uestion 1. Find: (i) 64^(1/2) (ii) 32^(1/5) (iii) 125^(1/3)
Solution.
(i) 64^(1/2) = (8×8)^(1/2) = 8² × 1/2 = 8¹ = 8
(ii) 32^(1/5) = (2⁵)^(1/5) = 2^(5×1/5) = 2¹ = 2
(iii) 125^(1/3) = (5³)^(1/3) = 5^(3×1/3) = 5¹ = 5Ans.
Question 2. Find: (i) 9^(3/2) (ii) 32^(2/5) (iii) 16^(3/4) (iv) 125^(−1/3)
Solution.
(i) 9^(3/2) = (3²)^(3/2) = 3³ = 3×3×3 = 27
(ii) 32^(2/5) = (2⁵)^(2/5) = 2² = 2×2 = 4
(iii) 16^(3/4) = (2⁴)^(3/4) = 2³ = 2×2×2 = 8
(iv) 125^(−1/3) = 1/125^(1/3) = 1/(5³)^(1/3) = 1/5 = 1/5Ans.
Question 3. Simplify: (i) 2^(2/3) · 2^(1/5) (ii) (1/3²)⁷ (iii) 11^(1/2)/11^(1/4) (iv) 7^(1/2) · 8^(1/2)
Solution.
(i) 2^(2/3) · 2^(1/5) = 2^(2/3 + 1/5) = 2^(13/15) [∵ aᵐ·aⁿ = aᵐ⁺ⁿ]
(ii) (1/3²)⁷ = 1⁷/3^(2×7) = 1/3¹⁴ [∵ (a/b)ᵐ = aᵐ/bᵐ]
(iii) 11^(1/2)/11^(1/4) = 11^(1/2 − 1/4) = 11^(1/4) [∵ aᵐ/aⁿ = aᵐ⁻ⁿ]
(iv) 7^(1/2) × 8^(1/2) = (7×8)^(1/2) = 56^(1/2) [∵ aᵐ·bᵐ = (ab)ᵐ]