Question

If 3^{x + y} = 81 and 81^{x – y} = 3, then the value of x and y are

(a) \(\frac{17}{8},\frac98\)

(b) \(\frac{17}{8},\frac{15}{8}\)

(c) \(\frac{17}{8},\frac{11}8\)

(d) \(\frac{15}{8},\frac{11}8\)

Answer 1

(b) \(\frac{17}{8}\), \(\frac{15}{8}\)

3^{x + y} = 81 ⇒ 3^{x + y} = 3⁴ = x + y = 4     ... (i) 

81^{x – y} = 3 ⇒ (3⁴)^{x – y} = 3¹

⇒ 4x – 4y = 1                     ... (ii) 

Eqn (i) × 4 + Eqn (ii) gives 

4x + 4y + 4x – 4y = 16 + 1 

⇒ 8x = 17 ⇒ x = \(\frac{17}{8}\) 

Putting x = \(\frac{17}{8}\) in (i), we get \(\frac{17}{8}\) + y = 4

⇒ y = 4 - \(\frac{17}{8}\) = \(\frac{15}{8}\)

∴ x = \(\frac{17}{8}\), y = \(\frac{15}{8}\).