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Question Number 192851 by MATHEMATICSAM last updated on 29/May/23

$$\mathrm{Solve}:$$$$\frac{{\log }_{a^2\sqrt{x}}a}{{\log }_{2x}a}+{\log }_{ax}a.{\log }_{\frac{1}{a}}2x=0$$

Commented by Frix last updated on 29/May/23

$$x=\frac{1}{2}\vee x=a^2$$

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