Question Number 86515 by jagoll last updated on 29/Mar/20
$$\mathrm{log}_{\mathrm{2}} \:\left(\mathrm{x}\right)\:+\:\mathrm{log}_{\mathrm{3}} \:\left(\mathrm{x}\right)\:=\:\mathrm{1}\: \\ $$$$\mathrm{x}\:=\: \\ $$
Commented by john santu last updated on 29/Mar/20
$$\Rightarrow\:\:^{\mathrm{2}} \mathrm{log}\:\mathrm{x}\:=\:\:^{\mathrm{2}} \mathrm{log}\:\mathrm{3}\:.\:^{\mathrm{3}} \mathrm{log}\:\mathrm{x}\: \\ $$$$\Leftrightarrow\:\:^{\mathrm{3}} \mathrm{log}\:\left(\mathrm{x}\right)\:\left\{\:\:^{\mathrm{2}} \mathrm{log}\:\mathrm{3}\:+\mathrm{1}\:\right\}\:=\:\mathrm{1} \\ $$$$\:^{\mathrm{3}} \mathrm{log}\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\:^{\mathrm{2}} \mathrm{log}\:\mathrm{6}}\:=\:^{\mathrm{6}} \mathrm{log}\:\mathrm{2} \\ $$$$\mathrm{x}\:=\:\mathrm{3}^{\:^{\mathrm{6}} \mathrm{log}\:\mathrm{2}} \\ $$
Commented by jagoll last updated on 29/Mar/20
$$\mathrm{thank}\:\mathrm{you} \\ $$