Question Number 161215 by mathlove last updated on 14/Dec/21
$$\mathrm{log}\:\underset{\mathrm{2}} {{x}}+{log}\underset{{x}} {\mathrm{2}}=\frac{{e}^{\mathrm{2}} +\mathrm{1}}{{e}}\:\:\:\:\:\:\:\:\:\:{x}=? \\ $$
Commented by cortano last updated on 14/Dec/21
$$\:{let}\:\begin{cases}{\mathrm{log}\:_{\mathrm{2}} \left({x}\right)={u}}\\{\mathrm{log}\:_{{x}} \left(\mathrm{2}\right)=\frac{\mathrm{1}}{{u}}}\end{cases} \\ $$$$\:\Rightarrow{u}+\frac{\mathrm{1}}{{u}}\:=\:{e}+{e}^{−\mathrm{1}} \\ $$$$\Rightarrow{u}^{\mathrm{2}} −\left({e}+{e}^{−\mathrm{1}} \right){u}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{u}\:=\:\frac{{e}+{e}^{−\mathrm{1}} \pm\:\sqrt{{e}^{\mathrm{2}} +{e}^{−\mathrm{2}} −\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow{u}\:=\:\frac{{e}+{e}^{−\mathrm{1}} \:\pm\:\sqrt{\left({e}−{e}^{−\mathrm{1}} \right)^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\Rightarrow\begin{cases}{{u}_{\mathrm{1}} =\:{e}\Rightarrow\mathrm{log}\:_{\mathrm{2}} \left({x}\right)={e}\Rightarrow{x}=\mathrm{2}^{{e}} }\\{{u}_{\mathrm{2}} ={e}^{−\mathrm{1}} \Rightarrow\mathrm{log}\:_{\mathrm{2}} \left({x}\right)={e}^{−\mathrm{1}} \Rightarrow{x}=\mathrm{2}^{\frac{\mathrm{1}}{{e}}} }\end{cases} \\ $$
Answered by Rasheed.Sindhi last updated on 14/Dec/21
$$\mathrm{log}_{\mathrm{2}} {x}+\mathrm{log}_{{x}} \mathrm{2}=\frac{{e}^{\mathrm{2}} +\mathrm{1}}{{e}}\:\: \\ $$$$\mathrm{log}_{\mathrm{2}} {x}+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{2}} {x}}={e}+\frac{\mathrm{1}}{{e}} \\ $$$$\mathrm{log}_{\mathrm{2}} {x}={e},\frac{\mathrm{1}}{{e}}\Rightarrow{x}=\mathrm{2}^{{e}} ,\mathrm{2}^{\mathrm{1}/{e}} \\ $$$$ \\ $$
Commented by cortano last updated on 14/Dec/21
$${how}\:{about}\:\begin{cases}{\mathrm{log}\:_{\mathrm{2}} \left({x}\right)=\frac{\mathrm{1}}{{e}}}\\{\mathrm{log}\:_{{x}} \left(\mathrm{2}\right)={e}}\end{cases}\:? \\ $$
Commented by Rasheed.Sindhi last updated on 14/Dec/21
$${You}'{re}\:{right}!\:{I}\:{extended}\:{my}\:{answer}. \\ $$