Question

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

Answer 1

$log_x \frac{4x + 5}{6 -5x} < -1$
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 $I: 0 < x < 1 \,...(ii)$
$log_x (\frac{4x + 5}{6- 5x}) < -1$
$\Rightarrow \frac{ 4x + 5}{6 - 5x} > \frac{1}{x}$
$\Rightarrow \frac{4x + 5}{ 6 - 5x} -\frac{1}{x} > 0$
$\Rightarrow \frac{4x^2 + 5x + 5x -6}{x(6 - 5x)} > 0$
$\Rightarrow \frac{4x^2 + 10x - 6}{x(5x - 6)} < 0$
$\Rightarrow \frac{2(x + 3)(2x -1)}{x(5x - 6)} < 0$
$\Rightarrow x \in (-3, 0) \cup (\frac{1}{2}, \frac{6}{5}) \,...(iii)$
From (i), (ii) and (iii), we get
$\therefore x \in (\frac{1}{2} , 1)$
Case II :$ x > 1\,...(iv)$
$log_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
$x \in \phi$
Thus, $x \in ( \frac{1}{2}, 1)$