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

### Question 1 (i). Simplify 3(a4b3)10 Ã— 5(a2b2)3

Solution:

Given 3(a4b3)10 Ã— 5(a2b2)3

= 3 Ã— a40 Ã— b30 Ã— 5 Ã— a6 Ã— b6

= 3 Ã— a46 Ã— b36 Ã— 5                    [am Ã— an = am+n]

= 15 Ã— a46 Ã— b36

= 15a46b36

Thus, 3(a4b3)10 Ã— 5(a2b2)3 = 15a46b36

### Question 1 (ii). Simplify (2x-2y3)3

Solution:

Given (2x-2y3)3

= 23 Ã— x-6 Ã— y9

= 8 Ã— x-6 Ã— y9                             [am Ã— an = am+n]

= 8x-6y9

Thus, (2x-2y3)3 = 8x-6y9

### Question 1 (iii). Simplify

Solution:

Given

=

=                  [am Ã— an = am+n]

=

= 3/102

= 3/100

Thus,

### Question 1 (iv). Simplify

Solution:

Given

[am Ã— an = am+n]

= -2Ã—a2Ã—b5Ã—a-2Ã—b-2

= -2Ã—a2+(-2)Ã—b5+(-2)                       [am Ã— an = am+n]

= -2Ã—a0Ã—b3

= -2b3                                             [a0=1]

Thus, =-2b3

### Question 1 (v). Simplify

Solution:

Given

[am Ã— an = am+n]

Thus,

### Question 1 (vi). Simplify

Solution:

Given

[(am)n = amn]

= a18n-54 Ã— a-(2n-4)                                            [am Ã— an = am+n]

= a18n-54-2n+4

= a16n-50

Thus, = a16n-50

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

Solution:

Given a = 3 and b = -2

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

aa + bb = 33 + (-2)-2

= 27 + 1/4

= (108 + 1)/4

= 109/4

Thus, aa + bb = 109/4

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

Solution:

Given a = 3 and b = -2

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

ab + ba = 3-2 + (-2)3

= 1/9 + (-8)

= (1 – 72)/9

= -71/9

Thus, ab + ba = -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))3Ã—-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 (am)n = amn, we get

By using the formula am/an = am-n, we get

By using the formula am Ã— an = am+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, (am)n = amn, we get

[am Ã— an = am+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 (am)n = amn, we get

By using the formula am/an = am-n, we get

By using the formula am Ã— an = am+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

â‡’ 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

[am Ã— an = am+n]

= 33n+2-(3n-3)                                                    [am/an = am-n]

= 35

= 243

Thus, = 243

### Question 7 (ii). Simplify

Solution:

Given

[am Ã— an = am+n]

= 4/24

= 1/6

Thus, = 1/6

Solution:

Given,

=

= (19 Ã— 5)/5

= 19

Thus,

Solution:

Given

= (48 + 4)/13

= 52/13

= 4

Thus,

### Question 8 (i). Solve the equation 72x+3 = 1 for x.

Solution:

Given equation 72x+3 = 1

We know that, for any aâˆˆ Real numbers, a0 = 1

Let a = 7

â‡’ 72x+3 = 70

Since the bases are equal, let us equate the exponents

â‡’ 2x + 3 = 0

â‡’ x = -3/2

Thus, the value of x is -3/2

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

Solution:

Given 2x+1 = 4x-3

We can write 4 = 22

â‡’ 2x+1 = 22(x-3)

â‡’ 2x+1 = 22x-6

Since the bases are equal, let us equate the exponents

â‡’ x + 1 = 2x – 6

â‡’ x = 7

Thus, the value of x is 7

### Question 8 (iii). Solve the equation 25x+3 = 8x+3 for x.

Solution:

Given 25x+3 = 8x+3

We know that 8 = 23

â‡’ 25x+3 = 23(x+3)

â‡’ 25x+3 = 23x+9

Since the bases are equal, let us equate the exponents

â‡’ 5x + 3 = 3x + 9

â‡’ 5x – 3x = 9 – 3

â‡’ 2x = 6

â‡’ x = 3

Thus, the value of x is 3

### Question 8 (iv). Solve the equation 42x = 1/32 for x.

Solution:

Given 42x = 1/32

â‡’ 22(2x) = 1/32

â‡’ 22(2x) Ã— 32 = 1

â‡’ 24x Ã— 25 = 1

â‡’ 24x+5 = 20

Since the bases are equal, let us equate the exponents

â‡’ 4x + 5 = 0

â‡’ x = -5/4

Thus, the value of x is -5/4

### Question 8 (v). Solve the equation 4x – 1 Ã— (0.5)3-2x = (1/8)x for x.

Solution:

Given 4x – 1 Ã— (0.5)3-2x = (1/8)x

â‡’

â‡’

â‡’ 22(x-1) Ã— 2-(3-2x) = 2-3x

â‡’ 22x-2-3+2x = 2-3x

â‡’ 24x-5 = 2-3x

Since the bases are equal, let us equate the exponents

â‡’ 4x – 5 = -3x

â‡’ 7x = 5

â‡’ x = 5/7

Thus, the value of x is 5/7

### Question 8 (vi). Solve the equation 23x-7 = 256 for x.

Solution:

Given 23x-7 = 256

â‡’ 23x-7 = 28

Since the bases are equal, let us equate the exponents

â‡’ 3x – 7 = 8

â‡’ x = 15/3

â‡’ x = 5

Thus, the value of x is 5

### Question 9 (i). Solve the equation 22x – 2x+3 + 24 = 0 for x.

Solution:

Given 22x – 2x+3 + 24 = 0

â‡’ (2x)2 – 2 Ã— 2x Ã— 22 + (22)2 = 0

â‡’ (2x – 22)2 = 0

â‡’ 2x – 22 = 0

â‡’ 2x = 22

Since the bases are equal, let us equate the exponents

â‡’ x = 2

Thus, the value of x is 2

### Question 9 (ii). Solve the equation 32x+4 + 1 = 2.3x+2 for x.

Solution:

Given 32x+4 + 1 = 2.3x+2

â‡’

â‡’

â‡’ (3x+2 – 1)2 = 0

â‡’ 3x+2 – 1 = 0

â‡’ 3x+2 = 30

Since the bases are equal, let us equate the exponents

â‡’ x + 2 = 0

â‡’ 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 Ã— 32 Ã— 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 = 2a3b7c, find a, b and c.

Solution:

Given 1176 = 2a3b7c

Let us first find out prime factorization of 1176

Thus, 1176 = 23 Ã— 31 Ã— 72

1176 = 233172 = 2a3b7c

Thus, on comparing, we get

a = 3, b = 1, c = 2

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

### (ii) the value of 2-a3b7c

Solution:

Given 4725 = 3a5b7c

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

Thus, 4725 = 33 Ã— 52 Ã— 71

4725 = 335271 = 3a5b7c

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-a3b7c

2-a3b7c= 2-3Ã—32Ã—71

= 1/8 Ã— 9 Ã— 7 = 63/8

Thus, the value of 2-a3b7c is 63/8

### Question 13. If a = xyp-1, b = xyq-1, c = xyr-1, prove that aq-rbr-pcp-q = 1.

Solution:

Given a = xyp-1, b = xyq-1, c = xyr-1

aq-rbr-pcp-q=

= xq-r+r-p+p-q y(p-1)(q-r)+(r-p)(q-1)+(p-q)(r-1)

= xq-r+r-p+p-q ypq-q-pr+r+rq-r-pq+p+pr-p-qr+q

= x0y0

= 1

Thus, we proved that aq-rbr-pcp-q = 1

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