Question

Solve

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

Answer 1

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

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

$\Rightarrow [{\log }_{2}4+{\log }_2(4^x+1)]{\log }_2(4^x+1)=3{\log }_{2}2$

$\Rightarrow [2+{\log }_2(4^x+1)]{\log }_2(4^x+1)=3$

$\Rightarrow {\left[{\log }_2(4^x+1)\right]}^2+2{\log }_2(4^x+1)-3=0$

$\Rightarrow \left[{\log }_2(4^x+1)+3\right][{\log }_2(4^x+1)-1]=0$

$\Rightarrow {\log }_2(4^x+1)+3=0$

$\Rightarrow {\log }_2(4^x+1)=-3$

$\Rightarrow (4^x+1)=\frac{1}{8}$

$\Rightarrow xϵImaginary$

And,

$\Rightarrow {\log }_2(4^x+1)=1$

$\Rightarrow (4^x+1)=2$

$\Rightarrow 4^x=1$

$\Rightarrow x=0$