Question Number 193391 by DAVONG last updated on 12/Jun/23
Answered by aba last updated on 12/Jun/23
$$\left(\mathrm{1}\right)\:\mathrm{log}_{\mathrm{a}} \left(\mathrm{6}\right)−\mathrm{log}_{\mathrm{a}} \left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\mathrm{log}_{\mathrm{a}} \left(\frac{\mathrm{6}}{\mathrm{x}}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{ln}\left(\frac{\mathrm{6}}{\mathrm{x}}\right)=\mathrm{ln}\left(\sqrt{\mathrm{a}}\right)\:\Rightarrow\:\mathrm{x}=\frac{\mathrm{6}}{\:\sqrt{\mathrm{a}}}\:\checkmark \\ $$
Commented by MATHEMATICSAM last updated on 12/Jun/23
$$\left.\mathrm{1}\right)\:\mathrm{log}_{\mathrm{a}} \mathrm{6}\:−\:\mathrm{log}_{\mathrm{a}} \mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{log}_{\mathrm{a}} \left(\frac{\mathrm{6}}{\mathrm{x}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{a}^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\frac{\mathrm{6}}{\mathrm{x}} \\ $$$$\Rightarrow\:\mathrm{x}\:=\:\frac{\mathrm{6}}{\mathrm{a}^{\frac{\mathrm{1}}{\mathrm{2}}} }\:=\:\frac{\mathrm{6}}{\:\sqrt{\mathrm{a}}} \\ $$
Answered by aba last updated on 12/Jun/23
$$\alpha+\beta=\frac{\pi}{\mathrm{4}\:}\:\Rightarrow\:\mathrm{tg}\left(\alpha+\beta\right)=\mathrm{1}\:\Rightarrow\:\frac{\mathrm{tg}\left(\alpha\right)+\mathrm{tg}\left(\beta\right)}{\mathrm{1}−\mathrm{tg}\left(\alpha\right)\mathrm{tg}\left(\beta\right)}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{tg}\left(\alpha\right)+\mathrm{tg}\left(\beta\right)=\mathrm{1}−\mathrm{tg}\left(\alpha\right)\mathrm{tg}\left(\beta\right) \\ $$$$\Rightarrow\mathrm{tg}\left(\alpha\right)\left(\mathrm{1}+\mathrm{tg}\left(\beta\right)\right)+\mathrm{tg}\left(\beta\right)+\mathrm{1}=\mathrm{2} \\ $$$$\Rightarrow\left(\mathrm{1}+\mathrm{tg}\left(\alpha\right)\right)\left(\mathrm{1}+\mathrm{tg}\left(\beta\right)\right)=\mathrm{2}\:\checkmark \\ $$