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

• A. 250
• B. 25
• C. 100
• D. 150

Answer 1

The Correct Option is D

We are given: \[ \log_5(x+y) + \log_5(x-y) = 3 \]

Using the property \( \log_b(m) + \log_b(n) = \log_b(mn) \), we get: \[ \log_5[(x+y)(x-y)] = 3 \Rightarrow (x+y)(x-y) = 5^3 = 125 \Rightarrow x^2 - y^2 = 125 \tag{1} \]

Also given: \[ \log_2 y - \log_2 x = 1 - \log_2 3 \] Using \( \log_b(m) - \log_b(n) = \log_b(m/n) \), we get: \[ \log_2\left(\frac{y}{x}\right) = \log_2 2 - \log_2 3 = \log_2\left(\frac{2}{3}\right) \Rightarrow \frac{y}{x} = \frac{2}{3} \]

Let: \[ x = 3k, \quad y = 2k \] Substitute into Equation (1): \[ x^2 - y^2 = (3k)^2 - (2k)^2 = 9k^2 - 4k^2 = 5k^2 = 125 \Rightarrow k^2 = 25 \Rightarrow k = 5 \]

Thus: \[ x = 3k = 15, \quad y = 2k = 10 \Rightarrow x \times y = 15 \times 10 = \boxed{150} \]

Final Answer: \( \boxed{150} \)