Question

Solve the following equation:
${\log }_2(4^{x+1}+4)\cdot {\log }_2(4^x+1)={\log }_{1/\sqrt{2}}\sqrt{\frac{1}{8}}$

Answer 1

${\log }_2(4^{x+1}+4){\log }_2(4^x+1)={\log }_{\sqrt{\frac{1}{2}}}\sqrt{\frac{1}{8}}{\Rightarrow \log }_2(4^{x+1}+4){\log }_2(4^x+1)=3$

Let ${log}_2(4^x+1)=a.$

$(a+2)a=3\Rightarrow a^2+2a-3=0\Rightarrow a=-3,1$

Therefore, ${log}_2{4}^x+1=-3\Rightarrow 4^x+1=\frac{1}{8}\Rightarrow 4^x=\frac{-7}{3}$ and ${log}_2{4}^x+1=1\Rightarrow 4^x+1=2\Rightarrow 4^x=1\Rightarrow x=0.$

$4^x$ cannot be negative solution.

Therefore $x=0$ is only solution.