Question

$log^2 _{2} x+log _{2} 4x=4$

Answer 1

$\log_{\big2}^{\big2}x + \log_{\big 2}(4x) = 4 \\ \\ \log_{\big2}^{\big2}x +\log_{\big 2}4 + \log_{\big 2}x = 4 \\ \\ \log_{\big2}^{\big2}x+2+\log_{\big2}x = 4 \\ \\ \log_{\big2}^{\big2}x+\log_{\big2}x +2-4 =0\\ \\ \log_{\big2}^{\big2}x+\log_{\big2}x-2 = 0\\ \\ Conditie de existenta: x \ \textgreater \ 0 \Rightarrow D=(0,+\infty) \\ \\ Notam \log_{\big2}x = t.$

$\left\| \begin{array}{c} t^2+t-2 = 0\quad \quad \quad \quad \quad \quad \quad \quad \\ \Delta = 1^2-4\cdot 1\cdot (-2) = 1+8=9 \end{array} \right| \Rightarrow t_{1,2} = \dfrac{-1\pm \sqrt9}{2} =\dfrac{-1\pm3}{2} \\ \\\ \bullet t = \dfrac{-1+3}{2} \Rightarrow t = \dfrac{2}{2}\Rightarrow t = 1 \Rightarrow \log_{\big2}x=1 \Rightarrow x = 2^1 \Rightarrow x = 2\in D$

$\bullet t = \dfrac{-1-3}{2}\Rightarrow t = \dfrac{-4}{2} \Rightarrow t = -2 \Rightarrow \log_{\big2}x = -2 \Rightarrow x = 2^{-2}\Rightarrow \\ \\ \Rightarrow x = \dfrac{1}{2^2}\Rightarrow x = \dfrac{1}{4}\in D$

$\Rightarrow \boxed{S = \Big\{\dfrac{1}{4},2\Big\}}$

Answer 2

   
[tex]\displaystyle\\ log^2 _{2} x+log _{2} 4x=4\\\\ log^2 _{2} x+log _{2} x+log _{2}4=4\\\\ log^2 _{2}x+log _{2}x+2=4\\\\ log^2 _{2}x+log _{2}x+2-4=0\\\\ log^2 _{2}x+log _{2}x-2=0\\\\ \texttt{Substitutie: }~~\boxed{log _{2}x = y}\\\\ y^2 + y -2=0\\\\ y_{12}= \frac{-b\pm\sqrt{b^2-4ac} }{2a}=\frac{-1\pm\sqrt{1+8}}{2} =\frac{-1\pm\sqrt{9}}{2}= \frac{-1\pm3}{2}\\\\ y_1 = \frac{-1+3}{2}=\frac{2}{2}=\boxed{1}\\\\ y_1 = \frac{-1-3}{2}=\frac{-4}{2}=\boxed{-2}\\\\ \texttt{Revenim la substitutie:}\\\\ [/tex]


[tex]log_2 x = y\\\\ \text{Solutia 1:}\\\\ log_2 x = 1\\\\ x_1 = 2^1 = \boxed{2}\\\\ \text{Solutia 2:}\\\\ log_2 x = -2\\\\ x_2 = 2^{-2} = \frac{1}{2^2} = \boxed{\frac{1}{4}}[/tex]