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

Exercise 12.1 of Chapter 12 (Exponents and Powers) in the Class 8 NCERT Mathematics textbook focuses on the fundamental concepts of exponents and their properties. This exercise introduces students to the laws of exponents and how to apply them in various mathematical expressions. Students will practice simplifying expressions with exponents, understanding negative and zero exponents, and applying the laws of exponents to solve problems.

Question 1. Evaluate:

Solution:

(i) 3^–2 

3^-2 = \frac{1}{(3^2)} = \frac{1}{9}                 (Property used: a^-n = \frac{1}{a^n} )

(ii) (– 4)^{– 2} 

(-4)^-2 = \frac{1}{(-4)^2} = \frac{1}{16}         (Property used: a^-n = \frac{1}{a^n} )

(iii) (\frac{1}{2} ) ^-5 

(\frac{1}{2} )^-5  = (2)⁵ = 32           (Property used: (\frac{b}{a})^{-n} = \frac{a^n}{b^n} )

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: a^m ÷ aⁿ= a^m-n)

= \frac{1}{(-4)^3}  

= \mathbf{(\frac{1}{(-4)})^3}

(ii) (\frac{1}{2^3})^2

= \frac{(1)^2}{(2^3)^2}                             (Property used: (a^m)ⁿ = a^m×n) 

= \frac{1}{2^6}

= \mathbf{\frac{1}{2^6}}

(iii) (-3)⁴ × (\frac{5}{3} )⁴

= ((3)⁴ × \frac{5^4}{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: a^m ÷ aⁿ= a^m-n)

= 3 ^(-7+10) × 3^-5

= 3³ × 3^-5

= 3 ^(3+(-5))                                                   (Property used: a^m × aⁿ = a ^{m + n})

= 3^-2                                                             (Property used: a^-m  =\frac{1}{a^m} )

= \frac{1}{3^2}

= \mathbf{\frac{1}{3^2}}

(v) 2^-3 × (-7)^-3

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

= (-14)^-3                                                     (Property used: a^-m  =\frac{1}{a^m} )

= \frac{1}{(-14)^3}

=\mathbf{\frac{1}{(-14)^3}}

Question 3. Find the value of

Solution:

(i) (3⁰ + 4^-1) × 2²

= (1 + (\frac{1}{4} )) × 4                     (a⁰ = 1  a ≠ 0)

= (\frac{5}{4} ) × 4

= 5

(ii) (2^-1 × 4^-1) ÷ 2^-2

= (2 × 4)^-1 ÷ \frac{1}{2^2}                                (Property used: a^m × b^m = (a×b)^m)

= (8)-1 ÷ \frac{1}{4}

= (\frac{1}{8} ) ÷ \frac{1}{4}

=(\frac{1}{8} ) × 4

= (\frac{1}{2} )

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

= 2² + 3² + 4²                                        (Property used: (\frac{1}{a})^{-m}   =a^m)

= 4 + 9 + 16

= 29

(iv) (3^-1 + 4^-1 + 5^-1)⁰

= (\frac{1}{3} + \frac{1}{4} + \frac{1}{5} )⁰                             (a⁰ = 1 (a ≠ 0)

= 0 

(v) {(\frac{-2}{3} )^-2}²

= (\frac{-2}{3} ) ^-2×2                                      (Property used: (a^m)ⁿ = a^m×n) 

= (\frac{-2}{3} )^-4    = (\frac{-3}{2} )⁴                     (Property used: (b/a)^-n = aⁿ/bⁿ)

= \frac{3^4}{2^4}

= \frac{81}{16}

Question 4. Evaluate 

Solution:

(i) (8^-1 × 5³) / 2^-4

= (\frac{1}{8}  × 125) / (2^-4)                      (Property used: (b/a)^-n = aⁿ/bⁿ)

= (\frac{1}{8} ) × 125 × 2⁴

= 250

(ii) (5^-1 × 2^-1) × 6^-1

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

= 10^-1 × 6^-1

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

= 60^-1

= \frac{1}{60}

Question 5. Find the value of m for which 5^m ÷ 5^{– 3} = 5⁵ 

Solution:

5^{m-(– 3)} = 5⁵                       (Property used: a^m ÷ aⁿ= a^m-n)

5^m+3 = 5⁵ 

m+3 = 5

m = 5-3

m = 2

Question 6. Evaluate 

Solution:

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

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

= (-1)^-1

= (1/(-1))¹

= (-1)

(ii) (\frac{5}{8} )^-7 × (\frac{8}{5} )^-4

= (\frac{5}{8} )⁷ × (\frac{8}{5} )^-4                       (Property used: (b/a)^-n = (a/b)ⁿ)

= (\frac{8}{5} ) ^{7+ (-4)}                               (Property used: a^m × aⁿ = a ^{m + n})

= (\frac{8}{5} )³  = 8³/5³

= \frac{512}{125}

Question 7. Simplify

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 10 \times t^{-8})}          (Property used: a^m ÷ aⁿ= a^m-n ) (25 = 5²)

=\frac{(5^{2-(-3)} \times t^{-4-(-8)})}{ 10}

= \frac{(5^{5} \times t^{4})}{ 10}

= \frac{(625 \times t^{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 125)}{(5^{-7} \times (2 \times 3)^{-5})}

= \frac{(3^{-5} \times 2^{-5}\times 5^{-5} \times 125)}{(5^{-7} \times 2^{-5} \times 3^{-5})}                   (Property used: (a×b)^m = a^m × b^m)

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

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

= 5⁵

Summary

Exercise 12.1 of Chapter 12 on Exponents and Powers provides a comprehensive introduction to the concept of exponents and their applications in mathematics. This exercise covers a wide range of topics, including positive, negative, and zero exponents, as well as the fundamental laws of exponents such as the product rule, quotient rule, and power of a power rule. Students are challenged to apply these rules to simplify complex expressions, convert between standard and exponential forms of numbers, and solve equations involving exponents. The problems are designed to build a strong foundation in working with exponents, which is crucial for understanding more advanced mathematical concepts in algebra, calculus, and scientific notation.