Question

If ${\log }_{1/\sqrt{2}}\left(\frac{1}{\sqrt{8}}\right)={\log }_2(4^x+1)\times {\log }_2(4^{x+1}+4)$, then $x$ is equal to

• A. $0$
• B. $1$
• C. $2$
• D. $-1$

Answer 1

Answer: (A)

$0$

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

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

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

$\Rightarrow (2+t)t=3$, where $t={\log }_2(4^x+1)$

$\Rightarrow t=-3,1$

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

$4^x=-\frac{7}{8}$ (not possible)

If ${\log }_2(4^x+1)=1$, then

$4^x+1=2$

$\Rightarrow 4^x=1$

$\Rightarrow x=0$