Question Number 171847 by Mikenice last updated on 21/Jun/22
$${solve}: \\ $$$${log}_{{x}} \mathrm{10}+{log}_{{x}^{\mathrm{2}} } \mathrm{10}=\mathrm{6} \\ $$
Commented by kaivan.ahmadi last updated on 21/Jun/22
$${log}_{{x}} \mathrm{10}+\frac{\mathrm{1}}{\mathrm{2}}{log}_{{x}} \mathrm{10}=\mathrm{6}\Rightarrow\frac{\mathrm{3}}{\mathrm{2}}{log}_{{x}} \mathrm{10}=\mathrm{6}\Rightarrow \\ $$$${log}_{{x}} \mathrm{10}=\mathrm{4}\Rightarrow{x}^{\mathrm{4}} =\mathrm{10}\Rightarrow{x}=\sqrt[{\mathrm{4}}]{\mathrm{10}} \\ $$
Answered by cherokeesay last updated on 21/Jun/22
$$\Leftrightarrow\frac{{ln}\mathrm{10}}{{lnx}}+\frac{{ln}\mathrm{10}}{\mathrm{2}{lnx}}=\mathrm{6}\:\:\:\:\:\:\:\:\left({x}>\mathrm{0}\right) \\ $$$$\frac{\mathrm{3}{ln}\mathrm{10}}{\mathrm{2}{lnx}}=\mathrm{6}\Leftrightarrow\mathrm{10}={x}^{\mathrm{4}} \Leftrightarrow \\ $$$${x}_{\mathrm{1}} =\sqrt[{\mathrm{4}}]{\mathrm{10}},\:\:{x}_{\mathrm{2}} =\sqrt[{\mathrm{4}}]{\mathrm{10}}{i} \\ $$