[tex]\log_{2x}4x+\log_{4x}16x =4 \\ \\ \log_{2x}(2x)^2-\log_{2x}x+\log_{4x}(4x)^2 - \log_{4x}x = 4 \\ \\ \log_{2x}x+\log_{4x}x = 0 \\ \\ \dfrac{\ln x}{\ln 2x}+\dfrac{\ln x}{\ln 4x} = 0\\ \\ \ln x\cdot \Big(\dfrac{1}{\ln 2x}+\dfrac{1}{\ln 4x}\Big) = 0 \\ \\ (1)\quad \ln x = 0 \Rightarrow \boxed{x = 1} \\ \\ (2)\quad \dfrac{1}{\ln 2x}+\dfrac{1}{\ln 4x} = 0[/tex]
[tex]\dfrac{1}{\ln 2x}+\dfrac{1}{\ln 2+\ln 2x} = 0\\ \\ \ln 2x = t,\quad t\neq 0\\ \\ \dfrac{1}{t}+\dfrac{1}{\ln 2+t} = 0 \\ \\ t+\ln 2+t = 0 \Rightarrow 2t = -\ln 2 \Rightarrow t = -\dfrac{\ln 2}{2} \Rightarrow \\ \\ \Rightarrow \ln 2x = - \dfrac{\ln 2}{2}\Rightarrow 2x = e^{-\dfrac{\ln 2}{2}} \Rightarrow x = \dfrac{(e^{\ln 2})^{-\dfrac{1}{2}}}{2} \Rightarrow \\ \\ \Rightarrow x = \dfrac{2^{-\dfrac{1}{2}}}{2} \Rightarrow \boxed{x = \dfrac{1}{2\sqrt 2}} \\ \\ \\\Rightarrow \text{Raspuns corect C)}[/tex]
