Solve  ${\log }_2\left(4^{x+1}+4\right)·{\log }_2\left(4^x+1\right)={\log }_{1/\sqrt{2}}\left(\frac{1}{\sqrt{8}}\right)$

${\log }_2\left(4^{x+1}+4\right)·{\log }_2\left(4^x+1\right)={\log }_{1/\sqrt{2}}\left(\frac{1}{\sqrt{8}}\right)$

Eq. (i) defined, for $4^x+1>0$ which is true for all $x\in R.$

$$\begin{gathered} {\log }_2\left[4\left(4^x+1\right)\right]·{\log }_2\left(4^x+1\right)={\log }_{\sqrt{2}}\sqrt{8}=3 \\ \left[2+{\log }_2\left(4^x+1\right)\right]{\log }_2\left(4^x+1\right)=3 \end{gathered}$$

Let ${\log }_2\left(4^x+1\right)=y$

$$\begin{gathered} \therefore (y+2)y=3 \\ {y}^2+2y-3=0 \\ (y+3)(y-1)=0 \\ y=1 or y=-3 \\ \therefore {\log }_2\left(4^x+1\right)=1 or {\log }_2\left(4^x+1\right)=-3 \\ {4}^x+1=2 or 4^x+1=\frac{1}{8} \\ {4}^x=1 or 4^x=\frac{1}{8}-1 \end{gathered}$$

$x=0 or 4^x=-\frac{7}{8}$ which is not possible. $x=0$