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 = (3²)^3/2 [Since, 9 = 3²]

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

= 3³

= 27

ii) 32^2/5

32^2/5 = (2⁵)^2/5 [Since, 32 = 2⁵]

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

= 2²

= 4

iii) 16^3/4

16^3/4 = (2⁴)^3/4 [ Since, 16 = 2⁴]

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

= 2³

= 8

iv) 125^-1/3

125^-1/3 = (5³)^-1/3 [Since, 125 = 5³ ]

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

= 5^-1

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

☛ Check: NCERT Solutions for Class 9 Maths Chapter 1

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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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