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

Commented by 1549442205 last updated on 28/Jun/20
$one other way:the conditions for the inequality is defined as:((x−5)/5)>0⇔x>5 Since log_8 ((x−5)/5)=log_2 (((x−5)/5))^(1/3) ,so we get x.(1/3)log_2 ((x−5)/5)≥3log_2 ((x−5)/5) ⇔((x−9)/3).log_2 ((x−5)/5)≥0⇔]_( { ((x−9≤0)),((log_2 ((x−5)/5)≤0)) :} ⇔ { ((5<x≤9)),((x≤10)) :}⇔5<x≤9) ^( { ((x−9≥0)),((log_2 ((x−5)/5)≥0)) :} ⇔ { ((x≥9)),((x≥10)) :}⇔x≥10) Combining both cases we get solution set of the given inequality is (5;9]∪[10;+∞)$
$one other way : the conditions for the$ $inequality is defined as : \frac{x - 5}{5} > 0 \Leftrightarrow x > 5$ $Since \log_{8} \frac{x - 5}{5} = \log_{2} {(\frac{x - 5}{5})}^{\frac{1}{3}}, so we get$ $x . \frac{1}{3} \log_{2} \frac{x - 5}{5} ⩾ {3 \log}_{2} \frac{x - 5}{5}$ ${\Leftrightarrow \frac{x - 9}{3} . \log_{2} \frac{x - 5}{5} ⩾ 0 \Leftrightarrow]}_{{\begin{cases} x - 9 ⩽ 0 \\ \log_{2} \frac{x - 5}{5} ⩽ 0 \end{cases} \Leftrightarrow {\begin{cases} 5 < x ⩽ 9 \\ x ⩽ 10 \end{cases} \Leftrightarrow 5 < x ⩽ 9}^{{\begin{cases} x - 9 ⩾ 0 \\ \log_{2} \frac{x - 5}{5} ⩾ 0 \end{cases} \Leftrightarrow {\begin{cases} x ⩾ 9 \\ x ⩾ 10 \end{cases} \Leftrightarrow x ⩾ 10}$ $Combining both cases we get solution set$ $of the given inequality is (5; 9] \cup [10; + \infty)$
Commented by Rasheed.Sindhi last updated on 28/Jun/20

$x \log_{8} (\frac{x - 5}{5}) ⩾ 3 \log_{2} (\frac{x - 5}{5})$ $x (\frac{\log_{2} (\frac{x - 5}{5}}{\log_{2} 8}) - 3 \log_{2} (\frac{x - 5}{5}) ⩾ 0$ $x (\frac{\log_{2} (\frac{x - 5}{5}}{3}) - 3 \log_{2} (\frac{x - 5}{5}) ⩾ 0$ $\log_{2} (\frac{x - 5}{5}) (\frac{x}{3} - 3) ⩾ 0$ $. . . . .$
Commented by bramlex last updated on 28/Jun/20
$⇔(1/3)x log _2 (((x−5)/5))−3log _2 (((x−5)/5)) ≥ 0 ((1/3)x−3)(log _2 (((x−5)/5))) ≥ 0 (((x−9)/3))(log _2 (((x−5)/5))) ≥ 0 case(1) x ≥ 9 ∧ log _2 (((x−5)/5))≥0 x ≥ 10 ⇒ x ≥ 10 case(2) x ≤9 ∧ log _2 (((x−5)/5))≤0 x ≤10 ⇒x ≤9 inequality defined on x>5 hence the solution set is {x/ 5<x≤9 ∪ x ≥10 }$
$\Leftrightarrow \frac{1}{3} x \log_{2} (\frac{x - 5}{5}) - 3 \log_{2} (\frac{x - 5}{5}) ⩾ 0$ $(\frac{1}{3} x - 3) (\log_{2} (\frac{x - 5}{5})) ⩾ 0$ $(\frac{x - 9}{3}) (\log_{2} (\frac{x - 5}{5})) ⩾ 0$ $c a s e (1) x ⩾ 9 \land \log_{2} (\frac{x - 5}{5}) ⩾ 0$ $x ⩾ 10 \Rightarrow x ⩾ 10$ $c a s e (2) x ⩽ 9 \land \log_{2} (\frac{x - 5}{5}) ⩽ 0$ $x ⩽ 10 \Rightarrow x ⩽ 9$ $i n e q u a l i t y d e f i n e d o n x > 5$ $h e n c e t h e s o l u t i o n s e t i s$ ${x / 5 < x ⩽ 9 \cup x ⩾ 10}$
	Answered by mahdi last updated on 28/Jun/20
	$log_8 ((x−5)/5)=(1/3)log_2 ((x−5)/5)=u t>0 x.u≥9u I { ((if u>0⇒x≥9)),((log_8 ((x−5)/5)>0 ⇒((x−5)/5)>1⇒x>10)) :}⇒x>10 II { ((if u<0⇒x≤9)),((log_8 ((x−5)/5)<0 ⇒0<((x−5)/5)<1⇒5<x<10)) :}⇒5<x≤9 III { ((if u=0⇒∀x,x×0≥9×0)),((log_8 ((x−5)/5)=0 ⇒((x−5)/5)=1⇒x=10)) :}⇒x=10 I∪II∪III=x∈(5,9]∪[10,+∞)$
	$\log_{8} \frac{x - 5}{5} = \frac{1}{3} \log_{2} \frac{x - 5}{5} = u t > 0$ $x . u ⩾ 9 u$ $I {\begin{cases} if u > 0 \Rightarrow x ⩾ 9 \\ \log_{8} \frac{x - 5}{5} > 0 \Rightarrow \frac{x - 5}{5} > 1 \Rightarrow x > 10 \end{cases} \Rightarrow x > 10$ $II {\begin{cases} if u < 0 \Rightarrow x ⩽ 9 \\ \log_{8} \frac{x - 5}{5} < 0 \Rightarrow 0 < \frac{x - 5}{5} < 1 \Rightarrow 5 < x < 10 \end{cases} \Rightarrow 5 < x ⩽ 9$ $III {\begin{cases} if u = 0 \Rightarrow \forall x, x \times 0 ⩾ 9 \times 0 \\ \log_{8} \frac{x - 5}{5} = 0 \Rightarrow \frac{x - 5}{5} = 1 \Rightarrow x = 10 \end{cases} \Rightarrow x = 10$ $I \cup II \cup III = x \in (5, 9] \cup [10, + \infty)$
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