[tex]\log_{\big2}\text{x}+\log_{\big{\text{x}}}2 = 2 \\ \\ \log_{\big2}\text{x}+\dfrac{1}{\log_{\big2}\text{x}} = 2 \\ \\ $Conditie de existenta: \quad $ x\ \textgreater \ 0 \Rightarrow D = (0,+\infty) \\ \\ $Notam \log_{\big2}\text{x} = t. \\ \\ t+\dfrac{1}{t}=2\Big|\cdot t \\ \\ t^2+1 = 2t \\ \\ t^2-2t+1 = 0 \\ \\ (t-1)^2 = 0 \\ \\ t-1=0 \\ \\ t=1 \\ \\ \log_{\big2}\text{x} = 1 \\ \\ \text{x} = 2^1 \\ \\ \text{x} = 2\in D \\ \\ \boxed{S = \Big\{2\Big\}}[/tex]
