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

**Solution:**

Let's use the concept of exponents and laws of exponents to solve the following.

i) 9^{3/2}

9^{3/2 }= (3^{2})^{3/2} [Since, 9 = 3^{2}]

= (3)^{2 × 3/2} [Using the exponent rule (a^{p})^{q }= a^{pq}]

= 3^{3}

= 27

ii) 32^{2/5}

32^{2/5} = (2^{5})^{2/5} [Since, 32 = 2^{5}]

= (2)^{5 × 2/5} [Using the exponent rule (a^{p})^{q }= a^{pq}]

= 2^{2}

= 4

iii) 16^{3/4}

16^{3/4} = (2^{4})^{3/4} [ Since, 16 = 2^{4}]

= (2)^{4 × 3/4 }[Using the exponent rule (a^{p})^{q }= a^{pq}]

= 2^{3}

= 8

iv) 125^{-1/3}

125^{-1/3} = (5^{3})^{-1/3} [Since, 125 = 5^{3 }]

= (5)^{3 × (-1/3) }^{ }[Using the exponent rule (a^{p})^{q }= a^{pq}]

= 5^{-1}

= 1/5 [Using the exponent rule a^{-n} = 1/a^{n}]

**☛ Check: **NCERT Solutions for Class 9 Maths Chapter 1

**Video Solution:**

## Find i) 9³/²^{ }ii) 32²/⁵^{ }iii) 16³/⁴^{ }iv) 125⁻¹/³

NCERT Solutions Class 9 Maths Chapter 1 Exercise 1.6 Question 2:

**Summary:**

Thus, the values of 9^{3/2}, 32 ^{2/5}, 16 ^{3/4 }and 125^{-1/3} are 27, 4, 8, and 1/5 respectively.

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