Question Number 100720 by bemath last updated on 28/Jun/20

Commented by1549442205 last updated on 28/Jun/20

  one other way:the conditions for the  inequality is defined as:((x−5)/5)>0⇔x>5  Since log_8 ((x−5)/5)=log_2 (((x−5)/5))^(1/3) ,so we get  x.(1/3)log_2 ((x−5)/5)≥3log_2 ((x−5)/5)  ⇔((x−9)/3).log_2 ((x−5)/5)≥0⇔]_( { ((x−9≤0)),((log_2 ((x−5)/5)≤0)) :}  ⇔ { ((5<x≤9)),((x≤10)) :}⇔5<x≤9) ^( { ((x−9≥0)),((log_2 ((x−5)/5)≥0)) :}    ⇔ { ((x≥9)),((x≥10)) :}⇔x≥10)   Combining both cases we get solution set  of the given inequality is (5;9]∪[10;+∞)

Commented byRasheed.Sindhi last updated on 28/Jun/20

xlog_8 (((x−5)/5) )≥3log _2 (((x−5)/5))  x(((log_2 (((x−5)/5) )/(log_2 8 )))−3log _2 (((x−5)/5))≥0  x(((log_2 (((x−5)/5) )/(3 )))−3log _2 (((x−5)/5))≥0  log _2 (((x−5)/5))((x/3)−3)≥0  .....

Commented bybramlex last updated on 28/Jun/20

⇔(1/3)x log _2 (((x−5)/5))−3log _2 (((x−5)/5)) ≥ 0  ((1/3)x−3)(log _2 (((x−5)/5))) ≥ 0  (((x−9)/3))(log _2 (((x−5)/5))) ≥ 0  case(1) x ≥ 9 ∧ log _2 (((x−5)/5))≥0  x ≥ 10 ⇒ x ≥ 10  case(2) x ≤9 ∧ log _2 (((x−5)/5))≤0  x ≤10 ⇒x ≤9  inequality defined on x>5   hence the solution set is  {x/ 5<x≤9 ∪ x ≥10 }

Answered by mahdi last updated on 28/Jun/20

log_8 ((x−5)/5)=(1/3)log_2 ((x−5)/5)=u    t>0  x.u≥9u  I { ((if   u>0⇒x≥9)),((log_8 ((x−5)/5)>0 ⇒((x−5)/5)>1⇒x>10)) :}⇒x>10  II { ((if   u<0⇒x≤9)),((log_8 ((x−5)/5)<0 ⇒0<((x−5)/5)<1⇒5<x<10)) :}⇒5<x≤9  III { ((if   u=0⇒∀x,x×0≥9×0)),((log_8 ((x−5)/5)=0 ⇒((x−5)/5)=1⇒x=10)) :}⇒x=10  I∪II∪III=x∈(5,9]∪[10,+∞)