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Question Number 4615 by love math last updated on 14/Feb/16
2^(log_(0.5) ^2 x) +x^(log_(0.5) x) >2.5
$$\mathrm{2}^{{log}_{\mathrm{0}.\mathrm{5}} ^{\mathrm{2}} {x}} +{x}^{{log}_{\mathrm{0}.\mathrm{5}} {x}} >\mathrm{2}.\mathrm{5} \\ $$
Answered by Yozzii last updated on 14/Feb/16
Rearranging the given inequality leads to x^(log_(0.5) x) >2.5−2^(log_(0.5) ^2 x) . Taking logaritms to base 0.5 on both sides leads to log_(0.5) x^(log_(0.5) x) >log_(0.5) (2.5−2^(log_(0.5) ^2 x) ). By the power rule⇒(log_(0.5) x)(log_(0.5) x)>log_(0.5) (2.5−2^(log_(0.5) ^2 x) ) ⇒log_(0.5) ^2 x>log_(0.5) (2.5−2^(log_(0.5) ^2 x) ) Let u=log_(0.5) ^2 x. ∴ u>log_(0.5) (2.5−2^u ) ⇒0.5^u >0.5^(log_(0.5) (2.5−2^u )) (1/2^u )>(5/2)−2^u ⇒1>(5/2)×2^u −(2^u )^2 (2^u )^2 −(5/2)(2^u )+1>0 2(2^u )^2 −5(2^u )+2>0 (2(2^u )−1)(2^u −2)>0 (∗) (∗) is true iff (1) 2^(u+1) −1>0 and 2^u −2>0 or (2) 2^(u+1) −1<0 and 2^u −2<0. From (1) 2^(u+1) −1>0⇒u+1>0⇒u>−1 and 2^u −2>0⇒u>1. Hence,u>1. Since u=log_(0.5) ^2 x⇒log_(0.5) ^2 x>1 (log_(0.5) x+1)(log_(0.5) x−1)>0 ⇒ (3) log_(0.5) x+1>0 and log_(0.5) x−1>0 or (4) log_(0.5) x+1<0 and log_(0.5) x−1<0. From (3),x>0.5^(−1) =2 and x>0.5^1 =0.5 ∴ x>2. From (4) 0<x<0.5^(−1) =2 and 0<x<0.5 ∴ 0<x<0.5. From (2) 2^(u+1) −1<0⇒u+1<0⇒u<−1 and 2^u −2<0⇒u<1 ∴ u<−1. Since u=log_(0.5) ^2 x⇒log_(0.5) ^2 x<−1. But for x∈R, log_(0.5) ^2 x≮−1<0. ∴ for x∈R, (2) has no solutions. For real x, the solution set Υ of the inequality given is Υ={x∈R∣0<x<1/2 or x>2}.
$${Rearranging}\:{the}\:{given}\:{inequality} \\ $$$${leads}\:{to}\:{x}^{{log}_{\mathrm{0}.\mathrm{5}} {x}} >\mathrm{2}.\mathrm{5}−\mathrm{2}^{{log}_{\mathrm{0}.\mathrm{5}} ^{\mathrm{2}} {x}} . \\ $$$${Taking}\:{logaritms}\:{to}\:{base}\:\mathrm{0}.\mathrm{5}\:{on}\:{both} \\ $$$${sides}\:{leads}\:{to}\:{log}_{\mathrm{0}.\mathrm{5}} {x}^{{log}_{\mathrm{0}.\mathrm{5}} {x}} >{log}_{\mathrm{0}.\mathrm{5}} \left(\mathrm{2}.\mathrm{5}−\mathrm{2}^{{log}_{\mathrm{0}.\mathrm{5}} ^{\mathrm{2}} {x}} \right). \\ $$$${By}\:{the}\:{power}\:{rule}\Rightarrow\left({log}_{\mathrm{0}.\mathrm{5}} {x}\right)\left({log}_{\mathrm{0}.\mathrm{5}} {x}\right)>{log}_{\mathrm{0}.\mathrm{5}} \left(\mathrm{2}.\mathrm{5}−\mathrm{2}^{{log}_{\mathrm{0}.\mathrm{5}} ^{\mathrm{2}} {x}} \right) \\ $$$$\Rightarrow{log}_{\mathrm{0}.\mathrm{5}} ^{\mathrm{2}} {x}>{log}_{\mathrm{0}.\mathrm{5}} \left(\mathrm{2}.\mathrm{5}−\mathrm{2}^{{log}_{\mathrm{0}.\mathrm{5}} ^{\mathrm{2}} {x}} \right) \\ $$$${Let}\:{u}={log}_{\mathrm{0}.\mathrm{5}} ^{\mathrm{2}} {x}.\:\therefore\:{u}>{log}_{\mathrm{0}.\mathrm{5}} \left(\mathrm{2}.\mathrm{5}−\mathrm{2}^{{u}} \right) \\ $$$$\Rightarrow\mathrm{0}.\mathrm{5}^{{u}} >\mathrm{0}.\mathrm{5}^{{log}_{\mathrm{0}.\mathrm{5}} \left(\mathrm{2}.\mathrm{5}−\mathrm{2}^{{u}} \right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{{u}} }>\frac{\mathrm{5}}{\mathrm{2}}−\mathrm{2}^{{u}} \Rightarrow\mathrm{1}>\frac{\mathrm{5}}{\mathrm{2}}×\mathrm{2}^{{u}} −\left(\mathrm{2}^{{u}} \right)^{\mathrm{2}} \\ $$$$\left(\mathrm{2}^{{u}} \right)^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{2}}\left(\mathrm{2}^{{u}} \right)+\mathrm{1}>\mathrm{0} \\ $$$$\mathrm{2}\left(\mathrm{2}^{{u}} \right)^{\mathrm{2}} −\mathrm{5}\left(\mathrm{2}^{{u}} \right)+\mathrm{2}>\mathrm{0} \\ $$$$\left(\mathrm{2}\left(\mathrm{2}^{{u}} \right)−\mathrm{1}\right)\left(\mathrm{2}^{{u}} −\mathrm{2}\right)>\mathrm{0}\:\:\:\:\:\:\:\:\left(\ast\right) \\ $$$$\left(\ast\right)\:{is}\:{true}\:{iff}\:\left(\mathrm{1}\right)\:\mathrm{2}^{{u}+\mathrm{1}} −\mathrm{1}>\mathrm{0}\:{and}\:\mathrm{2}^{{u}} −\mathrm{2}>\mathrm{0} \\ $$$${or}\:\left(\mathrm{2}\right)\:\mathrm{2}^{{u}+\mathrm{1}} −\mathrm{1}<\mathrm{0}\:{and}\:\mathrm{2}^{{u}} −\mathrm{2}<\mathrm{0}. \\ $$$$ \\ $$$${From}\:\left(\mathrm{1}\right)\:\mathrm{2}^{{u}+\mathrm{1}} −\mathrm{1}>\mathrm{0}\Rightarrow{u}+\mathrm{1}>\mathrm{0}\Rightarrow{u}>−\mathrm{1} \\ $$$${and}\:\mathrm{2}^{{u}} −\mathrm{2}>\mathrm{0}\Rightarrow{u}>\mathrm{1}.\:{Hence},{u}>\mathrm{1}. \\ $$$${Since}\:{u}={log}_{\mathrm{0}.\mathrm{5}} ^{\mathrm{2}} {x}\Rightarrow{log}_{\mathrm{0}.\mathrm{5}} ^{\mathrm{2}} {x}>\mathrm{1} \\ $$$$\left({log}_{\mathrm{0}.\mathrm{5}} {x}+\mathrm{1}\right)\left({log}_{\mathrm{0}.\mathrm{5}} {x}−\mathrm{1}\right)>\mathrm{0} \\ $$$$\Rightarrow\:\left(\mathrm{3}\right)\:{log}_{\mathrm{0}.\mathrm{5}} {x}+\mathrm{1}>\mathrm{0}\:{and}\:{log}_{\mathrm{0}.\mathrm{5}} {x}−\mathrm{1}>\mathrm{0} \\ $$$${or}\:\left(\mathrm{4}\right)\:{log}_{\mathrm{0}.\mathrm{5}} {x}+\mathrm{1}<\mathrm{0}\:{and}\:{log}_{\mathrm{0}.\mathrm{5}} {x}−\mathrm{1}<\mathrm{0}. \\ $$$${From}\:\left(\mathrm{3}\right),{x}>\mathrm{0}.\mathrm{5}^{−\mathrm{1}} =\mathrm{2}\:{and}\:{x}>\mathrm{0}.\mathrm{5}^{\mathrm{1}} =\mathrm{0}.\mathrm{5} \\ $$$$\therefore\:{x}>\mathrm{2}.\: \\ $$$${From}\:\left(\mathrm{4}\right)\:\mathrm{0}<{x}<\mathrm{0}.\mathrm{5}^{−\mathrm{1}} =\mathrm{2}\:{and}\:\mathrm{0}<{x}<\mathrm{0}.\mathrm{5} \\ $$$$\therefore\:\mathrm{0}<{x}<\mathrm{0}.\mathrm{5}. \\ $$$$ \\ $$$${From}\:\left(\mathrm{2}\right)\:\mathrm{2}^{{u}+\mathrm{1}} −\mathrm{1}<\mathrm{0}\Rightarrow{u}+\mathrm{1}<\mathrm{0}\Rightarrow{u}<−\mathrm{1} \\ $$$${and}\:\mathrm{2}^{{u}} −\mathrm{2}<\mathrm{0}\Rightarrow{u}<\mathrm{1}\:\therefore\:{u}<−\mathrm{1}.\:{Since} \\ $$$${u}={log}_{\mathrm{0}.\mathrm{5}} ^{\mathrm{2}} {x}\Rightarrow{log}_{\mathrm{0}.\mathrm{5}} ^{\mathrm{2}} {x}<−\mathrm{1}.\:{But}\:{for}\:{x}\in\mathbb{R}, \\ $$$${log}_{\mathrm{0}.\mathrm{5}} ^{\mathrm{2}} {x}\nless−\mathrm{1}<\mathrm{0}.\:\therefore\:{for}\:{x}\in\mathbb{R},\:\left(\mathrm{2}\right)\:{has}\:{no} \\ $$$${solutions}.\: \\ $$$$ \\ $$$${For}\:{real}\:{x},\:{the}\:{solution}\:{set}\:\Upsilon\:{of}\:{the}\: \\ $$$${inequality}\:{given}\:{is}\:\Upsilon=\left\{{x}\in\mathbb{R}\mid\mathrm{0}<{x}<\mathrm{1}/\mathrm{2}\:{or}\:{x}>\mathrm{2}\right\}. \\ $$

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