Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $-1$

Answer 1

Answer: (A)

$0$

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

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

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

Since product of two no is $3$, so one number will be $1$ and other number will be $3$.

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

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

$\Rightarrow 4^x+1=2\Rightarrow 4^x=1\Rightarrow x=0$

Now for $x=0$; second number should be $3$.

So,

${\log }_2(4^{x+1}+4)={\log }_2(4^{0+1}+4)={\log }_{2}8=3$

Hence, $x=0$ is satisfy the above equation.