If (x4 – 2x2y2 + y2)a –1 = (x – y)2a (x + y)–2, then the value

1.	If (x4 – 2x2y2 + y2)a –1 = (x – y)2a (x + y)–2, then the value of a is(a) x2 – y2 (b) log (xy) (c) \(\frac{log(x-y)}{log(x+y)}\)(d) log (x – y)
Answer» (c) \(\frac{log(x-y)}{log(x+y)}\) Given, (x⁴ – 2x²y² + y²)^{a –1}= (x – y)^2a (x + y)^–2 ⇒ [(x² – y²)²]^{a –1} = (x – y)^2a (x + y)^–2 ⇒ (x – y)^{2(a – 1)} (x + y)^{2(a –1)} = (x – y)^2a (x + y)^–2 ⇒ \(\frac{(x-y)^{2(a-1)}}{(x-y)^{2a}}\) . \(\frac{(x+y)^{2(a-1)}}{(x+y)^{-2}}\) = 1 ⇒ (x-y)^-2 (x+y)^2a = 1 ⇒ log [(x – y)^–2 (x + y)^2a] = log 1 ⇒ –2 log (x – y) + 2a log(x + y) = log 1 ⇒ 2a log (x + y) = 2 log (x – y) ⇒ a = \(\frac{log(x-y)}{log(x+y)}\). [Since log 1 = 0]

If (x4 – 2x2y2 + y2)a –1 = (x – y)2a (x + y)–2, then the value of a is(a) x2 – y2 (b) log (xy) (c) \(\frac{log(x-y)}{log(x+y)}\)(d) log (x – y)

Answer»

(c) \(\frac{log(x-y)}{log(x+y)}\)

Given, (x⁴ – 2x²y² + y²)^{a –1}= (x – y)^2a (x + y)^–2

⇒ [(x² – y²)²]^{a –1} = (x – y)^2a (x + y)^–2

⇒ (x – y)^{2(a – 1)} (x + y)^{2(a –1)} = (x – y)^2a (x + y)^–2

⇒ \(\frac{(x-y)^{2(a-1)}}{(x-y)^{2a}}\) . \(\frac{(x+y)^{2(a-1)}}{(x+y)^{-2}}\) = 1 ⇒ (x-y)^-2 (x+y)^2a = 1

⇒ log [(x – y)^–2 (x + y)^2a] = log 1 ⇒ –2 log (x – y) + 2a log(x + y) = log 1

⇒ 2a log (x + y) = 2 log (x – y) ⇒ a = \(\frac{log(x-y)}{log(x+y)}\). [Since log 1 = 0]