Question

If $2{\log }_{3}x-4{\log }_{x}27\le 5(x>1)$, then the number of integral values of $x$ is

Answer 1

$2{\log }_{3}x-4{\log }_{x}27\le 5$
$\Rightarrow 2{\log }_{3}x-4\cdot 3{\log }_{x}3\le 5$
Let ${\log }_{3}x=y$
$\Rightarrow 2y-\frac{12}{y}\le 5\Rightarrow 2y^2-5y-12\le 0[\text{As}x>1\Rightarrow y>0]$
$\Rightarrow (2y+3)(y-4)\le 0\Rightarrow y\in [-\frac{3}{2},4]$
But $y>0$
$\therefore 0<y\le 4$
$\Rightarrow 0<{\log }_{3}x\le 4\Rightarrow 1<x\le 81$
Thus, talking about integral values of $x,$ we get:
$x\in \{2,3,4,5\cdots ,80\}$
Or $80$ integral values of $x.$