Question

Solve $2{\log }_{3}x-4{\log }_{x}27\le 5\left((x>1)\right)$
What is the max possible value of x?

Answer 1

Let ${\log }_{3}x=y\Rightarrow x=3y$
Therefor, the given inequality become $2{\log }_{3}x-12{\log }_{x}3\le 5$
or $2y-\frac{12}{y}\le 5$
or $2y^2-5y-12\le 0$ (as $x>1\Rightarrow y>0$)
or $(2y+3)(y-4)\le 0$
$\Rightarrow$ $yϵ[-\frac{3}{2},4]\Rightarrow -\frac{3}{2}\le {\log }_{3}x\le 4$
$\Rightarrow$ $3^{-3/2}\le x\le 81$
Ans: 81