Question Number 170669 by solomonwells last updated on 28/May/22
Commented by kaivan.ahmadi last updated on 28/May/22
$${y}={log}_{\mathrm{6}} {x}\Rightarrow{x}=\mathrm{6}^{{y}} \\ $$$$\mathrm{6}^{{y}^{\mathrm{2}} } +\left(\mathrm{6}^{{y}} \right)^{{y}} =\mathrm{12}\Rightarrow\mathrm{2}×\mathrm{6}^{{y}^{\mathrm{2}} } =\mathrm{12}\Rightarrow\mathrm{6}^{{y}^{\mathrm{2}} } =\mathrm{6}\Rightarrow \\ $$$${y}^{\mathrm{2}} =\mathrm{1}\Rightarrow{y}=\pm\mathrm{1} \\ $$$${if}\:{y}=\mathrm{1}\Rightarrow{log}_{\mathrm{6}} {x}=\mathrm{1}\Rightarrow{x}=\mathrm{6} \\ $$$${if}\:{y}=−\mathrm{1}\Rightarrow{log}_{\mathrm{6}} {x}=−\mathrm{1}\Rightarrow{x}=\mathrm{6}^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Commented by solomonwells last updated on 28/May/22
$$\boldsymbol{{i}}\:\boldsymbol{{really}}\:\boldsymbol{{appreciate}}\:\boldsymbol{{this}}\:\boldsymbol{{sir}} \\ $$