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Question Number 153950 by liberty last updated on 12/Sep/21

Answered by EDWIN88 last updated on 13/Sep/21

  log _(∣x−(7/4)∣) (log _(1/2) x)≤ 0  ⇒log _(∣x−(7/4)∣) (log _(1/2) x) ≤ log _(∣x−(7/4)∣) (1)  ⇒(log _(1/2) x−1)(∣x−(7/4)∣−1)≤0  ⇒(2x−1)((1/2)−1)(x−(3/4))(x−((11)/4))≤0  ⇒(2x−1)(x−(3/4))(x−((11)/4))≥0  ⇒ (1/2)<x≤(3/4) ∪ x≥((11)/4)   domain 0<x<1 ;x≠(3/4);x≠((11)/4)  ⇒ x∈ ((1/2),(3/4))

$$\:\:\mathrm{log}\:_{\mid{x}−\frac{\mathrm{7}}{\mathrm{4}}\mid} \left(\mathrm{log}\:_{\frac{\mathrm{1}}{\mathrm{2}}} {x}\right)\leqslant\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{log}\:_{\mid{x}−\frac{\mathrm{7}}{\mathrm{4}}\mid} \left(\mathrm{log}\:_{\frac{\mathrm{1}}{\mathrm{2}}} {x}\right)\:\leqslant\:\mathrm{log}\:_{\mid{x}−\frac{\mathrm{7}}{\mathrm{4}}\mid} \left(\mathrm{1}\right) \\ $$$$\Rightarrow\left(\mathrm{log}\:_{\frac{\mathrm{1}}{\mathrm{2}}} {x}−\mathrm{1}\right)\left(\mid{x}−\frac{\mathrm{7}}{\mathrm{4}}\mid−\mathrm{1}\right)\leqslant\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2}{x}−\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)\left({x}−\frac{\mathrm{3}}{\mathrm{4}}\right)\left({x}−\frac{\mathrm{11}}{\mathrm{4}}\right)\leqslant\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2}{x}−\mathrm{1}\right)\left({x}−\frac{\mathrm{3}}{\mathrm{4}}\right)\left({x}−\frac{\mathrm{11}}{\mathrm{4}}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}<{x}\leqslant\frac{\mathrm{3}}{\mathrm{4}}\:\cup\:{x}\geqslant\frac{\mathrm{11}}{\mathrm{4}}\: \\ $$$${domain}\:\mathrm{0}<{x}<\mathrm{1}\:;{x}\neq\frac{\mathrm{3}}{\mathrm{4}};{x}\neq\frac{\mathrm{11}}{\mathrm{4}} \\ $$$$\Rightarrow\:{x}\in\:\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{4}}\right)\: \\ $$

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