Solve for x and y; if x > 0 and y > 0 :

log xy = log $\dfrac{x}{y}$ + 2 log 2 = 2.

Given,

⇒ log xy = log $\dfrac{x}{y}$ + 2 log 2 = 2 ……..(1)

Solving L.H.S. of the equation :

$\Rightarrow \text{log xy} = \text{log }\dfrac{x}{y} + \text{2 log 2} \\[1em] \Rightarrow \text{log xy} = \text{log }\dfrac{x}{y} + \text{log 2}^2 \\[1em] \Rightarrow \text{log xy} = \text{log }\dfrac{x}{y} + \text{log 4} \\[1em] \Rightarrow \text{log xy} = \text{log} \Big(\dfrac{x}{y} \times 4\Big) \\[1em] \Rightarrow xy = \dfrac{4x}{y} \\[1em] \Rightarrow y^2 = \dfrac{4x}{x} \\[1em] \Rightarrow y^2 = 4 \\[1em] \Rightarrow y = \sqrt{4} = 2.$

From equation (1), we get :

⇒ log xy = 2

⇒ log x(2) = 2

⇒ log 2x = 2

⇒ log 2x = 2log 10

⇒ log 2x = log 10²

⇒ 2x = 10²

⇒ 2x = 100

⇒ x = $\dfrac{100}{2}$ = 50.

Hence, x = 50 and y = 2.