Question Number 193438 by Mingma last updated on 14/Jun/23
Answered by qaz last updated on 14/Jun/23
$${log}_{\mathrm{3}} \left(\mathrm{9}{x}−\mathrm{3}\right)=\mathrm{1}+{log}_{\mathrm{3}} \left(\mathrm{3}{x}−\mathrm{1}\right)\:\:\:\:\:,{log}_{\mathrm{3}} \left({x}−\frac{\mathrm{1}}{\mathrm{3}}\right)={log}_{\mathrm{3}} \left(\mathrm{3}{x}−\mathrm{1}\right)−\mathrm{1} \\ $$$${log}_{\mathrm{3}} \left(\mathrm{3}{x}−\mathrm{1}\right)={y} \\ $$$$\sqrt{\mathrm{1}+{y}}\leqslant{y}−\mathrm{1}\:\:\:\:\Rightarrow{y}\geqslant\mathrm{3}\:\:\:\:\:\Rightarrow{x}\in\left[\frac{\mathrm{28}}{\mathrm{3}},+\infty\right) \\ $$
Commented by Mingma last updated on 14/Jun/23
Perfect ��