If log₂ (x + y) = log₃ (x - y) = $\dfrac{\text{log 25}}{\text{log 0.2}}$, find the values of x and y.

Given,

log₂ (x + y) = log₃ (x - y) = $\dfrac{\text{log 25}}{\text{log 0.2}}$

Solving, equation :

$\Rightarrow \text{log}$2 \space (x + y) = \dfrac{\text{log 25}}{\text{log 0.2}} \\[1em] \Rightarrow \text{log}2 \space (x + y) = \text{log}{0.2} \space 25 \\[1em] \Rightarrow \text{log}2 \space (x + y) = \text{log}{\dfrac{2}{10}} \space 25 \\[1em] \Rightarrow \text{log}2 \space (x + y) = \text{log}{\dfrac{1}{5}} \space 5^2 \\[1em] \Rightarrow \text{log}2 \space (x + y) = \text{log}{5^{-1}} \space 5^2 \\[1em] \Rightarrow \text{log}2 \space (x + y) = \dfrac{2}{-1}\text{log}{5} \space 5 \\[1em] \Rightarrow \text{log}2 \space (x + y) = -2 \times 1 \\[1em] \Rightarrow \text{log}_2 \space (x + y) = -2 \\[1em] \Rightarrow x + y = 2^{-2} \\[1em] \Rightarrow x + y = \dfrac{1}{2^2} \\[1em] \Rightarrow x + y = \dfrac{1}{4}\text{ ……(1)}

Solving, equation :

$\Rightarrow \text{log}$3 \space (x - y) = \dfrac{\text{log 25}}{\text{log 0.2}} \\[1em] \Rightarrow \text{log}3 \space (x - y) = \text{log}{0.2} \space 25 \\[1em] \Rightarrow \text{log}3 \space (x - y) = \text{log}{\dfrac{2}{10}} \space 25 \\[1em] \Rightarrow \text{log}3 \space (x - y) = \text{log}{\dfrac{1}{5}} \space 5^2 \\[1em] \Rightarrow \text{log}3 \space (x - y) = \text{log}{5^{-1}} \space 5^2 \\[1em] \Rightarrow \text{log}3 \space (x - y) = \dfrac{2}{-1}\text{log}{5}5 \\[1em] \Rightarrow \text{log}3 \space (x - y) = -2 \times 1 \\[1em] \Rightarrow \text{log}_3 \space (x - y) = -2 \\[1em] \Rightarrow x - y = 3^{-2} \\[1em] \Rightarrow x - y = \dfrac{1}{3^2} \\[1em] \Rightarrow x - y = \dfrac{1}{9}\text{ ……(2)}

Adding equation (1) and (2), we get :

$\Rightarrow (x + y) + (x - y) = \dfrac{1}{4} + \dfrac{1}{9} \\[1em] \Rightarrow x + x + y - y = \dfrac{9 + 4}{36} \\[1em] \Rightarrow 2x = \dfrac{13}{36} \\[1em] \Rightarrow x = \dfrac{13}{36 \times 2} = \dfrac{13}{72}.$

Substituting value of x in equation (1), we get :

$\Rightarrow x + y = \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{13}{72} + y = \dfrac{1}{4} \\[1em] \Rightarrow y = \dfrac{1}{4} - \dfrac{13}{72} \\[1em] \Rightarrow y = \dfrac{18 - 13}{72} \\[1em] \Rightarrow y = \dfrac{5}{72}.$

Hence, x = $\dfrac{13}{72}\text{ and y} = \dfrac{5}{72}$.