(x^2 /(log_((5−x)) (x))) ≤ (5x−4) log


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Question Number 83786 by jagoll last updated on 06/Mar/20

$\frac{x^{2}}{\log_{(5 - x)} (x)} ⩽ (5 x - 4) \log_{x} (5 - x)$
	Answered by john santu last updated on 06/Mar/20

	$x^{2} \log_{x} (5 - x) ⩽ (5 x - 4) \log_{x} (5 - x)$ $(x - 1) ({(5 - x)}^{x^{2}} - {(5 - x)}^{5 x - 4}) ⩽ 0$ $(i) x > 0 \land x \neq 1 \land x \neq 4 \land x < 5$ $\Rightarrow for x < 1 \Rightarrow x^{2} - 5 x + 4 ⩾ 0$ $(x - 1) (x - 4) ⩾ 0, \Rightarrow x < 1$ $\Rightarrow for x > 1 \Rightarrow x^{2} - 5 x + 4 ⩽ 0$ $1 < x < 4$ $the solution is 0 < x < 1 \lor 1 < x < 4$
	Commented by jagoll last updated on 06/Mar/20

	$thank you sir, but the$ $answer is 0 < x < 1 \lor 1 < x < 4$ $\lor 4 < x < 5$
	Answered by MJS last updated on 07/Mar/20
	$x^2 ((ln (5−x))/(ln x))≤(5x−4)((ln (5−x))/(ln x)) 0<x<5 ∧ x≠1 ∧ x≠4 case 1 ((ln (5−x))/(ln x))>0 ∧ x^2 ≤5x−4 case 1.1 ln (5−x) <0 ∧ ln x <0 no solution case 1.2 ln (5−x) >0 ∧ ln x >0 1<x<4 x^2 −5x+4≤0 ⇒ 1≤x≤5 ⇒ 1<x<4 case 2 ((ln (5−x))/(ln x))<0 ∧ x^2 ≥5x−4 case 2.1 ln (5−x) <0 ∧ ln x >0 4<x<5 case 2.2 ln (5−x) >0 ∧ ln x <0 0<x<1 x^2 −5x+4≥0 ⇒ x≤1 ∨ x≥4 ⇒ 0<x<1 ∨ 4<x<5 answer x∈]0; 5[\{1; 4} but regarding the limits we could define for 0≤x≤5$
	$x^{2} \frac{\ln (5 - x)}{\ln x} ⩽ (5 x - 4) \frac{\ln (5 - x)}{\ln x}$ $0 < x < 5 \land x \neq 1 \land x \neq 4$ $case 1 \frac{\ln (5 - x)}{\ln x} > 0 \land x^{2} ⩽ 5 x - 4$ $case 1.1 \ln (5 - x) < 0 \land \ln x < 0$ $no solution$ $case 1.2 \ln (5 - x) > 0 \land \ln x > 0$ $1 < x < 4$ $x^{2} - 5 x + 4 ⩽ 0 \Rightarrow 1 ⩽ x ⩽ 5$ $\Rightarrow 1 < x < 4$ $case 2 \frac{\ln (5 - x)}{\ln x} < 0 \land x^{2} ⩾ 5 x - 4$ $case 2.1 \ln (5 - x) < 0 \land \ln x > 0$ $4 < x < 5$ $case 2.2 \ln (5 - x) > 0 \land \ln x < 0$ $0 < x < 1$ $x^{2} - 5 x + 4 ⩾ 0 \Rightarrow x ⩽ 1 \lor x ⩾ 4$ $\Rightarrow 0 < x < 1 \lor 4 < x < 5$ $answer x \in] 0; 5 [∖ {1; 4}$ $but regarding the limits we could define for$ $0 ⩽ x ⩽ 5$
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