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Question Number 83786 by jagoll last updated on 06/Mar/20
(x^2 /(log_((5−x)) (x))) ≤ (5x−4) log_x (5−x)
x2log(5−x)(x)⩽(5x−4)logx(5−x)
Answered by john santu last updated on 06/Mar/20
x^2 log_x (5−x) ≤ (5x−4) log_x (5−x) (x−1)((5−x)^x^2 −(5−x)^(5x−4) ) ≤ 0 (i) x > 0 ∧ x ≠ 1 ∧x ≠ 4 ∧ x < 5 ⇒ for x < 1 ⇒ x^2 −5x+4 ≥ 0 (x−1)(x−4) ≥0 , ⇒ x < 1 ⇒ for x > 1 ⇒ x^2 −5x+4 ≤ 0 1 < x < 4 the solution is 0 < x < 1∨ 1 < x < 4
x2logx(5−x)⩽(5x−4)logx(5−x)(x−1)((5−x)x2−(5−x)5x−4)⩽0(i)x>0∧x≠1∧x≠4∧x<5⇒forx<1⇒x2−5x+4⩾0(x−1)(x−4)⩾0,⇒x<1⇒forx>1⇒x2−5x+4⩽01<x<4thesolutionis0<x<1∨1<x<4
Commented by jagoll last updated on 06/Mar/20
thank you sir , but the answer is 0 < x < 1 ∨ 1 < x < 4 ∨ 4 < x < 5
thankyousir,buttheansweris0<x<1∨1<x<4∨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
x2ln(5−x)lnx⩽(5x−4)ln(5−x)lnx0<x<5∧x≠1∧x≠4case1ln(5−x)lnx>0∧x2⩽5x−4case1.1ln(5−x)<0∧lnx<0nosolutioncase1.2ln(5−x)>0∧lnx>01<x<4x2−5x+4⩽0⇒1⩽x⩽5⇒1<x<4case2ln(5−x)lnx<0∧x2⩾5x−4case2.1ln(5−x)<0∧lnx>04<x<5case2.2ln(5−x)>0∧lnx<00<x<1x2−5x+4⩾0⇒x⩽1∨x⩾4⇒0<x<1∨4<x<5answerx∈]0;5[∖{1;4}butregardingthelimitswecoulddefinefor0⩽x⩽5

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