Question

If ${\log }_2(4^{x+1}+4){\log }_2(4^x+1)={\log }_2\left(8\right)$, then $x$ is equal to

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is B $0$

${\log }_2(4^{x+1}+4){\log }_2(4^x+1)={\log }_2\left(8\right)$

${\log }_2({4(4}^x+1)){\log }_2(4^x+1)={\log }_2\left(8\right)$

$[{\log }_2\left(4\right)+{\log }_2{(4}^x+1)]{\log }_2(4^x+1)={\log }_2\left(8\right)\dots (\log ab=\log a+\log b)$

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

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

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

$\Rightarrow y^2+2y-3=0$

$\Rightarrow (y-1)(y+3)=0$

$\Rightarrow y=1,-3$

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

$\Rightarrow 4^x=1,\frac{-7}{8}$

Since,$4^x>0$

Therefore, $4^x=1$

$\Rightarrow x=0$

Ans: B