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Question

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


Solution

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

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