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JEE Advance - Mathematics (1978 - No. 15)

Solve the following equation for $$x:\,\,2\,{\log _x}a + {\log _{ax}}a + 3\,\,{\log _{{a^2}x}}\,a = 0,a > 0$$
$$x = a^{-1/2}$$
$$x = a^{1/2}$$
$$x = a^{-4/3}$$
$$x = a^{4/3}$$
$$x = a^2$$

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