Introduction to linear inequalities

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Question

Number of integers less than or equal to 10 satisfying the $inequality 2 \log_{1 / 2} (x - 1) \leq \frac{1}{3} - \frac{1}{\log_{x^{2} - x} 8} is$

Moderate

Solution

$\begin{array}{l} We must have x > 1 \\ 2 \log_{1 / 2} (x - 1) \leq \frac{1}{3} - \frac{1}{\log_{x^{2} - x} 8} \end{array}$
$\begin{array}{l} \Rightarrow \frac{1}{3} - \frac{\log_{2} (x^{2} - x)}{3} + 2 \log_{2} (x - 1) \geq 0 \\ \Rightarrow \log_{2} 2 - \log_{2} (x^{2} - x) + 6 \log_{2} (x - 1) \geq 0 \\ \Rightarrow \log_{2} \frac{2 (x - 1)^{6}}{x (x - 1)} \geq 0 \\ \Rightarrow \frac{2 (x - 1)^{5}}{x} \geq 1 \end{array}$
$\begin{array}{l} Putting x - 1 = y \Rightarrow y > 0 \\ \Rightarrow \frac{2 y^{5}}{y + 1} - 1 \geq 0 \\ \Rightarrow \frac{2 y^{5} - y - 1}{y + 1} \geq 0 \\ \Rightarrow \frac{2 y^{5} - 2 y + y - 1}{y + 1} \geq 0 \\ \Rightarrow \frac{2 y (y^{4} - 1) + y - 1}{y + 1} \geq 0 \\ \Rightarrow \frac{(y - 1) [2 y (y + 1) (y^{2} + 1) + 1]}{y + 1} \geq 0 \\ \Rightarrow \frac{y - 1}{y + 1} \geq 0 \end{array}$
$\begin{aligned} \Rightarrow y \geq 1 \\ \Rightarrow x \geq 2 \end{aligned}$

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Introduction to linear inequalities

Remember concepts with our Masterclasses.

Number of integers less than or equal to 10 satisfying the inequality 2log1/2⁡(x−1)≤13−1logx2−x⁡8 is

Ready to Test Your Skills?

free study material

Number of integers less than or equal to 10 satisfying the $inequality 2 \log_{1 / 2} (x - 1) \leq \frac{1}{3} - \frac{1}{\log_{x^{2} - x} 8} is$