Question

If ${\log }_{x}2+{\log }_{x^2}2>1$, then set of all values of $x$ is

• A. $(1,2\sqrt{2})$
• B. $(2\sqrt{2},\infty )$
• C. $(2,\infty )-\left\{2\sqrt{2}\right\}$
• D. $(2\sqrt{2},5)$

Answer 1

The correct option is A $(1,2\sqrt{2})$

${\log }_{x}2+{\log }_{x^2}2$ is valid when $x>0,x\ne \pm 1$
Now, ${\log }_{x}2+{\log }_{x^2}2>1$
$\Rightarrow {\log }_{x}2+\frac{1}{2}{\log }_{x}2>1[\because {\log }_{a^m}b=\frac{1}{m}{\log }_{a}b]$
$\Rightarrow {\log }_{x}2+{\log }_x\sqrt{2}>1,[\because m\log a=\log a^m]$
$\Rightarrow {\log }_{x}2\sqrt{2}>1,[\because \log a+\log b=\log (ab\left)\right]$
Above valid for for $x>1$ and $x<2\sqrt{2}$
Therefore, $x\in (1,2\sqrt{2})$
Ans: A