1. Find:
(i) ` \64^\frac{1}{2}\ `
(ii) ` \32^\frac{1}{5}\ `
(iii) ` \125^\frac{1}{3}\ `
solution :
(i) ` \64^\frac{1}{2}\ `
` \=\(8×8)^\frac{1}{2}\ `
` \=\(8^2)^\frac{1}{2}\ `
` \=8^1\ `` \[ ∵2 × \frac{1}{2} = 1 ]\ `
` \=8\ `
(ii) ` \32^\frac{1}{5}\ `
` \=\(2×2×2×2×2)^\frac{1}{5}\ `
` \=\(2^5)^\frac{1}{5}\ `
` \=2^1\ ` ` \[ ∵ 5 × \frac{1}{5}= 1]\ `
` \=2\ `
(iii) ` \125^\frac{1}{3}\ `
` \=\(5×5×5)^\frac{1}{3}\ `
` \=\(5^3)^\frac{1}{3}\ `
` \=\5^1\ ` ` \[ ∵3 ×\frac{1}{3}=1 ]\ `
` \=5\ `
2. Find:
(i) ` \9^\frac{3}{2}\ `
(ii) ` \32^\frac{2}{5}\ `
(iii) ` \16^\frac{3}{4}\ `
(iv) ` \125^\frac{-1}{3}\ `
solution :
(i) ` \9^\frac{3}{2}\ `
` \=\(3×3)^\frac{3}{2}\ `
` \=\(3^2)^\frac{3}{2}\ `
` \=\3^3\ `` \[ ∵ 2 ×\frac{3}{2}= 3 ]\ `
` \=27\ `
(ii) ` \32^\frac{2}{5}\ `
` \=\(2×2×2×2×2)^\frac{2}{5}\ `
` \=\(2^5)^\frac{2}{5}\ `
` \=\2^2\ `` \[ ∵ 5 ×\frac{2}{5}=2 ]\ `
`\=4\ `
(iii) ` \16^\frac{3}{4}\ `
` \=\(2×2×2×2)^\frac{3}{4}\ `
` \=\(2^4)^\frac{3}{4}\ `
` \=\2^3\ `` \[ ∵ 4 ×\frac{3}{4}= 3 ]\ `
` \=8\ `
(iv) ` \125^\frac{-1}{3}\ `
` \=\(5×5×5)^\frac{-1}{3}\ `
` \=\(5^3)^\frac{-1}{3}\ `
` \=5^1\ `` \[ ∵ 3 ×\frac{-1}{3}= -1 ]\ `
` \=1/5\ `
3. Simplify:
(i) ` \2^\frac{2}{3}×2^\frac{1}{5}\ `
(ii) ` \(\frac{1}{3^3})^7\ `
(iii) ` \frac{11^\frac{1}{2}}{11^\frac{1}{4}}\ `
(iv) ` \7^\frac{1}{2}×8^\frac{1}{2}\ `
solution :
(i) ` \2^\frac{2}{3}×2^\frac{1}{5}\ `
` \=\2^(\frac{2}{3}+\frac{1}{5})\ `
` \[ ∵ a^n×a^m=(a^(n+m)) ]\ `
` \=2^\frac{10+3}{5×3}\ `
` \=2^\frac{13}{15}\ `
(ii) ` \(\frac{1}{3^3})^7\ `
` \=\(3^-3)^7\ `
` \[ ∵(a^n)^m=a^{n×m}]\ `
` \=3^(-21)\ `
(iii) ` \frac{11^\frac{1}{2}}{11^\frac{1}{4}}\ `
` \=11^(\frac{1}{2}-\frac{1}{4})\ `
` \[ ∵\frac{a^n}{a^m}=a^{n-m}]\ `
` \=11^\frac{2-1}{4}\ `
` \=11^\frac{1}{4}\ `
(iv) ` \7^\frac{1}{2}×8^\frac{1}{2}\ `
` \=(7×8)^\frac{1}{2}\ `
` \[ ∵a^m×b^m=(a×b)^m]\ `
` \=56^\frac{1}{2}\ `
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