Solve: 2 log₂(log₂x) + log_1/2(log₂ 2sq.root2x)=1

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$$\begin{gathered} 2{\log }_2\left({\log }_2\mathrm{x}\right)+{\log }_{\frac{1}{2}}\left\{{\log }_2\left(2\sqrt{2\mathrm{x}}\right)\right\}=1 \\ \Rightarrow 2{\log }_2\left({\log }_2\mathrm{x}\right)+{\log }_{2^{-1}}\left\{{\log }_2\left(2\sqrt{2\mathrm{x}}\right)\right\}=1 \\ \mathrm{Using} \mathrm{identities}. \\ {\log }_{\mathrm{a}}\left({\mathrm{m}}^{\mathrm{n}}\right)=\mathrm{n}\times {\log }_{\mathrm{a}}\left(\mathrm{m}\right) \\ {\log }_{{\mathrm{a}}^{\mathrm{n}}}\left(\mathrm{m}\right)=\frac{1}{\mathrm{n}}{\log }_{\mathrm{a}}\left(\mathrm{m}\right), \mathrm{we} \mathrm{get}, \\ {\log }_2{\left({\log }_2\mathrm{x}\right)}^2-{\log }_2\left\{{\log }_2\left(2\sqrt{2\mathrm{x}}\right)\right\}=1 \\ \mathrm{Using} \mathrm{identit}y \\ {\log }_{\mathrm{a}}\left(\frac{\mathrm{m}}{\mathrm{n}}\right)={\log }_{\mathrm{a}}\left(\mathrm{m}\right)-{\log }_{\mathrm{a}}\left(\mathrm{n}\right), \mathrm{we} \mathrm{get} \\ {\log }_2\left\{\frac{{\left({\log }_2\mathrm{x}\right)}^2}{{\log }_2\left(\sqrt{2\mathrm{x}}\right)}\right\}=1 \\ \Rightarrow \frac{{\left({\log }_2\mathrm{x}\right)}^2}{{\log }_2\left(\sqrt{2\mathrm{x}}\right)}=2^1 \\ \Rightarrow {\left({\log }_2\mathrm{x}\right)}^2=2{\log }_2\left(2\sqrt{2\mathrm{x}}\right) \\ \Rightarrow {\left({\log }_2\mathrm{x}\right)}^2=2{\log }_2\left(2\sqrt{2}\sqrt{\mathrm{x}}\right) \\ \Rightarrow {\left({\log }_2\mathrm{x}\right)}^2=2{\log }_2\left(\sqrt{8}\sqrt{\mathrm{x}}\right) \\ \Rightarrow {\left({\log }_2\mathrm{x}\right)}^2=2\left\{{\log }_2\left(\sqrt{\mathrm{x}}\right)+{\log }_2\left(\sqrt{8}\right)\right\} \\ \Rightarrow {\left({\log }_2\mathrm{x}\right)}^2=2\left\{{\log }_2{\left(\mathrm{x}\right)}^{\frac{1}{2}}+{\log }_2\left(2^{\frac{3}{2}}\right)\right\} \\ \Rightarrow {\left({\log }_2\mathrm{x}\right)}^2=2\left\{\frac{1}{2}{\log }_2\left(\mathrm{x}\right)+\frac{3}{2}{\log }_{2}2\right\} \\ \mathrm{We} \mathrm{know} \mathrm{that} {\log }_{\mathrm{a}}\left(\mathrm{a}\right)=1 \\ \Rightarrow {\left({\log }_2\mathrm{x}\right)}^2={\log }_2\left(\mathrm{x}\right)+3 \\ \Rightarrow {\left({\log }_2\mathrm{x}\right)}^2-{\log }_2\left(\mathrm{x}\right)-3=0 \\ \mathrm{which} \mathrm{is} \mathrm{quadratic} \mathrm{in} {\log }_2\mathrm{x} \\ \Rightarrow {\log }_2\mathrm{x}=\frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2-4\times 1\times \left(-3\right)}}{2} \\ \Rightarrow {\log }_2\mathrm{x}=\frac{1\pm \sqrt{13}}{2} \\ \Rightarrow {\log }_2\mathrm{x}=\frac{1+\sqrt{13}}{2} \mathrm{or} \Rightarrow {\log }_2\mathrm{x}=\frac{1-\sqrt{13}}{2} \\ \mathbf{\Rightarrow }\mathbf{x}\mathbf{=}{\mathbf{2}}^{\left(\frac{\mathbf{1}\mathbf{+}\sqrt{\mathbf{13}}}{\mathbf{2}}\right)}\mathbf{ }\mathbf{or}\mathbf{ }\mathbf{x}\mathbf{=}{\mathbf{2}}^{\left(\frac{\mathbf{1}\mathbf{-}\sqrt{\mathbf{13}}}{\mathbf{2}}\right)} \end{gathered}$$

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