Question

If 1/2 log x+ 1/2 log y + log 2 = log( x+ y), then

Answer 1

Answer :

Given,

1/2 logx + 1/2 logy + log2 = log(x + y)

⇒ 1/2 (logx + logy) + log2 = log(x + y)

⇒ 1/2 log(xy) + log2 = log(x + y),
since logx + logy = log(xy)

⇒ log(xy) + 2 log2 = 2 log(x + y)

⇒ log(xy) + log(2²) = log(x + y)²,
since a logb = log (b^a)

⇒ log(xy) + log4 = log(x + y)²

⇒ log(xy × 4) = log(x + y)²

⇒ log(4xy) = log(x + y)²

Taking both sides as exponential power, we get

4xy = (x + y)²,
since log (e^a) = a

⇒ 4xy = x² + 2xy + y²

⇒ x² + (2 - 4)xy + y² = 0

⇒ x² - 2xy + y² = 0

⇒ (x - y)² = 0

⇒ x - y = 0

⇒ x = y

∴ The required relation between x and y is x = y.

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Answer 2

Answer:

previous answer is absolutely crct!!

y did u even think of scrolling down man?!