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Let x and y be positive real numbers such that
log5 (x + y) + log5 (x – y) = 3, and log2 y – log2 x = 1 – log2 3. Then xy equals
log5 (x + y) + log5 (x – y) = 3, and log2 y – log2 x = 1 – log2 3. Then xy equals
(2019)
- a)150
- b)100
- c)250
- d)25
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Let x and y be positive real numbers such thatlog5 (x + y) + log5 (x &...
Given, log5(x + y) + log5(x – y) = 3
⇒ log5[(x + y)(x – y)] = 3
⇒ (x + y)(x – y) = 53 = 125
⇒ x2 – y2 = 125 ...(i)
And log2y – log2x = 1 – log2 3.
Then if x = 3k, then y = 2k
Now putting this value of x and y in (i), we get
(3k)2 – (2k)2 = 125
⇒ 5k2 = 125, ∴ k = 5
∴ x × y = 3k × 2k = 6 × 25 = 150
⇒ log5[(x + y)(x – y)] = 3
⇒ (x + y)(x – y) = 53 = 125
⇒ x2 – y2 = 125 ...(i)
And log2y – log2x = 1 – log2 3.
Then if x = 3k, then y = 2k
Now putting this value of x and y in (i), we get
(3k)2 – (2k)2 = 125
⇒ 5k2 = 125, ∴ k = 5
∴ x × y = 3k × 2k = 6 × 25 = 150
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