Integral value of x which satisfies the equation ${\log }_{6}54+{\log }_{\mathrm{x}}16={\log }_{\sqrt{2}}\mathrm{x}-{\log }_{36}(4/9)$ is

$\begin{array}{l}{\log }_{6}54+{\log }_{\mathrm{x}}16={\log }_{\sqrt{2}}\mathrm{x}-{\log }_{36}\frac{4}{9}\\ \text{ or }1+2{\log }_{6}3+{\log }_{\mathrm{x}}16=2{\log }_2\mathrm{x}-{\log }_6\frac{2}{3}\\ \text{ or }1+2{\log }_{6}3+{\log }_{\mathrm{x}}16=2{\log }_2\mathrm{x}-{\log }_{6}2+{\log }_{6}3\\ \text{ or }1+{\log }_{\mathrm{x}}16=2{\log }_2\mathrm{x}-\left({\log }_{6}2+{\log }_{6}3\right)\\ \text{ or }1+{\log }_{\mathrm{x}}16=2{\log }_2\mathrm{x}-1\\ \text{ or }\frac{4}{{\log }_2\mathrm{x}}=2{\log }_2\mathrm{x}-2\end{array}$

$\begin{array}{l}\Rightarrow \mathrm{t}=2\text{ or }\mathrm{t}=-1\\ \Rightarrow \mathrm{x}=4\text{ or }1/2\end{array}$