RD Sharma Solutions | Class 9 | Chapter 2 Exponents Of Real Numbers - GMS - Learning Simply
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RD Sharma Solutions | Class 9 | Chapter 2 Exponents Of Real Numbers

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  • RD Sharma Solutions for Class 9 Maths Chapter 2 Exponents of Real Numbers

    RD Sharma Solutions Class 9 Chapter 2 helps students to understand concepts like integral exponents of a real number, laws of exponents and rational powers. To facilitate easy learning and help students understand the concepts of exponents of Real Numbers, free RD Sharma solutions are provided here which can be further downloaded in the form of a PDF. Practice questions related to exponents by solving the RD Sharma textbook questions for class 9 Maths Chapter 2. To download pdf file of this material click on the below link.

    If a is a positive real number and n is a positive integer, then the principal nth root of a is the unique positive real number x such that xn = a. The principal nth root of a is denoted by a(1/n).


    Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 2 Exponents of Real Numbers

    Exercise 2.1

    Question 1: Simplify the following

    (i) 3(a4 b3)10 x 5 (a2 b2)3

    (ii) (2x -2 y3)3

    RD Sharma Class 9 Chapter 2 Exponents of Real Numbers Ex 2.1 Question 1

    Solution:

    Using laws: (am)n = amn , a0 = 1, a-m = 1/a and am x an = am+n]

    (i) 3(a4 b3)10 x 5 (a2 b2)3

    On simplifying the given equation, we get;

    = 3(a40 b30) x 5 (a6 b6)

    = 15 (a46 b36)

    [using laws: (am)n = amn and am x an = am+n]

    (ii) (2x -2 y3)3

    On simplifying the given equation, we get;

    = (23 x -2 × 3 y3×3)

    = 8 x -6 y9

    (iii)

    RD Sharma Class 9 Chapter 2 Exponents of Real Numbers Ex 2.1 Question 1 solution
    RD Sharma Class 9 Chapter 2 Exponents of Real Numbers Ex 2.1 Question 1 solution

    Question 2: If a = 3 and b =-2, find the values of:

    (i) aa+ bb

    (ii) ab + ba

    (iii) (a+b)ab

    Solution:

    (i) aa+ bb

    Now putting the values of ‘a’ and ‘b’, we get;

    = 3+ (−2)−2

    = 33 + (−1/2)2

    = 27 + 1/4

    = 109/4

    (ii) ab + ba

    Now putting the values of ‘a’ and ‘b’, we get;

    = 3−2 + (−2)3

    = (1/3)2 + (−2)3

    = 1/9 – 8

    = −71/9

    (iii) (a+b)ab

    Now putting the values of ‘a’ and ‘b’, we get;

    = (3 + (−2))3(−2)

    = (3–2))−6

    = 1−6

    = 1

    Question 3: Prove that

    RD Sharma Class 9 Chapter 2 Maths Ex 2.1 Question 3

    Solution:

    (i) L.H.S. =

    RD Sharma Class 9 Chapter 2 Maths Ex 2.1 Question 3 solution

    = R.H.S.

    (ii) We have to prove here;

    RD Sharma Class 9 Chapter 2 Maths Ex 2.1 Question 3 solution

    L.H.S. =

    RD Sharma Class 9 Chapter 2 Maths Ex 2.1 Question 3 solution

    =R.H.S.

    (iii) L.H.S. =

    RD Sharma Class 9 Chapter 2 Maths Ex 2.1 Question 3 solution

    Question 4: Prove that

    RD Sharma Class 9 Chapter 2 Maths Ex 2.1 Question 4

    Solution:

    (i) L.H.S

    RD Sharma Class 9 Chapter 2 Maths Ex 2.1 Question 4 solution

    = R.H.S.

    (ii) L.H.S

    RD Sharma Class 9 Chapter 2 Maths Ex 2.1 Question 4 solution

    = R.H.S.

    Question 5: Prove that

    RD Sharma Class 9 Chapter 2 Maths Ex 2.1 Question 5

    Solution:

    (i) L.H.S.

    RD Sharma Class 9 Chapter 2 Maths Ex 2.1 Question 5 solution

    = R.H.S.

    (ii)

    L.H.S.

    RD Sharma Class 9 Chapter 2 Maths Ex 2.1 Question 5 solution

    = R.H.S.

    Question 6: If abc = 1, show that

    RD Sharma Class 9 Chapter 2 Maths Ex 2.1 Question 6

    Solution:

    RD Sharma Class 9 Chapter 2 Maths Ex 2.1 Question 6 solution

    Exercise 2.2

    Question 1: Assuming that x, y, z are positive real numbers, simplify each of the following:

    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers

    Solution:

    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers
    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers
    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers
    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers

    Question 2: Simplify

    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers

    Solution:

    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers
    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers
    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers
    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers

    Question 3: Prove that

    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers

    Solution:

    (i) L.H.S.

    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers

    =R.H.S.

    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers
    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers
    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers
    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers
    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers
    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers

    Question 4.

    Show that:

    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers

    Solution:

    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers
    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers
    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers
    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers
    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers
    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers
    RD Sharma Solutions Chapter 2 Exercise 2.2 Exponents of Real Numbers

    Exercise-VSAQs

    Question 1: Write (625)–1/4 in decimal form.

    Solution:

    (625)–1/4 = (54)-1/4 = 5-1 = 1/5 = 0.2

    Question 2: State the product law of exponents:

    Solution:

    To multiply two parts having same base, add the exponents.

    Mathematically: xm x xn = xm +n

    Question 3: State the quotient law of exponents.

    Solution:

    To divide two exponents with the same base, subtract the powers.

    Mathematically: xm ÷ xn = xm – n

    Question 4: State the power law of exponents.

    Solution:

    Power law of exponents :

    (xm)n = xm x n = xmn

    Question 5: For any positive real number x, find the value of

    RD Sharma Solutions for Class 9 Maths Chapter 2

    Solution:

    RD Sharma Solutions for Class 9 Maths Chapter 2

    Question 6: Write the value of {5(81/3 + 271/3 ) 3}1/4 .

    Solution:

    {5(81/3 + 271/3 ) 3}1/4

    = {5(23×1/3 + 33×1/3 ) 3}1/4

    = { 5(2 + 3)^3}1/4

    = (5) 1/4

    = 5

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