NCERT Solutions for Class 9 Maths Chapter 1 - Number Systems Exercise 1.6

NCERT Solutions Class 9 Maths Chapter 1 Number Systems Exercise 1.6, given here, provides easy reference while solving the exercise problems. The sixth exercise in the NCERT Solutions for Class 9 Maths Chapter 1 explains the laws of exponents for Real Numbers. They provide a detailed and stepwise explanation of each answer to the questions given in the exercises in the NCERT textbook.

The answers can be downloaded in the form of a PDF, which can be used by the students at their own pace. The NCERT solutions are always prepared by following NCERT guidelines so that they should cover the whole syllabus accordingly. These are very helpful in scoring well in the CBSE examinations.

NCERT Solutions for Class 9 Maths Chapter 1- Number Systems Exercise 1.6

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Access Other Exercise Solutions of Class 9 Maths Chapter 1 – Number Systems

Students can access the Chapter 1 exercise-wise links of Class 9 Maths NCERT Solutions below.

Exercise 1.1 Solutions 4 Questions ( 2 long, 2 short)
Exercise 1.2 Solutions 4 Questions ( 3 long, 1 short)
Exercise 1.3 Solutions 9 Questions ( 9 long)
Exercise 1.4 Solutions 2 Questions ( 2 long)
Exercise 1.5 Solutions 5 Questions ( 4 long 1 short)

Access Answers of Maths NCERT class 9 Chapter 1 – Number Systems Exercise 1.6

1. Find:

(i)64^1/2

Solution:

64^1/2 = (8×8)^1/2

= (8²)^½

= 8¹ [⸪2×1/2 = 2/2 =1]

= 8

(ii)32^1/5

Solution:

32^1/5 = (2⁵)^1/5

= (2⁵)^⅕

= 2¹ [⸪5×1/5 = 1]

= 2

(iii)125^1/3

Solution:

(125)^1/3 = (5×5×5)^1/3

= (5³)^⅓

= 5¹ (3×1/3 = 3/3 = 1)

= 5

2. Find:

(i) 9^3/2

Solution:

9^3/2 = (3×3)^3/2

= (3²)^3/2

= 3³ [⸪2×3/2 = 3]

=27

(ii) 32^2/5

Solution:

32^2/5 = (2×2×2×2×2)^2/5

= (2⁵)^2⁄5

= 2² [⸪5×2/5= 2]

= 4

(iii)16^3/4

Solution:

16^3/4 = (2×2×2×2)^3/4

= (2⁴)^3⁄4

= 2³ [⸪4×3/4 = 3]

= 8

(iv) 125^-1/3

125^-1/3 = (5×5×5)^-1/3

= (5³)^-1⁄3

= 5^-1 [⸪3×-1/3 = -1]

= 1/5

3. Simplify:

(i) 2^2/3×2^1/5

Solution:

2^2/3×2^1/5 = 2^(2/3)+(1/5) [⸪Since, a^m×aⁿ=a^m+n____ Laws of exponents]

= 2^13/15 [⸪2/3 + 1/5 = (2×5+3×1)/(3×5) = 13/15]

(ii) (1/3³)⁷

Solution:

(1/3³)⁷ = (3^-3)⁷ [⸪Since,(a^m)ⁿ = a^{m x n}____ Laws of exponents]

= 3^-21

(iii) 11^1/2/11^1/4

Solution:

11^1/2/11^1/4 = 11^(1/2)-(1/4)

= 11^1/4 [⸪(1/2) – (1/4) = (1×4-2×1)/(2×4) = 4-2)/8 = 2/8 = ¼ ]

(iv) 7^1/2×8^1/2

Solution:

7^1/2×8^1/2 = (7×8)^1/2 [⸪Since, (a^m×b^m = (a×b)^m ____ Laws of exponents]

= 56^1/2