Class 9 RD Sharma Solutions - Chapter 2 Exponents of Real Numbers- Exercise 2.1

Chapter 2 of Class 9 RD Sharma focuses on the concept of exponents of real numbers. This chapter introduces students to the fundamental principles of the exponents and their application in simplifying mathematical expressions. Understanding exponents is crucial for grasping more advanced topics in algebra and calculus.

Exponents of Real Numbers

Exponents also known as indices represent the power to which a number or variable is raised. They are essential in expressing large numbers in the compact form and in solving equations involving exponential terms. The general form is a^b where a is the base and b is the exponent. The Exponents follow specific rules and properties such as the product rule, quotient rule, and power rule which are used to simplify and manipulate expressions involving powers.

Question 1 (i). Simplify 3(a⁴b³)¹⁰ × 5(a²b²)³

Solution:

Given 3(a⁴b³)¹⁰ × 5(a²b²)³

= 3 × a⁴⁰ × b³⁰ × 5 × a⁶ × b⁶

= 3 × a⁴⁶ × b³⁶ × 5                    [a^m × aⁿ = a^m+n]

= 15 × a⁴⁶ × b³⁶

= 15a⁴⁶b³⁶

Thus, 3(a⁴b³)¹⁰ × 5(a²b²)³ = 15a⁴⁶b³⁶

Question 1 (ii). Simplify (2x^-2y³)³

Solution:

Given (2x^-2y³)³

= 2³ × x^-6 × y⁹

= 8 × x^-6 × y⁹                             [a^m × aⁿ =a^m+n]

= 8x^-6y⁹

Thus, (2x^-2y³)³ = 8x^-6y⁹

Question 1 (iii). Simplify 

Solution:

Given 

=

=                  [a^m × aⁿ = a^m+n]

=

= 3/10²

= 3/100

Thus, 

Question 1 (iv). Simplify 

Solution:

Given 

= 

= 

=                      [a^m × aⁿ = a^m+n]

= -2×a²×b⁵×a^-2×b^-2

= -2×a^2+(-2)×b^5+(-2)                       [a^m × aⁿ = a^m+n]

= -2×a⁰×b³

= -2b³                                             [a⁰=1]

Thus, =-2b³

Question 1 (v). Simplify 

Solution:

Given 

= 

=                            [a^m × aⁿ = a^m+n]

Thus, 

Question 1 (vi). Simplify 

Solution:

Given 

=                               [(a^m)ⁿ = a^mn]

= 

= a^18n-54 × a^-(2n-4)                                            [a^m × aⁿ = a^m+n]

= a^18n-54-2n+4

= a^16n-50

Thus, = a^16n-50

Question 2 (i) If a = 3 and b = -2,find the value of a^a + b^b

Solution:

Given a = 3 and b = -2

On substituting the value of a and b in a^a + b^b, we get

a^a + b^b = 3³ + (-2)^-2

= 27 + 1/4

= (108 + 1)/4

= 109/4

Thus, a^a + b^b = 109/4

Question 2 (ii). If a = 3 and b = -2,find the value of a^b + b^a

Solution:

Given a = 3 and b = -2

On substituting the value of a and b in a^b + b^a, we get

a^b + b^a = 3^-2 + (-2)³

= 1/9 + (-8)

= (1 - 72)/9

= -71/9

Thus, a^b + b^a = -71/9

Question 2 (iii). If a = 3 and b = -2,find the value of (a + b)^ab.

Solution:

Given a = 3 and b = -2

On substituting the value of a and b in (a + b)^ab, we get

(a + b)^ab = (3 + (-2))^3×-2

= (1)^-6

= 1

Thus, (a + b)^ab = 1

Question 3 (i). Prove that 

Solution:

Let us first solve left-hand side of the given equation

By using the formula (a^m)ⁿ = a^mn, we get

= 

By using the formula a^m/aⁿ = a^m-n, we get

= 

= 

= 

By using the formula a^m × aⁿ = a^m+n , we get

=

= x

= 1

= Right-hand side of the given equation

Thus, we proved that 

Question 3 (ii). Prove that 

Solution:

Let us consider the left-hand side of the given equation

By using the formula, (a^m)ⁿ = a^mn, we get

= 

= 

=                                    [a^m × aⁿ = a^m+n]

= 1

= Right-hand side of the given equation

Thus, we proved that 

Question 3 (iii). Prove that 

Solution:

Let us first solve left-hand side of the given equation

By using the formula (a^m)ⁿ = a^mn, we get

= 

By using the formula a^m/aⁿ = a^m-n, we get

= 

= 

= 

By using the formula a^m × aⁿ = a^m+n , we get

= 

= 

= Right-hand side of the given equation

Thus, we proved that 

Question 4 (i). Prove that 

Solution:

Let us first consider the left-hand side of given equation

= 

= 

= 

= 

= 1

= Right-hand side of the given equation

Thus, we proved that 

Question 4 (ii). Prove that 

Solution:

Let us first consider the left-hand side of given equation

= 

= 

= 

= 

= 1 

= Right-hand side of the given equation

Thus, we proved that 

Question 5 (i). Prove that 

Solution:

Let us first consider the left-hand side of given equation

= 

= 

= abc

= Right hand side of the given equation

Thus, we proved that 

Question 5 (ii). Prove that 

Solution:

Let us first consider the left hand side of given equation

= 

= 

= 

= 

= Right hand side of the given equation

Thus, we proved that 

Question 6. If abc = 1, show that 

Solution:

Given abc = 1

⇒ c = 1/ab

Let us first consider the left-hand side of given equation

= 

= 

= 

By substituting the value of c in above equation, we get

= 

= 

= 

= 

= 

= 1

= Right hand side of the given equation

Thus, we have shown that if abc = 1, 

Question 7 (i). Simplify 

Solution:

Given 

= 

= 

=                           [a^m × aⁿ = a^m+n]

= 

= 3^3n+2-(3n-3)                                                    [a^m/aⁿ = a^m-n]

= 3⁵

= 243

Thus, = 243

Question 7 (ii). Simplify 

Solution:

Given 

= 

= 

=                          [a^m × aⁿ = a^m+n]

= 

= 4/24

= 1/6

Thus, = 1/6

Question 7 (iii). Simplify 

Solution:

Given, 

= 

=

= (19 × 5)/5

= 19

Thus, 

Question 7 (iv). Simplify 

Solution:

Given 

= 

= 

= 

= 

= 

= (48 + 4)/13

= 52/13 

= 4

Thus, 

Question 8 (i). Solve the equation 7^2x+3 = 1 for x.

Solution:

Given equation 7^2x+3 = 1

We know that, for any a∈ Real numbers, a⁰ = 1

Let a = 7

⇒ 7^2x+3 = 7⁰

Since the bases are equal, let us equate the exponents

⇒ 2x + 3 = 0

⇒ x = -3/2

Thus, the value of x is -3/2

Question 8 (ii). Solve the equation 2^x+1 = 4^x-3 for x.

Solution:

Given 2^x+1 = 4^x-3

We can write 4 = 2²

⇒ 2^x+1 =2^2(x-3)

⇒ 2^x+1 = 2^2x-6

Since the bases are equal, let us equate the exponents

⇒ x + 1 = 2x - 6

⇒ x = 7

Thus, the value of x is 7

Question 8 (iii). Solve the equation 2^5x+3 = 8^x+3 for x.

Solution:

Given 2^5x+3 = 8^x+3

We know that 8 = 2³

⇒ 2^5x+3 = 2^3(x+3)

⇒ 2^5x+3 = 2^3x+9

Since the bases are equal, let us equate the exponents

⇒ 5x + 3 = 3x + 9

⇒ 5x - 3x = 9 - 3

⇒ 2x = 6

⇒ x = 3

Thus, the value of x is 3

Question 8 (iv). Solve the equation 4^2x = 1/32 for x.

Solution:

Given 4^2x = 1/32

⇒ 2^2(2x) = 1/32

⇒ 2^2(2x) × 32 = 1

⇒ 2^4x × 2⁵ = 1

⇒ 2^4x+5 = 2⁰

Since the bases are equal, let us equate the exponents

⇒ 4x + 5 = 0

⇒ x = -5/4

Thus, the value of x is -5/4

Question 8 (v). Solve the equation 4^{x - 1} × (0.5)^3-2x = (1/8)^x for x.

Solution:

Given 4^{x - 1} × (0.5)^3-2x = (1/8)^x

⇒ 

⇒ 

⇒ 2^2(x-1) × 2^-(3-2x) = 2^-3x

⇒ 2^2x-2-3+2x = 2^-3x

⇒ 2^4x-5 = 2^-3x

Since the bases are equal, let us equate the exponents

⇒ 4x - 5 = -3x

⇒ 7x = 5

⇒ x = 5/7

 Thus, the value of x is 5/7

Question 8 (vi). Solve the equation 2^3x-7 = 256 for x.

Solution:

Given 2^3x-7 = 256

⇒ 2^3x-7 = 2⁸

Since the bases are equal, let us equate the exponents

⇒ 3x - 7 = 8

⇒ x = 15/3 

⇒ x = 5

Thus, the value of x is 5

Question 9 (i). Solve the equation 2^2x - 2^x+3 + 2⁴ = 0 for x.

Solution:

Given 2^2x - 2^x+3 + 2⁴ = 0

⇒ (2^x)² - 2 × 2^x × 2² + (2²)² = 0

⇒ (2^x - 2²)² = 0

⇒ 2^x - 2² = 0

⇒ 2^x = 2²

Since the bases are equal, let us equate the exponents

⇒ x = 2

Thus, the value of x is 2

Question 9 (ii). Solve the equation 3^2x+4 + 1 = 2.3^x+2 for x.

Solution:

Given 3^2x+4 + 1 = 2.3^x+2

⇒ 

⇒ 

⇒ (3^x+2 - 1)² = 0

⇒ 3^x+2 - 1 = 0

⇒ 3^x+2 = 3⁰

Since the bases are equal, let us equate the exponents

⇒ x + 2 = 0

⇒ x = -2

Thus, the value of x is -2

Question 10. If 49392 = a⁴b²c³, find the values of a, b, and c where a, b and c are different positive primes.

Solution:

 Let us first find out prime factorization of 49392

👁 Image

Thus, 49392 = 2⁴ × 3² × 7³

Where 2, 3 and 7 are positive primes

49392 = 2⁴3²7³ = a⁴b²c³

Thus, on comparing, we get

a = 2,b = 3 and c = 7

Thus, the values of a, b and c are 2, 3, 7 respectively.

Question 11. If 1176 = 2^a3^b7^c, find a, b and c.

Solution:

Given 1176 = 2^a3^b7^c

 Let us first find out prime factorization of 1176

👁 Image

Thus, 1176 = 2³ × 3¹ × 7²

1176 = 2³3¹7² = 2^a3^b7^c

Thus, on comparing, we get

a = 3, b = 1, c = 2

Thus, the values of a, b and c are 3, 1, 2 respectively.

Question 12. Given 4725 = 3^a5^b7^c, find

(i) the integral values of a, b and c

(ii) the value of 2^-a3^b7^c

Solution:

Given 4725 = 3^a5^b7^c

(i) Let us first find out prime factorization of 4725

👁 Image

Thus, 4725 = 3³ × 5² × 7¹

4725 = 3³5²7¹ = 3^a5^b7^c

Thus, on comparing, we get

a = 3,b = 2,c = 1

Thus, the values of a, b and c are 3,2,1 respectively.

(ii) Here a = 3, b = 2, c = 1

On substituting these values in 2^-a3^b7^c

2^-a3^b7^c= 2^-3×3²×7¹

= 1/8 × 9 × 7 = 63/8

Thus, the value of 2^-a3^b7^c is 63/8

Question 13. If a = xy^p-1, b = xy^q-1, c = xy^r-1, prove that a^q-rb^r-pc^p-q = 1.

Solution:

Given a = xy^p-1, b = xy^q-1, c = xy^r-1

a^q-rb^r-pc^p-q=

= 

= x^q-r+r-p+p-q y^{(p-1)(q-r)+(r-p)(q-1)+(p-q)(r-1)}

= x^q-r+r-p+p-q y^{pq-q-pr+r+rq-r-pq+p+pr-p-qr+q}

= x⁰y⁰

= 1

Thus, we proved that a^q-rb^r-pc^p-q = 1

Conclusion

Mastering exponents is foundational for the advancing in algebra and higher mathematics. By understanding the rules and properties of the exponents students can efficiently simplify expressions and solve complex equations. This knowledge not only aids in the mathematical problem-solving but also in applying these concepts in the real-world scenarios and other scientific disciplines.