Question Number 5965 by love math last updated on 07/Jun/16
$${log}_{\frac{\mathrm{1}}{\mathrm{2}}} \left({x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{12}\right)>{log}_{{x}+\mathrm{5}} \left({x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{12}\right) \\ $$
Commented by Yozzii last updated on 07/Jun/16
![((lnu)/(ln0.5))>((lnu)/(ln(x+5))) (change of base/u=x^2 +7x+12) lnu((1/(ln0.5))−(1/(ln(x+5))))>0 −−−−−−−−−−−−−−−−−−−−−−− (1) lnu>0 and (1/(ln0.5))>(1/(ln(x+5))) u>e^0 =1 x^2 +7x+12>1 x^2 +7x+11>0 (x+((7+(√5))/2))(x+((7−(√5))/2))>0 ⇒x∈[(−∞,−((7+(√5))/2)≈−4.62)∪((((√5)−7)/2)≈−2.38,+∞)]...(i) Also, (1/(ln0.5))>(1/(ln(x+5))) ((ln(x+5)−ln0.5)/((ln0.5)ln(x+5)))>0 ((ln(2(x+5)))/(ln(x+5)))<0 (ln0.5<0) ⇒(1) ln(2(x+5))>0 & ln(x+5)<0 ⇒2(x+5)>1 & 0<x+5<1 x>−4.5 & −5<x<−4 ⇒x∈(−4.5,−4). ⇒(2) ln(2(x+5))<0 & ln(x+5)>0 ⇒−5<x<−4.5 & x>−4 (impossible) ∴ x∈(−4.5,−4)....(ii) The region of overlap of (i) and (ii) is empty. −−−−−−−−−−−−−−−−−−−−−−−−− lnu<0 and (1/(ln0.5))−(1/(ln(x+5)))<0 lnu<0⇒0<(x+4)(x+3)<1⇒x∈[(−((7+(√5))/2),−4)∪(−3,(((√5)−7)/2))]........ (i) (1/(ln0.5))−(1/(ln(x+5)))<0⇒((ln(2(x+5)))/(ln(x+5)))>0 ⇒(1) ln(2(x+5))>0 & ln(x+5)>0 x>−4.5 & x>−4⇒x>−4 (2)ln(2(x+5))<0 & ln(x+5)<0 0<2(x+5)<1 &0<x+5<1 −5<x<−4.5 &−5<x<−4⇒ x∈(−5,−4.5) ∴ x∈[(−4,+∞)∪(−5,−4.5)]....(ii) Region of overlap of (i) and (ii) is is x∈[(−((5+(√7))/2),−(9/2))∪(−3,(((√5)−7)/2))]. −−−−−−−−−−−−−−−−−−−−−−−− Answer: x∈[(((−5−(√7))/2),−(9/2))∪(−3,(((√5)−7)/2))]](https://www.tinkutara.com/question/Q5967.png)
$$\frac{{lnu}}{{ln}\mathrm{0}.\mathrm{5}}>\frac{{lnu}}{{ln}\left({x}+\mathrm{5}\right)}\:\:\:\:\left({change}\:{of}\:{base}/{u}={x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{12}\right) \\ $$$${lnu}\left(\frac{\mathrm{1}}{{ln}\mathrm{0}.\mathrm{5}}−\frac{\mathrm{1}}{{ln}\left({x}+\mathrm{5}\right)}\right)>\mathrm{0} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−− \\ $$$$\left(\mathrm{1}\right)\:{lnu}>\mathrm{0}\:{and}\:\frac{\mathrm{1}}{{ln}\mathrm{0}.\mathrm{5}}>\frac{\mathrm{1}}{{ln}\left({x}+\mathrm{5}\right)} \\ $$$${u}>{e}^{\mathrm{0}} =\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$${x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{12}>\mathrm{1} \\ $$$${x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{11}>\mathrm{0} \\ $$$$\left({x}+\frac{\mathrm{7}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{7}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)>\mathrm{0} \\ $$$$\Rightarrow{x}\in\left[\left(−\infty,−\frac{\mathrm{7}+\sqrt{\mathrm{5}}}{\mathrm{2}}\approx−\mathrm{4}.\mathrm{62}\right)\cup\left(\frac{\sqrt{\mathrm{5}}−\mathrm{7}}{\mathrm{2}}\approx−\mathrm{2}.\mathrm{38},+\infty\right)\right]…\left({i}\right) \\ $$$$ \\ $$$${Also},\:\frac{\mathrm{1}}{{ln}\mathrm{0}.\mathrm{5}}>\frac{\mathrm{1}}{{ln}\left({x}+\mathrm{5}\right)} \\ $$$$\frac{{ln}\left({x}+\mathrm{5}\right)−{ln}\mathrm{0}.\mathrm{5}}{\left({ln}\mathrm{0}.\mathrm{5}\right){ln}\left({x}+\mathrm{5}\right)}>\mathrm{0} \\ $$$$\frac{{ln}\left(\mathrm{2}\left({x}+\mathrm{5}\right)\right)}{{ln}\left({x}+\mathrm{5}\right)}<\mathrm{0}\:\:\left({ln}\mathrm{0}.\mathrm{5}<\mathrm{0}\right) \\ $$$$\Rightarrow\left(\mathrm{1}\right)\:{ln}\left(\mathrm{2}\left({x}+\mathrm{5}\right)\right)>\mathrm{0}\:\&\:{ln}\left({x}+\mathrm{5}\right)<\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}\left({x}+\mathrm{5}\right)>\mathrm{1}\:\&\:\mathrm{0}<{x}+\mathrm{5}<\mathrm{1} \\ $$$${x}>−\mathrm{4}.\mathrm{5}\:\:\:\&\:−\mathrm{5}<{x}<−\mathrm{4}\:\Rightarrow{x}\in\left(−\mathrm{4}.\mathrm{5},−\mathrm{4}\right). \\ $$$$\Rightarrow\left(\mathrm{2}\right)\:{ln}\left(\mathrm{2}\left({x}+\mathrm{5}\right)\right)<\mathrm{0}\:\&\:{ln}\left({x}+\mathrm{5}\right)>\mathrm{0} \\ $$$$\Rightarrow−\mathrm{5}<{x}<−\mathrm{4}.\mathrm{5}\:\&\:{x}>−\mathrm{4}\:\:\left({impossible}\right) \\ $$$$\therefore\:{x}\in\left(−\mathrm{4}.\mathrm{5},−\mathrm{4}\right)….\left({ii}\right) \\ $$$${The}\:{region}\:{of}\:{overlap}\:{of}\:\left({i}\right)\:{and}\:\left({ii}\right) \\ $$$${is}\:{empty}. \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${lnu}<\mathrm{0}\:{and}\:\frac{\mathrm{1}}{{ln}\mathrm{0}.\mathrm{5}}−\frac{\mathrm{1}}{{ln}\left({x}+\mathrm{5}\right)}<\mathrm{0} \\ $$$${lnu}<\mathrm{0}\Rightarrow\mathrm{0}<\left({x}+\mathrm{4}\right)\left({x}+\mathrm{3}\right)<\mathrm{1}\Rightarrow{x}\in\left[\left(−\frac{\mathrm{7}+\sqrt{\mathrm{5}}}{\mathrm{2}},−\mathrm{4}\right)\cup\left(−\mathrm{3},\frac{\sqrt{\mathrm{5}}−\mathrm{7}}{\mathrm{2}}\right)\right]……..\:\left({i}\right) \\ $$$$ \\ $$$$\frac{\mathrm{1}}{{ln}\mathrm{0}.\mathrm{5}}−\frac{\mathrm{1}}{{ln}\left({x}+\mathrm{5}\right)}<\mathrm{0}\Rightarrow\frac{{ln}\left(\mathrm{2}\left({x}+\mathrm{5}\right)\right)}{{ln}\left({x}+\mathrm{5}\right)}>\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}\right)\:{ln}\left(\mathrm{2}\left({x}+\mathrm{5}\right)\right)>\mathrm{0}\:\&\:{ln}\left({x}+\mathrm{5}\right)>\mathrm{0} \\ $$$${x}>−\mathrm{4}.\mathrm{5}\:\:\&\:{x}>−\mathrm{4}\Rightarrow{x}>−\mathrm{4} \\ $$$$\left(\mathrm{2}\right){ln}\left(\mathrm{2}\left({x}+\mathrm{5}\right)\right)<\mathrm{0}\:\:\&\:{ln}\left({x}+\mathrm{5}\right)<\mathrm{0} \\ $$$$\mathrm{0}<\mathrm{2}\left({x}+\mathrm{5}\right)<\mathrm{1}\:\:\&\mathrm{0}<{x}+\mathrm{5}<\mathrm{1} \\ $$$$−\mathrm{5}<{x}<−\mathrm{4}.\mathrm{5}\:\:\&−\mathrm{5}<{x}<−\mathrm{4}\Rightarrow\:{x}\in\left(−\mathrm{5},−\mathrm{4}.\mathrm{5}\right) \\ $$$$\therefore\:{x}\in\left[\left(−\mathrm{4},+\infty\right)\cup\left(−\mathrm{5},−\mathrm{4}.\mathrm{5}\right)\right]….\left({ii}\right) \\ $$$${Region}\:{of}\:{overlap}\:{of}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{is} \\ $$$${is}\:{x}\in\left[\left(−\frac{\mathrm{5}+\sqrt{\mathrm{7}}}{\mathrm{2}},−\frac{\mathrm{9}}{\mathrm{2}}\right)\cup\left(−\mathrm{3},\frac{\sqrt{\mathrm{5}}−\mathrm{7}}{\mathrm{2}}\right)\right]. \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${Answer}:\:{x}\in\left[\left(\frac{−\mathrm{5}−\sqrt{\mathrm{7}}}{\mathrm{2}},−\frac{\mathrm{9}}{\mathrm{2}}\right)\cup\left(−\mathrm{3},\frac{\sqrt{\mathrm{5}}−\mathrm{7}}{\mathrm{2}}\right)\right] \\ $$$$ \\ $$
Commented by love math last updated on 07/Jun/16
$${Then}\:{what}\:{need}\:{to}\:{do}? \\ $$
Commented by prakash jain last updated on 08/Jun/16
$$\mathrm{Yozzi}\:\mathrm{considered}\:\mathrm{two}\:\mathrm{cases}\:\mathrm{where} \\ $$$$\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{12}\right)>\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{second}\:\mathrm{case}\:\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{12}\right)<\mathrm{0} \\ $$$${range}\:{of}\:{valid}\:{x}\:{is}\:{also}\:{in}\:{the}\:{answer}. \\ $$
Answered by Ashis last updated on 07/Jun/16
$${log}\left({x}+\mathrm{5}\right)>{log}\left(\mathrm{1}/\mathrm{2}\right) \\ $$$$=>{log}\left(\mathrm{2}{x}+\mathrm{10}\right)>\mathrm{0} \\ $$$$=>\mathrm{2}{x}+\mathrm{10}>\mathrm{1} \\ $$$$=>{x}>−\frac{\mathrm{9}}{\mathrm{2}} \\ $$
Commented by Yozzii last updated on 07/Jun/16
$${What}\:{if}\:{x}=−\mathrm{3}?\:{Does}\:{the}\:{inequality}\: \\ $$$${hold}\:{if}\:{we}\:{assume}\:{the}\:{logarithm}\:{is}\:{real}\:{valued}?\: \\ $$
Commented by prakash jain last updated on 08/Jun/16
![Just to add to Yozzi′s comment. Complex logarithm and logs of −ve numbers are complex number and inequality does not make sense. So real valued logs are only required. x∈[−3,−4],x^2 +7x+12≤0 so log cannot be taken.](https://www.tinkutara.com/question/Q5976.png)
$$\mathrm{Just}\:\mathrm{to}\:\mathrm{add}\:\mathrm{to}\:\mathrm{Yozzi}'\mathrm{s}\:\mathrm{comment}. \\ $$$$\mathrm{Complex}\:\mathrm{logarithm}\:\mathrm{and}\:\mathrm{logs}\:\mathrm{of}\:−\mathrm{ve}\:\mathrm{numbers} \\ $$$$\mathrm{are}\:\mathrm{complex}\:\mathrm{number}\:\mathrm{and}\:\mathrm{inequality}\:\mathrm{does} \\ $$$$\mathrm{not}\:\mathrm{make}\:\mathrm{sense}.\:\mathrm{So}\:\mathrm{real}\:\mathrm{valued}\:\mathrm{logs}\:\mathrm{are} \\ $$$$\mathrm{only}\:\mathrm{required}. \\ $$$${x}\in\left[−\mathrm{3},−\mathrm{4}\right],{x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{12}\leqslant\mathrm{0}\:\mathrm{so}\:\mathrm{log}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{taken}. \\ $$