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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;+∞)

$$ \\ $$$$\mathrm{one}\:\mathrm{other}\:\mathrm{way}:\mathrm{the}\:\mathrm{conditions}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{inequality}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{as}:\frac{\mathrm{x}−\mathrm{5}}{\mathrm{5}}>\mathrm{0}\Leftrightarrow\mathrm{x}>\mathrm{5} \\ $$$$\mathrm{Since}\:\mathrm{log}_{\mathrm{8}} \frac{\mathrm{x}−\mathrm{5}}{\mathrm{5}}=\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{x}−\mathrm{5}}{\mathrm{5}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} ,\mathrm{so}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{x}.\frac{\mathrm{1}}{\mathrm{3}}\mathrm{log}_{\mathrm{2}} \frac{\mathrm{x}−\mathrm{5}}{\mathrm{5}}\geqslant\mathrm{3log}_{\mathrm{2}} \frac{\mathrm{x}−\mathrm{5}}{\mathrm{5}} \\ $$$$\left.\Leftrightarrow\frac{\mathrm{x}−\mathrm{9}}{\mathrm{3}}.\mathrm{log}_{\mathrm{2}} \frac{\mathrm{x}−\mathrm{5}}{\mathrm{5}}\geqslant\mathrm{0}\Leftrightarrow\right]_{\begin{cases}{\mathrm{x}−\mathrm{9}\leqslant\mathrm{0}}\\{\mathrm{log}_{\mathrm{2}} \frac{\mathrm{x}−\mathrm{5}}{\mathrm{5}}\leqslant\mathrm{0}}\end{cases}\:\:\Leftrightarrow\begin{cases}{\mathrm{5}<\mathrm{x}\leqslant\mathrm{9}}\\{\mathrm{x}\leqslant\mathrm{10}}\end{cases}\Leftrightarrow\mathrm{5}<\mathrm{x}\leqslant\mathrm{9}} ^{\begin{cases}{\mathrm{x}−\mathrm{9}\geqslant\mathrm{0}}\\{\mathrm{log}_{\mathrm{2}} \frac{\mathrm{x}−\mathrm{5}}{\mathrm{5}}\geqslant\mathrm{0}}\end{cases}\:\:\:\:\Leftrightarrow\begin{cases}{\mathrm{x}\geqslant\mathrm{9}}\\{\mathrm{x}\geqslant\mathrm{10}}\end{cases}\Leftrightarrow\mathrm{x}\geqslant\mathrm{10}} \\ $$$$\mathrm{Combining}\:\mathrm{both}\:\mathrm{cases}\:\mathrm{we}\:\mathrm{get}\:\mathrm{solution}\:\mathrm{set} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{given}\:\mathrm{inequality}\:\mathrm{is}\:\left(\mathrm{5};\mathrm{9}\right]\cup\left[\mathrm{10};+\infty\right) \\ $$

Commented by Rasheed.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 .....

$${x}\mathrm{log}_{\mathrm{8}} \left(\frac{{x}−\mathrm{5}}{\mathrm{5}}\:\right)\geqslant\mathrm{3log}\:_{\mathrm{2}} \left(\frac{{x}−\mathrm{5}}{\mathrm{5}}\right) \\ $$$${x}\left(\frac{\mathrm{log}_{\mathrm{2}} \left(\frac{{x}−\mathrm{5}}{\mathrm{5}}\:\right.}{\mathrm{log}_{\mathrm{2}} \mathrm{8}\:}\right)−\mathrm{3log}\:_{\mathrm{2}} \left(\frac{{x}−\mathrm{5}}{\mathrm{5}}\right)\geqslant\mathrm{0} \\ $$$${x}\left(\frac{\mathrm{log}_{\mathrm{2}} \left(\frac{{x}−\mathrm{5}}{\mathrm{5}}\:\right.}{\mathrm{3}\:}\right)−\mathrm{3log}\:_{\mathrm{2}} \left(\frac{{x}−\mathrm{5}}{\mathrm{5}}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{log}\:_{\mathrm{2}} \left(\frac{{x}−\mathrm{5}}{\mathrm{5}}\right)\left(\frac{{x}}{\mathrm{3}}−\mathrm{3}\right)\geqslant\mathrm{0} \\ $$$$..... \\ $$

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 }

$$\Leftrightarrow\frac{\mathrm{1}}{\mathrm{3}}{x}\:\mathrm{log}\:_{\mathrm{2}} \left(\frac{{x}−\mathrm{5}}{\mathrm{5}}\right)−\mathrm{3log}\:_{\mathrm{2}} \left(\frac{{x}−\mathrm{5}}{\mathrm{5}}\right)\:\geqslant\:\mathrm{0} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{3}}{x}−\mathrm{3}\right)\left(\mathrm{log}\:_{\mathrm{2}} \left(\frac{{x}−\mathrm{5}}{\mathrm{5}}\right)\right)\:\geqslant\:\mathrm{0} \\ $$$$\left(\frac{{x}−\mathrm{9}}{\mathrm{3}}\right)\left(\mathrm{log}\:_{\mathrm{2}} \left(\frac{{x}−\mathrm{5}}{\mathrm{5}}\right)\right)\:\geqslant\:\mathrm{0} \\ $$$${case}\left(\mathrm{1}\right)\:{x}\:\geqslant\:\mathrm{9}\:\wedge\:\mathrm{log}\:_{\mathrm{2}} \left(\frac{{x}−\mathrm{5}}{\mathrm{5}}\right)\geqslant\mathrm{0} \\ $$$${x}\:\geqslant\:\mathrm{10}\:\Rightarrow\:{x}\:\geqslant\:\mathrm{10} \\ $$$${case}\left(\mathrm{2}\right)\:{x}\:\leqslant\mathrm{9}\:\wedge\:\mathrm{log}\:_{\mathrm{2}} \left(\frac{{x}−\mathrm{5}}{\mathrm{5}}\right)\leqslant\mathrm{0} \\ $$$${x}\:\leqslant\mathrm{10}\:\Rightarrow{x}\:\leqslant\mathrm{9} \\ $$$${inequality}\:{defined}\:{on}\:{x}>\mathrm{5}\: \\ $$$${hence}\:{the}\:{solution}\:{set}\:{is} \\ $$$$\left\{{x}/\:\mathrm{5}<{x}\leqslant\mathrm{9}\:\cup\:{x}\:\geqslant\mathrm{10}\:\right\} \\ $$

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,+∞)

$$\mathrm{log}_{\mathrm{8}} \frac{\mathrm{x}−\mathrm{5}}{\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{log}_{\mathrm{2}} \frac{\mathrm{x}−\mathrm{5}}{\mathrm{5}}=\mathrm{u}\:\:\:\:\mathrm{t}>\mathrm{0} \\ $$$$\mathrm{x}.\mathrm{u}\geqslant\mathrm{9u} \\ $$$$\mathrm{I\begin{cases}{\mathrm{if}\:\:\:\mathrm{u}>\mathrm{0}\Rightarrow\mathrm{x}\geqslant\mathrm{9}}\\{\mathrm{log}_{\mathrm{8}} \frac{\mathrm{x}−\mathrm{5}}{\mathrm{5}}>\mathrm{0}\:\Rightarrow\frac{\mathrm{x}−\mathrm{5}}{\mathrm{5}}>\mathrm{1}\Rightarrow\mathrm{x}>\mathrm{10}}\end{cases}}\Rightarrow\mathrm{x}>\mathrm{10} \\ $$$$\mathrm{II\begin{cases}{\mathrm{if}\:\:\:\mathrm{u}<\mathrm{0}\Rightarrow\mathrm{x}\leqslant\mathrm{9}}\\{\mathrm{log}_{\mathrm{8}} \frac{\mathrm{x}−\mathrm{5}}{\mathrm{5}}<\mathrm{0}\:\Rightarrow\mathrm{0}<\frac{\mathrm{x}−\mathrm{5}}{\mathrm{5}}<\mathrm{1}\Rightarrow\mathrm{5}<\mathrm{x}<\mathrm{10}}\end{cases}}\Rightarrow\mathrm{5}<\mathrm{x}\leqslant\mathrm{9} \\ $$$$\mathrm{III\begin{cases}{\mathrm{if}\:\:\:\mathrm{u}=\mathrm{0}\Rightarrow\forall\mathrm{x},\mathrm{x}×\mathrm{0}\geqslant\mathrm{9}×\mathrm{0}}\\{\mathrm{log}_{\mathrm{8}} \frac{\mathrm{x}−\mathrm{5}}{\mathrm{5}}=\mathrm{0}\:\Rightarrow\frac{\mathrm{x}−\mathrm{5}}{\mathrm{5}}=\mathrm{1}\Rightarrow\mathrm{x}=\mathrm{10}}\end{cases}}\Rightarrow\mathrm{x}=\mathrm{10} \\ $$$$\mathrm{I}\cup\mathrm{II}\cup\mathrm{III}=\mathrm{x}\in\left(\mathrm{5},\mathrm{9}\right]\cup\left[\mathrm{10},+\infty\right) \\ $$

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