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

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

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

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

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