RD Sharma Solutions Class 9 Mathematics Chapter 2 Exercise 2.1 Exponents of Real Numbers are provided here. Class 9 Chapter 2 solutions are solved by subject experts in accordance with the CBSE syllabus for 9th standard. This study material contains RD Sharma Solutions Class 9 Exercise 2.1.
RD Sharma Solutions for Class 9 Maths Chapter 2 Exponents of Real Numbers Exercise 2.1
Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 2 Exponents of Real Numbers Exercise 2.1 Page number 2.12
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:
RD Sharma Solutions for Class 9 Maths Chapter 2 Exponents of Real Numbers Exercise 2.1
RD Sharma Solutions Class 9 Maths Chapter 2 Exponents of Real Numbers Exercise 2.1 is based on the following topics and subtopics:
- Exponents of Real Numbers Introduction
- Integral exponents of a real number
- Positive integral power and negative integral power
- Laws of integral exponents
First law: If a is any real number and m, n are positive integers, then am x an = a(mn)
Second Law: If a is non-zero real number and m, n are positive integers, then am / an = a(m-n)
Third law: If a is any real number and m, n are positive integers, then (am)n = a(mn) = (an)m
Fourth Law: If a and b are real numbers and m, n are positive integers then (ab)n = an bn and (a/b)n = an/bn, where b is not equal to zero.