Question

Solve: ${\log }_{x+\frac{1}{x}}\left({\log }_2\frac{x-1}{x+2}\right)>0$

• A. $x\in (-5,-2)$
• B. $x>1$
• C. $x<1$
• D. no possible values

Answer 1

The correct option is D no possible values

${\log }_{x+\frac{1}{x}}\left({\log }_2\frac{x-1}{x+2}\right)>0$
$\therefore$ $\frac{x-1}{x+2}>0$
$\Rightarrow$ $x<-2$ or $x>1$ (i)
Also $x+\frac{1}{x}>0$. Therefore,
$\frac{x^2+1}{x}>0$
$\Rightarrow$ $x>0$ (ii)
From Eqs. (i) and (ii), $x>1$.
Now ${\log }_{x+\frac{1}{x}}\left({\log }_2\frac{x-1}{x+2}\right)>0$
$\Rightarrow$ ${\log }_2\frac{x-1}{x+2}>1$
or $\frac{x-1}{x+2}>2$
or $\frac{x+5}{x+2}<0$
Hence, $x\in (-5,-2)$, which is not possible as $x>1$