×22 

= (1 + \frac{1}{4}) × 22            ∵ a-m = \frac{1}{a^m}

= \frac{(4+1)}{4} × 22

= \frac{5}{4} × 22

= \frac{5}{2^2} × 22
= 5 × 2 (2-2)           ∵ (a)m ÷(a)n = am – n
= 5×20
= 5×1 = 5

(ii) (2-1×4-1)÷2-2

= \frac{1}{2}\times \frac{1}{4} \div \frac{1}{2^2}                            ∵ a-m = \frac{1}{a^m}

= \frac{1}{2}\times \frac{1}{2^2} \div \frac{1}{2^2}

= \frac{1}{2^3} \div \frac{1}{2^2}

= \frac{1}{2^3} \times 2^2

= \frac{1}{2}

(iii) \left ( \frac{1}{2} \right )^{-2} + \left ( \frac{1}{3} \right )^{-2} + \left ( \frac{1}{4} \right )^{-2}

= (2-1)-2 + (3-1)-2 + (4-1)-2              ∵ a-m = \frac{1}{a^m}

= 2(-1×-2) + 3(-1×-2) + 4(-1×-2)          ∵ (a^{m})^n = a^{m\times n}
= 22+32+42
= 4+9+16
= 29

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

= 1

(v) \left \{ \left ( \frac{-2}{3} \right )^{-2} \right \}^2

= \left ( \frac{-2}{3} \right )^{-4}                   ∵ (a^{m})^n = a^{m\times n}

= \left (\frac{-3}{2} \right )^4          ∵ a-m = \frac{1}{a^m}

= \frac{81}{16} 

4. Evaluate:
(i) \frac{8^{-1}\times 5^3}{2^{-4}}

(ii) (5-1×2-2)×6-1 

Solution –

(i) \frac{8^{-1}\times 5^3}{2^{-4}}

= \frac{(2^3)^{-1}\times 5^3}{2^{-4}}

= \frac{2^{-3}\times 5^3}{2^{-4}}                    ∵ (a^{m})^n = a^{m\times n}
= 2 -3-(-4) × 53                  ∵ (a)m ÷(a)n = am – n
= 2 × 125
= 250

(ii) (5-1×2-2)×6-1 

= \left ( \frac{1}{5} \times \frac{1}{2}\right )\times \frac{1}{6}                  ∵ a-m = \frac{1}{a^m}

= \frac{1}{10}\times \frac{1}{6}

= \frac{1}{60}

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

Solution –

⇒ 5m ÷ 5-3 = 55

⇒ 5(m-(-3) ) = 55            ∵ (a)m ÷(a)n = am – n

⇒ 5m+3 =55
Comparing exponents on both sides, we get
m+3 = 5
m = 5 – 3
m = 2

6. Evaluate:

(i)  \left \{ \left ( \frac{1}{3} \right )^{-1} - \left ( \frac{1}{4} \right )^{-1}\right \}^{-1}

(ii)  \left ( \frac{5}{8} \right )^{-7} \times \left ( \frac{8}{5} \right )^{-4}

Solution –

(i)  \left \{ \left ( \frac{1}{3} \right )^{-1} - \left ( \frac{1}{4} \right )^{-1}\right \}^{-1}

= \left \{ \left ( \frac{3}{1} \right )^1 - \left ( \frac{4}{1} \right )^1\right \}                    ∵ a-m = \frac{1}{a^m}
= 3 – 4
= -1

(ii) \left ( \frac{5}{8} \right )^{-7} \times \left ( \frac{8}{5} \right )^{-4}

= \frac{5^{-7}}{8^{-7}} \times \frac{8^{-4}}{5^{-4}}                    ∵ \left ( \frac{a}{b} \right )^m = \frac{a^m}{b^m}

= 5-7-(-4) × 8-4-(-7)Ncert solution class 8 chapter 12-40          ∵ (a)m ÷(a)n = am – n

= 5-7+4 × 8-4+7

= 5-3 × 83

= \frac{8^3}{5^3}               ∵ a-m = \frac{1}{a^m}

= \frac{512}{125}

7. Simplify the following:

(i) \frac{25\times t^{-4}}{5^{-3} \times 10 \times t^{-8}} (t ≠ 0)

(ii) \frac{3^{-5} \times 10^{-5}\times 125}{5^{-7}\times 6^{-5}}

Solution –

(i) \frac{25\times t^{-4}}{5^{-3} \times 10 \times t^{-8}} (t ≠ 0)

= \frac{5^2\times t^{-4}}{5^{-3} \times 5 \times2\times t^{-8}}

= \frac{5^{2-(-3)-1}\times t^{-4-(-8)}}{2}                  ∵ (a)m ÷(a)n = am – n

= \frac{5^{2+3-1}\times t^{-4+8}}{2}  

= \frac{5^{5}\times t^{4}}{2}

= \frac{625t^4}{2}

(ii) \frac{3^{-5} \times 10^{-5}\times 125}{5^{-7}\times 6^{-5}} 

= \frac{3^{-5} \times (2\times 5)^{-5}\times 5^3}{5^{-7}\times (2\times 3)^{-5}}  

= \frac{3^{-5} \times 2^{-5}\times 5^{-5}\times 5^3}{5^{-7}\times 2^{-5}\times 3^{-5}}                ∵ (ab)m = ambm

= \frac{3^{-5} \times 2^{-5}\times 5^{-5+3}}{5^{-7}\times 2^{-5}\times 3^{-5}}

= \frac{3^{-5} \times 2^{-5}\times 5^{-2}}{5^{-7}\times 2^{-5}\times 3^{-5}}                    ∵ a^m\times a^n = a^{m+n} 

= 3-5-(-5) × 2-5-(-5) × 5-2-(-7)      ∵ (a)m ÷(a)n = am – n

= 3 -5+5 ×2 -5+5 × 5-2+7
= 30 × 20 × 55                               ∵ a0 = 1
= 1 × 1 × 3125
= 3125

 

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