Question Number 171990 by Mikenice last updated on 22/Jun/22
$${solve}: \\ $$$${log}_{\mathrm{7}} \mathrm{2}\:+\:{log}_{\mathrm{49}} {x}\:={log}_{\mathrm{7}} \sqrt{\mathrm{3}} \\ $$
Answered by nurtani last updated on 23/Jun/22
$$\Leftrightarrow\:{log}_{\mathrm{7}} \:\mathrm{2}+{log}_{\mathrm{7}} \:{x}^{\frac{\mathrm{1}}{\mathrm{2}}} ={log}_{\mathrm{7}} \:\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Leftrightarrow\:{log}_{\mathrm{7}} \:\mathrm{2}{x}^{\frac{\mathrm{1}}{\mathrm{2}}} =\:{log}_{\mathrm{7}} \:\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Leftrightarrow\:\mathrm{2}{x}^{\frac{\mathrm{1}}{\mathrm{2}}} =\:\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Leftrightarrow\:\left(\mathrm{2}{x}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)^{\mathrm{2}} =\:\left(\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\:\mathrm{4}{x}\:=\:\mathrm{3} \\ $$$$\Leftrightarrow\:{x}\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$