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).
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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;
= 33 + (−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
= (54 ) 1/4
= 5