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).

### Exercise 2.1

Question 1: Simplify the following

(i) 3(a4 b3)10 x 5 (a2 b2)3

(ii) (2x -2 y3)3

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)

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

Solution:

(i) L.H.S. =

= R.H.S.

(ii) We have to prove here;

L.H.S. =

=R.H.S.

(iii) L.H.S. =

Question 4: Prove that

Solution:

(i) L.H.S

= R.H.S.

(ii) L.H.S

= R.H.S.

Question 5: Prove that

Solution:

(i) L.H.S.

= R.H.S.

(ii)

L.H.S.

= R.H.S.

Question 6: If abc = 1, show that

Solution:

### Exercise 2.2

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

Solution:

Question 2: Simplify

Solution:

Question 3: Prove that

Solution:

(i) L.H.S.

=R.H.S.

Question 4.

Show that:

Solution:

### 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

Solution:

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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