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} y^{3 × 3}) = 8x^-6y⁹ 

 

Question: 2

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

(i) a^a + b^b 

(ii) a^b + b^a 

(iii) a^b + b^a 

Solution:

(i) We have,

a^a + b^b 

= 3³ + (−2) ⁻²

 = 3³ + (−1/2)² 

= 27 +1/4 

= 109/4 

(ii) a^b + b^a

 = 3⁻² + (−2)³

 = (1/3)² + (−2)³

 = 1/9 – 8

 = −(71/9) 

(iii) We have, 

a^b + b^a 

= (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,

= x^ac−bc × x^ba−ca × x^bc−ab

= x^{ac − 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) 7^2x+3 = 1 

(ii) 2^x+1 = 4^x−3 

(iii) 2^5x+3 = 8^x+3 

(iv) 4^2x = 1/32 

(v) 4^x−1 × (0.5)^3−2x = (1/8)^x 

(vi) 2^3x−7 = 256 

Solution:

(i) We have,

⟹ 7^2x+3 = 1

⟹ 7^2x+3 = 7⁰ 

⟹ 2x + 3 = 0

⟹ 2x = -3

⟹ x = −3/2 

(ii) We have,

= 2^x+1 = 4^x−3

= 2^x+1 = 2^2x−6

= x + 1 = 2x - 6

= x = 7

(iii) We have,

= 2^5x+3 = 8^x+3

= 2^5x+3 = 2^3x+9

= 5x + 3 = 3x + 9

= 2x = 6

= x = 3

(iv) We have,

= 4^2x = 1/32

= 2^4x = 1/2⁵

= 2^4x = 2⁻⁵

= 4x = - 5

x = -5/4

(v) We have,

4^x−1 × (0.5)^3−2x = (1/8)^x

2^2x−2 × (1/2)^3−2x = (1/2)^3x

2^2x−2 × 2^2x−3 = (1/2)^3x

2^{2x−2+ 2x−3} = (1/2)^3x

2^4x−5 = 2^−3x

4x-5 = -3x

7x = 5

x = 5/7

(vi) 2^3x−7 = 256

2^3x−7 = 2⁸

3x - 7 = 8

3x = 15

x = 5

 

Question: 9

Solve the following equations for x: 

(i) 2^2x − 2^x+3 + 2⁴ = 0 

(ii) 3^2x+4 + 1 = 2 × 3^x+2 

Solution:

(i) We have, ⟹ 2^2x − 2^x+3 + 2⁴ = 0 

⟹ 2^2x + 2⁴ = 2^x.2³ 

⟹ Let 2^x = 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,

3^2x+4 + 1 = 2 × 3^x+2

(3^x+2)² + 1 = 2 × 3^x+2

Let 3^x+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 = 2^a × 3^b × 7^c, 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 = 2^a × 3^b × 7^c 

By comparing, we get

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

 

Question: 12

Given 4725 = 3^a × 5^b × 7^c, find 

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

(ii) The value of 2^−a × 3^b × 7^c 

Solution:

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

3³ × 5² × 7¹ = 3^a × 5^b × 7^c

By comparing, we get

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

(ii) The value of 2^−a × 3^b × 7^c 

Sol:

2^-a × 3^b × 7^c = 2⁻³ × 3² × 7¹

2⁻³ × 3² × 7¹ = 1/8 × 9 × 7

63/8

 

Question: 13

If a = xy^p−1, b = xy^q−1 and c = xy^r−1, prove that a^q−r b^r−p c^p−q = 1 

Solution:

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

To prove, a^q−rb^r−pc^p−q = 1

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

Considering LHS, = a^q−r b^r−p c^p−q ...... (i)

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

= (xy^p−1)^q−r(xy^q−1)^r−p(xy^r−1)^p−q 

= xy^{pq−pr−q+r}xy^{qr−pq−r+p}xy^{rp−rq−p+q} 

= xy^{pq−pr−q+ r+qr−pq−r+p+rp−rq−p+q} 

= xy⁰ 

= 1