DAV Maths Chapter 3 Enrichment Question Answers

Are you looking for accurate, step-by-step verified solutions for DAV Class 8 Maths Chapter 3 Enrichment Questions (Exponents and Radicals)? At NCERTBooksPDF, we provide complete, fully explained solutions for all enrichment and challenge questions in Unit 3.

Unit 3 Enrichment Questions: Step-by-Step Solutions

Question 1: Evaluate the following numerical expressions:

(i) (16)^3/4
Solution:
Express 16 in prime base 2: 16 = 2⁴.
(16)^3/4 = (2⁴)^3/4
= 2^{4 × (3/4)} = 2³ = 8.

(ii) 2 × (27)^-2/3
Solution:
Express 27 in base 3: 27 = 3³.
= 2 × (3³)^-2/3
= 2 × 3^{3 × (-2/3)}
= 2 × 3^-2
= 2 × (1 / 3²) = 2 × (1/9) = 2/9.

(iii) 2 × 9^3/2 × 9^-1/2
Solution:
Method 1 (Using Product Rule of Exponents):
= 2 × 9^{(3/2 + (-1/2))}
= 2 × 9^{(3 – 1)/2}
= 2 × 9^2/2 = 2 × 9¹ = 2 × 9 = 18.

Method 2 (Expressing in base 3):
9 = 3² → 9^3/2 = 3³ = 27; 9^-1/2 = 3^-1 = 1/3.
= 2 × 27 × (1/3) = 2 × 9 = 18.

(iv) (81/16)^-3/4
Solution:
Reciprocate fraction to make exponent positive:
= (16/81)^3/4
Notice that 16 = 2⁴ and 81 = 3⁴ → 16/81 = (2/3)⁴.
= [(2/3)⁴]^3/4
= (2/3)^{4 × (3/4)} = (2/3)³ = 2³ / 3³ = 8/27.

Question 2: Solve the exponential equation:

Problem: If 25^{x – 1} = 5^{2x – 1} – 100, find the value of x.
Solution:
Express 25 in base 5: 25 = 5².
(5²)^{x – 1} = 5^{2x – 1} – 100
5^{2(x – 1)} = 5^{2x – 1} – 100
5^{2x – 2} = 5^{2x – 1} – 100

Rearrange terms:
5^{2x – 1} – 5^{2x – 2} = 100
Notice that 5^{2x – 1} = 5^{2x – 2 + 1} = 5^{2x – 2} × 5¹.
Factor out 5^{2x – 2}:
5^{2x – 2} (5 – 1) = 100
5^{2x – 2} (4) = 100
5^{2x – 2} = 100 / 4 = 25
Since 25 = 5²:
5^{2x – 2} = 5²

Equating exponents:
2x – 2 = 2
2x = 4
x = 2.
Check: LHS = 25^2-1 = 25. RHS = 5^2(2)-1 – 100 = 5³ – 100 = 125 – 100 = 25. LHS = RHS!

Question 3: Simplify and express with positive exponents:

Problem: [√³(x⁴y) × 1 / √³(xy⁷)]^-4
Solution:
Convert radicals into fractional exponents (power 1/3):
√³(x⁴y) = (x⁴y)^1/3
√³(xy⁷) = (xy⁷)^1/3

Combine into a single fraction inside brackets:
= [(x⁴y) / (xy⁷)]^{(1/3) × (-4)}
= [x^{4 – 1} / y^{7 – 1}]^-4/3
= [x³ / y⁶]^-4/3
Apply the outer exponent -4/3 to numerator and denominator:
Numerator: (x³)^-4/3 = x^{3 × (-4/3)} = x^-4
Denominator: (y⁶)^-4/3 = y^{6 × (-4/3)} = y^-8

= x^-4 / y^-8
Express with positive indices by reciprocating:
= y⁸ / x⁴.

Question 4: Challenge Proof:

Problem: Prove that:
(x^a / x^b)^{a2 + ab + b2} × (x^b / x^c)^{b2 + bc + c2} × (x^c / x^a)^{c2 + ca + a2} = 1
Proof:
Using quotient law x^a / x^b = x^{a – b}:
Term 1: [x^{a – b}]^{(a2 + ab + b2)} = x^{(a – b)(a2 + ab + b2)} = x^{a3 – b3}
Term 2: [x^{b – c}]^{(b2 + bc + c2)} = x^{(b – c)(b2 + bc + c2)} = x^{b3 – c3}
Term 3: [x^{c – a}]^{(c2 + ca + a2)} = x^{(c – a)(c2 + ca + a2)} = x^{c3 – a3}

Multiplying all terms together by adding the exponents:
= x^{(a3 – b3) + (b3 – c3) + (c3 – a3)}
= x^{a3 – a3 – b3 + b3 – c3 + c3}
= x⁰
= 1 = RHS.
Hence Proved!

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