Solve for x and y, when (√243)x ÷ 3y+1 = 1 and 81(4) – (x/2) –

1.	Solve for x and y, when (√243)x ÷ 3y+1 = 1 and 81(4) – (x/2) – 27y = 01. 3, 42. 5, 43. 2, 44. 4, 3
Answer» Correct Answer - Option 3 : 2, 4 Given∶ (√243)^x ÷ 3^y+1 = 1, also 81^{(4) – (x/2)} – 27^y = 0 Formula Used∶ √a = (a)^1/m, (a^m)ⁿ = a^{m × n}, a^m ÷ aⁿ = a^{m – n} Calculation∶ (√243)^x ÷ 3^y+1 = 1 {(3⁵)^1/2}^x ÷ 3^y+1 = 1 (3^5/2)^x ÷ 3^y+1 = 1 3^5x/2 ÷ 3^y+1 = 1 3^{(5x/2) – (y + 1)} = 3⁰ (5x/2) – (y + 1) = 0 (5x/2) – y – 1 = 0 5x – 2y – 2 = 0 ----(1) Also, 81^{4 – (x/2)} – 27^y= 0 (3^{4)[4 - (x/2)]} – (3³)^y = 0 3^{16 – (4x/2)} – 3^3y = 0 3^16 – 2x = 3^3y 16 – 2x = 3y 2x + 3y – 16 = 0 ----(2) Multiply equation (1) by 3 and equation (2) by 2 15x – 6y – 6 = 0 Or, 15x – 6y = 6 ----(3) 4x + 6y – 32 = 0 ----(4) Add equation (3) and (4) (15x – 6y = 6) + (4x + 6y = 32) ⇒ 19x = 38 ⇒ x = 2 Put x = 2 in equation (1) 5x – 2y – 2 = 0 ⇒ 5(2) – 2y – 2 = 0 ⇒ 10 – 2y – 2 = 0 ⇒ 2y = 8 ⇒ y = 8/2 = 4 ⇒ x = 2, y = 4 ∴ The value of x and y is 2 and 4.

Solve for x and y, when (√243)x ÷ 3y+1 = 1 and 81(4) – (x/2) – 27y = 01. 3, 42. 5, 43. 2, 44. 4, 3

Answer» Correct Answer - Option 3 : 2, 4

Given∶

(√243)^x ÷ 3^y+1 = 1, also 81^{(4) – (x/2)} – 27^y = 0

Formula Used∶

√a = (a)^1/m, (a^m)ⁿ = a^{m × n}, a^m ÷ aⁿ = a^{m – n}

Calculation∶

(√243)^x ÷ 3^y+1 = 1

{(3⁵)^1/2}^x ÷ 3^y+1 = 1

(3^5/2)^x ÷ 3^y+1 = 1

3^5x/2 ÷ 3^y+1 = 1

3^{(5x/2) – (y + 1)} = 3⁰

(5x/2) – (y + 1) = 0

(5x/2) – y – 1 = 0

5x – 2y – 2 = 0 ----(1)

Also, 81^{4 – (x/2)} – 27^y= 0

(3^{4)[4 - (x/2)]} – (3³)^y = 0

3^{16 – (4x/2)} – 3^3y = 0

3^16 – 2x = 3^3y

16 – 2x = 3y

2x + 3y – 16 = 0 ----(2)

Multiply equation (1) by 3 and equation (2) by 2

15x – 6y – 6 = 0

Or, 15x – 6y = 6 ----(3)

4x + 6y – 32 = 0 ----(4)

Add equation (3) and (4)

(15x – 6y = 6) + (4x + 6y = 32)

⇒ 19x = 38

⇒ x = 2

Put x = 2 in equation (1)

5x – 2y – 2 = 0

⇒ 5(2) – 2y – 2 = 0

⇒ 10 – 2y – 2 = 0

⇒ 2y = 8

⇒ y = 8/2 = 4

⇒ x = 2, y = 4

∴ The value of x and y is 2 and 4.