Introduction

Exercise 1.5 of the CBSE Class 9th NCERT Maths book is based on the Laws of Exponents for Real Numbers. It introduces and applies various rules for simplifying expressions involving exponents. The key laws covered include:

Laws of Exponents for Real Numbers:

1. Product Law: a^m ⋅ aⁿ = a^(m+n)
2. Quotient Law: a^m/aⁿ = a^(m-n)
3. Power of power: (a^m)ⁿ = a^m⋅n
4. Power of Products: (a⋅b)^m = a^m⋅b^m
5. Power of Quotient:  (a/b)^m = a^m/b^m, b ≠ 0
6. Zero exponent: a⁰ = 1, a ≠ 0
7. Negative exponent: a^-m = 1/a^m, a ≠ 0

Exercise 1.5

1. Find: (i) 64^(1/2) (ii) 32^(1/5) (iii) 125^(1/3)

Sol (i): 64^(1/2)
= (2x2x2x2x2x2)^(1/2)
= (2⁶)^(1/2)
= 2^6x(1/2)
= 2³
= 8 Ans

Sol (ii): 32^(1/5)
= (2x2x2x2x2)^(1/5)
= (2⁵)^(1/5)
= 2^5x(1/5)
=2 Ans

Sol (iii) 125^(1/3)
= (5x5x5)^1/3
= (5³)^1/3
= 5^3x(1/3)
= 5 Ans

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

Sol (i) 9^3/2
= (3x3)^3/2
= (32)^3/2
= (3)^(2x3)/2
= (3)³
= 27 Ans

Sol (ii) (32)^2/5
= (2x2x2x2x2)^2/5
= (25)^2/5
= 4 Ans

Sol (iii) 16^3/4
= (2x2x2x2)^3/4
= (2⁴)^3/4
= 2³
= 8 Ans

3. Simplify: (i) 2^2/3 ⋅ 2^1/5 (ii) (1/3³)⁷ (iii) 11^1/2/11^1/4 (iv) 7^1/2⋅8^1/2

Sol (i) 2^2/3 ⋅ 2^1/5
Applying Product Law: a^m ⋅ aⁿ = a^(m+n)
= 2^{(2/3 + 1/5)}
= 2^{[(10 + 3)/15]}
= 2^13/15 Ans

Sol (ii) (1/3³)⁷
Applying Power of Quotient:  (a/b)^m = a^m/b^m, b ≠ 0
= 1⁷/(3³)⁷
= 1/3^(3x7)
= 1/3²¹
= 3^-21 Ans

Sol (iii) 11^1/2/11^1/4
Applying Quotient Law: a^m/aⁿ = a^(m-n)
= 11^{(1/2 - 1/4)}
= 11^{[(2 - 1)/4]}
= 11^1/4 Ans

Sol (iv) 7^1/2⋅8^1/2
Power of Products: (a⋅b)^m = a^m⋅b^m
=> a^m⋅b^m = (a⋅b)^mg(On applying this formula, we get)
Now,
= (7 ⋅ 8)^1/2
= 56^1/2 Ans

Important Questions Based On Ex. 1.5
( Ex. 1.5 पर आधारित महत्वपूर्ण प्रश्न )