Let $x$ and $y$ be positive real numbers such that $\log _{5}(x+y)+\log _{5}(x-y)=3$, and $\log _{2} y-\log _{2} x=1-\log _{2} 3$. Then $x y$ equals

Given:

$\log_5(x+y) + \log_5(x-y) = 3$

$\log_2 y - \log_2 x = 1 - \log_2 3$

$x, y > 0$ (positive real numbers)

Find: $xy$

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We have: $\log_5(x+y) + \log_5(x-y) = 3$

When we add two logarithms with the same base, we multiply what's inside:

$\log_a(m) + \log_a(n) = \log_a(m \times n)$

$\log_5(x+y) + \log_5(x-y) = \log_5[(x+y)(x-y)]$

$(x+y)(x-y) = x^2 - y^2$ (difference of squares formula)

$\log_5(x^2 - y^2) = 3$

Converting from logarithmic to exponential form:

If $\log_5(x^2 - y^2) = 3$, then $x^2 - y^2 = 5^3 = 125$

Therefore: $x^2 - y^2 = 125$ ... (Equation 1)

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We have: $\log_2 y - \log_2 x = 1 - \log_2 3$

When we subtract logarithms with the same base, we divide what's inside:

$\log_a(m) - \log_a(n) = \log_a\left(\dfrac{m}{n}\right)$

Left side: $\log_2 y - \log_2 x = \log_2\left(\dfrac{y}{x}\right)$

Right side: $1 - \log_2 3 = \log_2 2 - \log_2 3 = \log_2\left(\dfrac{2}{3}\right)$

Note: $1 = \log_2 2$ because $2^1 = 2$

$\log_2\left(\dfrac{y}{x}\right) = \log_2\left(\dfrac{2}{3}\right)$

Since the bases are equal, the arguments must be equal:

$\dfrac{y}{x} = \dfrac{2}{3}$

Cross-multiplying: $3y = 2x$

Therefore: $x = \dfrac{3y}{2}$ ... (Equation 2)

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We substitute Equation 2 into Equation 1:

$x^2 - y^2 = 125$

$\left(\dfrac{3y}{2}\right)^2 - y^2 = 125$

$\dfrac{9y^2}{4} - y^2 = 125$

Getting a common denominator:

$\dfrac{9y^2}{4} - \dfrac{4y^2}{4} = 125$

$\dfrac{5y^2}{4} = 125$

$5y^2 = 500$

$y^2 = 100$

$y = 10$ (since $y > 0$)

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Using $x = \dfrac{3y}{2}$:

$x = \dfrac{3 \times 10}{2} = 15$

Therefore: $xy = 15 \times 10 = 150$