Learn Find I 6412 Ii 3215 Iii 12513

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# Find: i) 64^{1/2 }ii) 32^{1/5 }iii) 125^{1/3}

**Solution:**

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

i) 64^{1/2}

64^{1/2 }= (8^{2})^{ 1/2} [Since, 64 = 8^{2}]

= 8^{2 × 1/2} [ Using (a^{p})^{q }= a^{pq}, where a > 0, p and q are rational numbers]

= 8^{1}

= 8

ii) 32^{1/5}

32^{1/5 }= (2^{5})^{1/5 }^{ }[Since, 32 = 2^{5}]

= (2)^{5 × 1/5} [ Using (a^{p})^{q }= a^{pq}]

= 2^{1}

= 2

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

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

= 5^{1}

= 5

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

**Video Solution:**

## Find: i) 64¹/²^{ }ii) 32¹/⁵^{ }iii) 125¹/³

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

**Summary:**

Hence, the values of 64^{1/2}, 32^{1/5}, and 125^{1/3} are 8, 2, and 5 respectively.

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