Question

Number of integers satisfying $lo{g}_x\frac{4x+5}{6-5x}<-1$ is ____.

Answer 1

Answer: 0

$lo{g}_x\frac{4x+5}{6-5x}<-1$
We must have
$\frac{4x+5}{6-5x}<0$
$\Rightarrow \frac{4x+5}{5x-6}<0$
$\Rightarrow x\in (\frac{-5}{4},\frac{6}{5})$
Also $x>0$ and $x\ne 1$
$\therefore x\in (0,\frac{6}{5})-\{1\}...(i)$
Case $I:0<x<1...(ii)$
$lo{g}_x(\frac{4x+5}{6-5x})<-1$
$\Rightarrow \frac{4x+5}{6-5x}>\frac{1}{x}$
$\Rightarrow \frac{4x+5}{6-5x}-\frac{1}{x}>0$
$\Rightarrow \frac{4x^2+5x+5x-6}{x(6-5x)}>0$
$\Rightarrow \frac{4x^2+10x-6}{x(5x-6)}<0$
$\Rightarrow \frac{2(x+3)(2x-1)}{x(5x-6)}<0$
$\Rightarrow x\in (-3,0)\cup (\frac{1}{2},\frac{6}{5})...(iii)$
From (i), (ii) and (iii), we get
$\therefore x\in (\frac{1}{2},1)$
Case II :$x>1...(iv)$
$lo{g}_x(\frac{4x+5}{6-5x})<-1$
$\Rightarrow \frac{4x+5}{6-5x}<\frac{1}{x}$
$\Rightarrow x\in (-\infty ,-3)\cup (0,\frac{1}{2})\cup (\frac{6}{5},\infty )...(v)$
From (i), (iv) and (v), we get
$x\in \varphi$
Thus, $x\in (\frac{1}{2},1)$