RS Aggarwal Solutions: Exercise 2A - Exponents | Mathematics (Maths) Class 8 PDF Download

Q.1. Evaluate:
(i) 4⁻³

(ii)
(iii)
(iv) (−3)⁻⁴
(v)
Ans.
(i)
(ii)
(iii)
(iv)
(v)

Q.2. Evaluate:
(i)
(ii)
(iii)
(iv)
Ans.
(i)

(ii)

(iii)

(iv)


Q.3. Evaluate:
(i)
(ii)
(iii)
Ans.

(i)


(ii)

(iii)


Q.4. Evaluate:
(i)
(ii)
(iii)
Ans.
(i)

(ii)
(iii)


Q.5. Evaluate

Ans.


Q.6. Evaluate

Ans.

The L.C.M. of 4 and 1 is 4.



Q.7.
Evaluate

Ans.

Q.8.
Find the value of:
(i) (2⁰ + 3⁻¹) × 3²
(ii) (2⁻¹ × 3⁻¹) ÷ 2⁻³
(iii)
Ans.
(i)
(2⁰ + 3⁻¹) × 3² = ×3²  (because 2⁰ = 1 and 3⁻¹ = 1/3)


(ii)

(iii)
=2² + 3² + 4² = 4 + 9 + 16 = 29

Q.9. Find the value of x for which
Ans.
Consider the left side:

Given:

Comparing the powers:
− 9 = 3x ⇒ x= −3

Q.10. Find the value of x for which
Ans.
Given:


⇒ 2x − 1 = −3
2x = −3 + 1 = −2
⇒ x = −1

Q.11. By what number should (−6)⁻¹ be multiplied so that the product becomes 9⁻¹?
Ans.

Let the required number be x.


The greatest common divisor for the numerator and the denominator is 3.

Q.12. By what number should be divided so that the quotient may be 

Ans.
Let the number be x.



Q.13. If 5^{2x + 1} ÷ 25 = 125, find the value of x.
Ans.
Given:
5^2x+1 ÷ 25 = 125

We know:
25 = 5 × 5 = 5²
125 = 5 × 5 × 5 = 5³

or 5^[(2x+1)−2] = 5^[2x−1] = 5³
⇒ 2x − 1 = 3
2x = 3 + 1 = 4

∴ x=2