Question Number 171828 by Mikenice last updated on 21/Jun/22
$${solve}\:{for}\:{x}: \\ $$$$\mathrm{2}{log}_{{x}} \mathrm{4}={log}_{\mathrm{2}} {x} \\ $$$$ \\ $$
Commented by kaivan.ahmadi last updated on 21/Jun/22
$${log}_{{x}} \mathrm{16}={log}_{\mathrm{2}} {x}\Rightarrow\frac{{log}_{\mathrm{2}} \mathrm{16}}{{log}_{\mathrm{2}} {x}}={log}_{\mathrm{2}} {x}\Rightarrow \\ $$$$\mathrm{4}=\left({log}_{\mathrm{2}} {x}\right)^{\mathrm{2}} \Rightarrow\left\{_{{log}_{\mathrm{2}} {x}=−\mathrm{2}\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{4}}} ^{{log}_{\mathrm{2}} {x}=\mathrm{2}\Rightarrow{x}=\mathrm{4}} \right. \\ $$$$ \\ $$
Commented by Mikenice last updated on 21/Jun/22
$${thanks}\:{sir} \\ $$