0 \\ \\ t^{2}-5t+6 = 0 \\ \\ D = 25 - 24 = 1 \\ \\ t_{1} = \dfrac{5+1}{2} = 3 \ ; \ t_{2} = 2 \\ \\ $\left[\begin{gathered} \log_{2}x = 3 \\ \log_{2} x = 2 \\ \end{gathered} \right.$ \ \ ; \ $\left[ \begin{gathered} x = 8 \\ x = 4 \\ \end{gathered} \right.$" alt="\log_{2}^{2}x - 5\log _{2}x + 6 = 0 \\ \\ \log _{2}x = t, x > 0 \\ \\ t^{2}-5t+6 = 0 \\ \\ D = 25 - 24 = 1 \\ \\ t_{1} = \dfrac{5+1}{2} = 3 \ ; \ t_{2} = 2 \\ \\ $\left[\begin{gathered} \log_{2}x = 3 \\ \log_{2} x = 2 \\ \end{gathered} \right.$ \ \ ; \ $\left[ \begin{gathered} x = 8 \\ x = 4 \\ \end{gathered} \right.$" align="absmiddle" class="latex-formula">
Ответ: x₁ = 8 (больший), x₂ = 4 (меньший)