Question:
Let x and y be positive real numbers such that log5(x+y) + log5(x-y) = 3, and log2y - log2x = 1 - log23. Then xy equals
Let x and y be positive real numbers such that log5(x+y) + log5(x-y) = 3, and log2y - log2x = 1 - log23. Then xy equals
Updated On: Aug 20, 2024
- 250
- 25
- 100
- 150
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The Correct Option is D
Solution and Explanation
log5(x+y) + log5(x-y) = 3
⇒ log5 [ (x+y)(x-y)] = 3
⇒ (x+y)(x-y) = 53 = 125
⇒ x2 - y2 = 125 ... (1)
And log2y - log2x = 1 - log23.
⇒ log2 (y/x) = log22 - log23
⇒ log2 (y/x) = log2(2/3)
⇒ y/x = 2/3
Let x=3k and y=2k .
By Putting the values in (1)
(3k)2 - (2k)2 = 125
⇒ 5k2 = 125
⇒ k = 5
Therefore, x × y = 3k × 2k = 6 × 25 = 150
⇒ log5 [ (x+y)(x-y)] = 3
⇒ (x+y)(x-y) = 53 = 125
⇒ x2 - y2 = 125 ... (1)
And log2y - log2x = 1 - log23.
⇒ log2 (y/x) = log22 - log23
⇒ log2 (y/x) = log2(2/3)
⇒ y/x = 2/3
Let x=3k and y=2k .
By Putting the values in (1)
(3k)2 - (2k)2 = 125
⇒ 5k2 = 125
⇒ k = 5
Therefore, x × y = 3k × 2k = 6 × 25 = 150
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