If 2 log 3 x -4 log x 27 - 5 x -1- then the number of integral values of x is

2log_3x 4log_x27≤5 ⇒ 2log_3x 4· 3 log_x 3 ≤ 5 Let log_3x=y ⇒ 2y 12y≤5 ⇒ 2y^2 5y 12≤0 [ As x gt;1⇒ y gt;0 ] nbsp; ⇒ (2y+3)(y 4)≤0 ⇒ y∈[ 32,4] But y 0 ...

Question

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

Solution

$2 {log}_{3} x - 4 {log}_{x} 27 \leq 5$
$\Rightarrow 2 {log}_{3} x - 4 \cdot 3 {log}_{x} 3 \leq 5$
Let ${log}_{3} x = y$
$\Rightarrow 2 y - \frac{12}{y} \leq 5 \Rightarrow 2 y^{2} - 5 y - 12 \leq 0 [As x > 1 \Rightarrow y > 0]$
$\Rightarrow (2 y + 3) (y - 4) \leq 0 \Rightarrow y \in [- \frac{3}{2}, 4]$
But $y > 0$
$∴ 0 < y \leq 4$
$\Rightarrow 0 < {log}_{3} x \leq 4 \Rightarrow 1 < x \leq 81$

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If 2log3x−4logx27≤5 (x>1), then the number of integral values of x is

2log3x−4logx27≤5 ⇒2log3x−4⋅3logx3≤5 Let log3x=y ⇒2y−12y≤5⇒2y2−5y−12≤0 [ As x>1⇒y>0 ] ⇒(2y+3)(y−4)≤0⇒y∈[−32,4] But y>0 ∴0<y≤4 ⇒0<log3x≤4⇒1<x≤81

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