If $\dfrac{\text{log x}}{\text{log 5}} = \dfrac{\text{log y}^2}{\text{log 2}} = \dfrac{\text{log 9}}{\text{log}\dfrac{1}{3}}$, find x and y.

Given,

$\dfrac{\text{log x}}{\text{log 5}} = \dfrac{\text{log y}^2}{\text{log 2}} = \dfrac{\text{log 9}}{\text{log}\dfrac{1}{3}}$

Considering,

$\Rightarrow \dfrac{\text{log x}}{\text{log 5}} = \dfrac{\text{log 9}}{\text{log}\dfrac{1}{3}} \\[1em] \Rightarrow \text{log x} = \dfrac{\text{log 9 × log 5}}{\text{log}\dfrac{1}{3}} \\[1em] \Rightarrow \text{log x} = \dfrac{\text{log 9 × log 5}}{\text{log 1 - log 3}} \\[1em] \Rightarrow \text{log x} = \dfrac{\text{log 3}^2 × \text{log } 5}{\text{0 - log 3}} \\[1em] \Rightarrow \text{log x} = \dfrac{\text{2log 3 × log 5}}{\text{-log 3}} \\[1em] \Rightarrow \text{log x} = \text{-2log 5} \\[1em] \Rightarrow \text{log x} = \text{log 5}^{-2} \\[1em] \Rightarrow x = 5^{-2} = \dfrac{1}{25}.$

Considering,

$\Rightarrow \dfrac{\text{log y}^2}{\text{log 2}} = \dfrac{\text{log 9}}{\text{log}\dfrac{1}{3}} \\[1em] \Rightarrow \text{log y}^2 = \dfrac{\text{log 9} \times \text{log 2}}{\text{log 1 - log 3}} \\[1em] \Rightarrow \text{log y}^2 = \dfrac{\text{log 3}^2 \times \text{log 2}}{\text{0 - log 3}} \\[1em] \Rightarrow \text{log y}^2 = \dfrac{\text{2log 3} \times \text{log 2}}{\text{-log 3}} \\[1em] \Rightarrow \text{log y}^2 = \text{-2log 2} \\[1em] \Rightarrow \text{log y}^2 = \text{log 2}^{-2} \\[1em] \Rightarrow y^2 = 2^{-2} \\[1em] \Rightarrow y = \sqrt{\dfrac{1}{2^2}} = \dfrac{1}{2}.$

Hence, $x = \dfrac{1}{25} \text{and y} = \dfrac{1}{2}.$