Question

log patrat in baza 2 din x +log in baza 2 din 4x=4

Answer 1

[tex]\displaystyle \log^2_{2}{x}+\log_{2}{4x}=4,x\ \textgreater \ 0\\ \\ \log^2_{2}{x}+\log_{2}4+\log_2x=4\\ \\ \log^2_{2}{x}+2+\log_2x=4\log^2_{2}{x}\\ \\ \log^2_{2}{x}+\log_2x=2\\ \\ t=\log_{2}{x}\\ \\ t^2+t=2\\ \\ t^2+t-2=0\\ \\ \left \{ {{t_1+t_2=-1} \atop {t_1\times t_2=-2}} \right. \Leftrightarrow \left \{ {{t_1=-2} \atop {t_2=1}} \right. \\ \\ \left \{ {{\log_2x=-2} \atop {{\log_2x=1}} \right. \Leftrightarrow \left \| {{x=\frac 14(\ \textgreater \ 0)} \atop {x=2(\ \textgreater \ 0)}} \right. \\ \\ S=\{\frac 14;2\}[/tex]

Answer 2

$\log^\big2_{\big2}x+\log_{\big2}(4x)=4 \\ \\ \log^\big2_{\big2}x+\log_{\big2}4+\log_{\big2}x = 4 \\ \\ \log^\big2_{\big2}x+2+\log_{\big2}x = 4\\ \\ \log^\big2_{\big2}x+\log_{\big2}x +2-4=0 \\ \\ (\log_{\big2}x)^2+\log_{\big2}x -2=0\\ \\ Notam \log_{\big2}x = t.\\ \\ t^2+t-2 = 0\\ \Delta =1^2-4\cdot 1\cdot (-2) = 1+8 = 9 \Rightarrow t_{1,2} = \dfrac{-1\pm\sqrt9}{2} \Rightarrow t_{1,2} = \dfrac{-1\pm3}{2}$

$\bullet t_1 = \dfrac{-1+3}{2} \Rightarrow t_1 = \dfrac{2}{2} \Rightarrow t_1 = 1 \Rightarrow \log_{\big2}x_1 = 1 \Rightarrow x_1 = 2^1 \Rightarrow x_1 = 2 \\ \\ \bullet t_2 = \dfrac{-1-3}{2} \Rightarrow t_2 = \dfrac{-4}{2} \Rightarrow t_2 = -2 \Rightarrow \log_{\big2}x_2 = -2 \Rightarrow x_2 = 2^{-2}\Rightarrow \\ \\ \Rightarrow x_2 = \dfrac{1}{2^2} \Rightarrow x_2 = \dfrac{1}{4} \\ \\ \Rightarrow \boxed{S = \Big\{\dfrac{1}{4},2\Big\}}$