User:Lmov/Sandbox


 * $$\log_{2x}{\frac{2}{x}}\log_2^2 x+\log_2^4 x=1$$


 * $$x\in\left(0;\frac{1}{2}\right)\cup\left(\frac{1}{2};+\infty\right)$$


 * $$\frac{\ln{\frac{2}{x}}\ln^2 x}{\ln{2x}\ln^2 2}+

\frac{\ln^4 x}{\ln^4 2}=1$$


 * $$\ln{\frac{2}{x}}\ln^2 x\ln^2 2+\ln^4 x\ln{2x}=\ln{2x}\ln^4 2$$


 * $$\left(\ln 2-\ln x\right)\ln^2 x\ln^2 2+\ln^4 x\left(\ln 2+\ln x\right)-\left(\ln 2+\ln x\right)\ln^4 2=0$$


 * $$\ln^5 x+\ln^4 x\ln 2-\ln^3 x \ln^2 2+\ln^2 x\ln^3 2-\ln x\ln^4 2-\ln^5 2=0$$


 * $$\left(\frac{\ln x}{\ln 2}\right)^5+\left(\frac{\ln x}{\ln 2}\right)^4-\left(\frac{\ln x}{\ln 2}\right)^3+\left(\frac{\ln x}{\ln 2}\right)^2-\left(\frac{\ln x}{\ln 2}\right)-1=0$$


 * $$\log_2^5 x+\log_2^4 x-\log_2^3 x+\log_2^2 x-\log_2 x-1=0$$

Using Ruffini's rule find that 1 is a root:


 * $$\begin{matrix} & 1 & 1 & -1 & 1 & -1 & -1 \\ 1 & \left[1\right. & 2 & 1 & 2 & \left.1\right] & 0 \end{matrix}$$


 * $$\left(\log_2 x - 1\right)\left(\log_2^4 x + 2\log_2^3 x + \log_2^2 x + 2\log_2 x + 1\right)=0$$


 * $$\log_2 x=1$$


 * $$x_1=2\mbox{ (1)}$$


 * $$\log_2^4 x + 2\log_2^3 x + \log_2^2 x + 2\log_2 x + 1=0$$


 * $$\log_2^2 x + 2\log_2 x + 1 + 2\frac{1}{\log_2 x} + \frac{1}{\log_2^2 x}=0$$


 * $$\left(\log_2^2 x + \frac{1}{\log_2^2 x}\right) + 2\left(\log_2 x + \frac{1}{\log_2 x}\right) + 1=0$$


 * $$\mbox{Let }y=\log_2 x+\frac{1}{\log_2 x}\Rightarrow y^2-2=\log_2^2 x+\frac{1}{\log_2^2 x}$$


 * $$y^2-2+2y+1=0$$


 * $$y^2+2y-1=0$$


 * $$y_{1,2}=-1\pm\sqrt{2}$$


 * $$\log_2 x+\frac{1}{\log_2 x}=-1-\sqrt{2}$$


 * $$\log_2^2 x+\left(\sqrt{2}+1\right)\log_2 x+1=0$$


 * $$\log_2 x=\frac{-\left(\sqrt{2}+1\right)\pm\sqrt{2\sqrt{2}-1}}{2}$$


 * $$x_{2,3}=2^{\frac{-\left(\sqrt{2}+1\right)\pm\sqrt{2\sqrt{2}-1}}{2}} \mbox{ (2)}$$


 * $$\log_2 x+\frac{1}{\log_2 x}=-1+\sqrt{2}$$


 * $$\log_2^2 x+\left(1-\sqrt{2}\right)\log_2 x+1=0$$


 * $$D=\left(1-\sqrt{2}\right)^2-4=-1-2\sqrt{2}<0\Rightarrow\mbox{roots are complex}$$

Thus, from (1) and (2):


 * $$x_1=2, x_{2,3}=2^{\frac{-\left(\sqrt{2}+1\right)\pm\sqrt{2\sqrt{2}-1}}{2}}$$