Question Number 155992 by cortano last updated on 07/Oct/21
$$\:\left(\mathrm{1}+\mathrm{log}\:_{\mathrm{3}} \:\mathrm{x}\right).\sqrt{\mathrm{log}\:_{\mathrm{3x}} \:\sqrt[{\mathrm{3}}]{\frac{\mathrm{x}}{\mathrm{3}}}}\:\leqslant\:\mathrm{2} \\ $$
1-log-3-x-log-3x-x-3-1-3-2-