Question

If $2^{|\ln x-2|}=\frac{4}{2^{|\ln x|}}$, then complete set of values of $x$ is

• A. $[1,e]$
• B. $\left[1,e^2\right]$
• C. $\left[e^2,\infty \right)$
• D. $(0,1]\cup \left[e^2,\infty \right)$

Answer 1

Answer: (B)

$2^{|\ln x-2|}=2^{2-|\ln x|}$
$\Rightarrow |\ln x-2|+|\ln x|=2$
Using property
$|a|+|b|=|a-b|\Rightarrow ab\le 0$
$\therefore \ln x(\ln x-2)\le 0$
$\Rightarrow 0\le \ln x\le 2\Rightarrow 1\le x\le e^2$