Question

The complete set of values of $x$ for which the inequality ${\log }_x\left(\frac{4x+5}{6-5x}\right)<-1$ holds good, is

• A. $(1,\frac{6}{5})$
• B. $(0,1)$
• C. $(\frac{1}{2},1)$
• D. $(0,1)\cup (1,\frac{6}{5})$

Answer 1

The correct option is C $(\frac{1}{2},1)$

For ${\log }_x\left(\frac{4x+5}{6-5x}\right)<-1$ to be defined,
$\frac{4x+5}{6-5x}>0,x>0,x\ne 1\Rightarrow x\in (-\frac{5}{4},\frac{6}{5}),x>0,x\ne 1\therefore x\in (0,\frac{6}{5})-\left\{1\right\}$

Case $1:$ When $x\in (0,1)$
${\log }_x\frac{4x+5}{6-5x}<-1\Rightarrow \frac{4x+5}{6-5x}>x^{-1}\Rightarrow \frac{4x+5}{6-5x}-\frac{1}{x}>0\Rightarrow \frac{4x^2+10x-6}{x(6-5x)}>0\Rightarrow \frac{2(2x-1)(x+3)}{x(5x-6)}<0\Rightarrow x\in (-3,0)\cup (\frac{1}{2},\frac{6}{5})$
$\Rightarrow x\in (\frac{1}{2},1)\cdots \left(1\right)$ as $x\in (0,1)$

Case $2:$ When $x\in (1,\frac{6}{5})$
${\log }_x\frac{4x+5}{6-5x}<-1$
$\Rightarrow \frac{4x+5}{6-5x}<x^{-1}$
$\Rightarrow \frac{4x+5}{6-5x}-\frac{1}{x}<0$
$\Rightarrow \frac{4x^2+10x-6}{x(6-5x)}<0$
$\Rightarrow \frac{2(2x-1)(x+3)}{x(5x-6)}>0$
$\Rightarrow x\in (-\infty ,-3)\cup (0,\frac{1}{2})\cup (\frac{6}{5},\infty )$
$\Rightarrow x\in \varphi \cdots \left(2\right)$ as $x\in (1,\frac{6}{5})$

From $\left(1\right)$ and $\left(2\right)$,
$x\in (\frac{1}{2},1)$