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2.Find :
i)\(9^{3/2}\)
ii)\(32^{2/5}\)
iii)\(16^{3/4}\)
iv)\(125^{-1/3}\)

Answer :

i) \(9^{3/2}\) = \((3 × 3)^{3/2}\) = \(3^{2 × 3/2}\) = \(3^{3}\) = \(27\)

ii) \(32^{2/5}\) = \((2 × 2 × 2 × 2 × 2)^{2/5}\) = \(2^{2/5 × 5}\) = \(2^2\) = \(4\)

iii) \(16^{3/4}\) = \((2 × 2 × 2 × 2)^{3/4}\) = \(2^{4 × 3/4}\) = \(2^{3}\) =\(8\)

iv) \(125^{-1/3}\) = \((5 × 5 × 5)^{-1/3}\) = \(5^{3 × (-1/3)}\) = \(5^{-1}\) = \(1/5\)