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log-x-x-2-1-1-




Question Number 146272 by mathdanisur last updated on 12/Jul/21
log_x (x^2  + 1) ≤ 1
logx(x2+1)1
Commented by iloveisrael last updated on 13/Jul/21
 log _x (x^2 +1)≤log _x (x)  ⇒(x−1)(x^2 −x+1)≤0  since for x^2 −x+1 >0 ∀x∈R  then x<1 , x≠1 ∩ x>0  solution is 0<x<1
logx(x2+1)logx(x)(x1)(x2x+1)0sinceforx2x+1>0xRthenx<1,x1x>0solutionis0<x<1
Commented by mathdanisur last updated on 13/Jul/21
thanks Ser, answer: ∅.?
thanksSer,answer:.?
Commented by iloveisrael last updated on 13/Jul/21
no
no
Answered by gsk2684 last updated on 12/Jul/21
x^2 +1 ≤ x^1  if x>1  x^2 −x+1 ≤ 0   (x−(1/2))^2 −(1/4)+1 ≤ 0  (x−(1/2))^2 +(3/4) ≤ 0 no solution  and   x^2 +1 ≥ x if 0<x<1  x^2 −x+1 ≥ 0  (x−(1/2))^2 +(3/4) ≥ 0   true for all x   solution set is (0 1)
x2+1x1ifx>1x2x+10(x12)214+10(x12)2+340nosolutionandx2+1xif0<x<1x2x+10(x12)2+340trueforallxsolutionsetis(01)
Commented by mathdanisur last updated on 12/Jul/21
cool thanks Ser
coolthanksSer

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