MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: Rs. 15,900
  • View Details

Chapter 2: Exponents Of Real Numbers Exercise – 2.1

Question: 1

Simplify the following: 

(i) 3(a4 b3)10 × 5 (a2 b2)3 

(ii) (2x-2 y3)3 

Solution:

(i) 3(a4 b3)10 × 5 (a2 b2)3 

= 3(a40 b30) × 5(a6 b6)

= 15 (a46 b36

(ii) (2x-2 y3)3 

(23 × -2 × 3 y3 × 3) = 8x-6y9 

 

Question: 2

If a = 3 and b = - 2, find the values of: 

(i) aa + bb 

(ii) ab + ba 

(iii) ab + ba 

Solution:

(i) We have,

aa + bb 

= 33 + (−2) −2

 = 33 + (−1/2)2 

= 27 +1/4 

= 109/4 

(ii) ab + ba

 = 3−2 + (−2)3

 = (1/3)2 + (−2)3

 = 1/9 – 8

 = −(71/9) 

(iii) We have, 

ab + ba 

= (3 + (−2))3(−2) 

= (3 − 2))−6

 = 1−6 = 1 

 

Question: 3

Prove that:

Solution:

(i) To prove

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

Or, Therefore, LHS = RHS Hence proved

(ii) To prove, 

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

Therefore, LHS = RHS Hence proved

(iii) To prove, 

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

= xac−bc × xba−ca × xbc−ab

= xac − bc + ba − ca + bc − ab

= x0

= 1

Therefore, LHS = RHS

Hence proved 

 

Question: 4

Prove that: 

Solution:

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

Therefore, LHS = RHS Hence proved

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

Therefore, LHS = RHS Hence proved

 

Question: 5

Prove that:

Solution:

(i) To prove,

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

= abc

Therefore, LHS = RHS Hence proved

(ii) To prove,

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

Therefore, LHS = RHS

Hence proved 

 

Question: 6

If abc = 1, show that

Solution:

To prove,

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

We know abc = 1

c = 1/ab

By substituting the value c in equation (1), we get

Therefore, LHS = RHS Hence proved 

 

Question: 7

Simplify: 

Solution:

 

Question: 8

Solve the following equations for x: 

(i) 72x+3 = 1 

(ii) 2x+1 = 4x−3 

(iii) 25x+3 = 8x+3 

(iv) 42x = 1/32 

(v) 4x−1 × (0.5)3−2x = (1/8)x 

(vi) 23x−7 = 256 

Solution:

(i) We have,

⟹ 72x+3 = 1

⟹ 72x+3 = 70 

⟹ 2x + 3 = 0

⟹ 2x = -3

⟹ x = −3/2 

(ii) We have,

= 2x+1 = 4x−3

= 2x+1 = 22x−6

= x + 1 = 2x - 6

= x = 7

(iii) We have,

= 25x+3 = 8x+3

= 25x+3 = 23x+9

= 5x + 3 = 3x + 9

= 2x = 6

= x = 3

(iv) We have,

= 42x = 1/32

= 24x = 1/25

= 24x = 2−5

= 4x = - 5

x = -5/4

(v) We have,

4x−1 × (0.5)3−2x = (1/8)x

22x−2 × (1/2)3−2x = (1/2)3x

22x−2 × 22x−3 = (1/2)3x

22x−2+ 2x−3 = (1/2)3x

24x−5 = 2−3x

4x-5 = -3x

7x = 5

x = 5/7

(vi) 23x−7 = 256

23x−7 = 28

3x - 7 = 8

3x = 15

x = 5

 

Question: 9

Solve the following equations for x: 

(i) 22x − 2x+3 + 24 = 0 

(ii) 32x+4 + 1 = 2 × 3x+2 

Solution:

(i) We have, ⟹ 22x − 2x+3 + 24 = 0 

⟹ 22x + 24 = 2x.23 

⟹ Let 2x = y 

⟹ y2 + 24 = y × 23 

⟹ y2 − 8y + 16 = 0 

⟹ y2 − 4y − 4y + 16 = 0 

⟹ y(y - 4) - 4(y - 4) = 0

⟹ y = 4

⟹ x2 = 22 

⟹ x = 2

(ii) We have,

32x+4 + 1 = 2 × 3x+2

(3x+2)2 + 1 = 2 × 3x+2

Let 3x+2 = y

y2 + 1 = 2y

y2 − 2y + 1 = 0

y2 − y − y + 1 = 0

y(y − 1) − 1(y − 1) = 0

(y − 1)(y − 1) = 0

y = 1

 

Question: 10

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

Solution:

Taking out the LCM, the factors are 24, 32 and 73 a4b2c3 = 24, 32 and 73

a = 2, b = 3 and c = 7 [Since, a, b and c are primes]

 

Question: 11

If 1176 = 2a × 3b × 7c, Find a, b, and c. 

Solution:

Given that 2, 3 and 7 are factors of 1176.

Taking out the LCM of 1176, we get 23 × 31 × 72 = 2a × 3b × 7c 

By comparing, we get

a = 3, b = 1 and c = 2.

 

Question: 12

Given 4725 = 3a × 5b × 7c, find 

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

(ii) The value of 2−a × 3b × 7c 

Solution:

(i) Taking out the LCM of 4725, we get

33 × 52 × 71 = 3a × 5b × 7c

By comparing, we get

a = 3, b = 2 and c = 1.

(ii) The value of 2−a × 3b × 7c 

Sol:

2-a × 3b × 7c = 2−3 × 32 × 71

2−3 × 32 × 71 = 1/8 × 9 × 7

63/8

 

Question: 13

If a = xyp−1, b = xyq−1 and c = xyr−1, prove that aq−r br−p cp−q = 1 

Solution:

Given, a = xyp−1, b = xyq−1 and c = xyr−1 

To prove, aq−rbr−pcp−q = 1

Left hand side (LHS) = Right hand side (RHS)

Considering LHS, = aq−r br−p cp−q ...... (i)

By substituting the value of a, b and c in equation (i), we get

= (xyp−1)q−r(xyq−1)r−p(xyr−1)p−q 

= xypq−pr−q+rxyqr−pq−r+pxyrp−rq−p+q 

= xypq−pr−q+ r+qr−pq−r+p+rp−rq−p+q 

= xy0 

= 1


Course Features

  • 728 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution