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 byYozzii last updated on 07/Jun/16 | ||
$$\frac{{lnu}}{{ln}\mathrm{0}.\mathrm{5}}>\frac{{lnu}}{{ln}\left({x}+\mathrm{5}\right)}\:\:\:\:\left({change}\:{of}\:{base}/{u}={x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{12}\right) \\ $$ $${lnu}\left(\frac{\mathrm{1}}{{ln}\mathrm{0}.\mathrm{5}}−\frac{\mathrm{1}}{{ln}\left({x}+\mathrm{5}\right)}\right)>\mathrm{0} \\ $$ $$−−−−−−−−−−−−−−−−−−−−−−− \\ $$ $$\left(\mathrm{1}\right)\:{lnu}>\mathrm{0}\:{and}\:\frac{\mathrm{1}}{{ln}\mathrm{0}.\mathrm{5}}>\frac{\mathrm{1}}{{ln}\left({x}+\mathrm{5}\right)} \\ $$ $${u}>{e}^{\mathrm{0}} =\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$ $${x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{12}>\mathrm{1} \\ $$ $${x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{11}>\mathrm{0} \\ $$ $$\left({x}+\frac{\mathrm{7}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{7}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)>\mathrm{0} \\ $$ $$\Rightarrow{x}\in\left[\left(−\infty,−\frac{\mathrm{7}+\sqrt{\mathrm{5}}}{\mathrm{2}}\approx−\mathrm{4}.\mathrm{62}\right)\cup\left(\frac{\sqrt{\mathrm{5}}−\mathrm{7}}{\mathrm{2}}\approx−\mathrm{2}.\mathrm{38},+\infty\right)\right]...\left({i}\right) \\ $$ $$ \\ $$ $${Also},\:\frac{\mathrm{1}}{{ln}\mathrm{0}.\mathrm{5}}>\frac{\mathrm{1}}{{ln}\left({x}+\mathrm{5}\right)} \\ $$ $$\frac{{ln}\left({x}+\mathrm{5}\right)−{ln}\mathrm{0}.\mathrm{5}}{\left({ln}\mathrm{0}.\mathrm{5}\right){ln}\left({x}+\mathrm{5}\right)}>\mathrm{0} \\ $$ $$\frac{{ln}\left(\mathrm{2}\left({x}+\mathrm{5}\right)\right)}{{ln}\left({x}+\mathrm{5}\right)}<\mathrm{0}\:\:\left({ln}\mathrm{0}.\mathrm{5}<\mathrm{0}\right) \\ $$ $$\Rightarrow\left(\mathrm{1}\right)\:{ln}\left(\mathrm{2}\left({x}+\mathrm{5}\right)\right)>\mathrm{0}\:\&\:{ln}\left({x}+\mathrm{5}\right)<\mathrm{0} \\ $$ $$\Rightarrow\mathrm{2}\left({x}+\mathrm{5}\right)>\mathrm{1}\:\&\:\mathrm{0}<{x}+\mathrm{5}<\mathrm{1} \\ $$ $${x}>−\mathrm{4}.\mathrm{5}\:\:\:\&\:−\mathrm{5}<{x}<−\mathrm{4}\:\Rightarrow{x}\in\left(−\mathrm{4}.\mathrm{5},−\mathrm{4}\right). \\ $$ $$\Rightarrow\left(\mathrm{2}\right)\:{ln}\left(\mathrm{2}\left({x}+\mathrm{5}\right)\right)<\mathrm{0}\:\&\:{ln}\left({x}+\mathrm{5}\right)>\mathrm{0} \\ $$ $$\Rightarrow−\mathrm{5}<{x}<−\mathrm{4}.\mathrm{5}\:\&\:{x}>−\mathrm{4}\:\:\left({impossible}\right) \\ $$ $$\therefore\:{x}\in\left(−\mathrm{4}.\mathrm{5},−\mathrm{4}\right)....\left({ii}\right) \\ $$ $${The}\:{region}\:{of}\:{overlap}\:{of}\:\left({i}\right)\:{and}\:\left({ii}\right) \\ $$ $${is}\:{empty}. \\ $$ $$−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$ $${lnu}<\mathrm{0}\:{and}\:\frac{\mathrm{1}}{{ln}\mathrm{0}.\mathrm{5}}−\frac{\mathrm{1}}{{ln}\left({x}+\mathrm{5}\right)}<\mathrm{0} \\ $$ $${lnu}<\mathrm{0}\Rightarrow\mathrm{0}<\left({x}+\mathrm{4}\right)\left({x}+\mathrm{3}\right)<\mathrm{1}\Rightarrow{x}\in\left[\left(−\frac{\mathrm{7}+\sqrt{\mathrm{5}}}{\mathrm{2}},−\mathrm{4}\right)\cup\left(−\mathrm{3},\frac{\sqrt{\mathrm{5}}−\mathrm{7}}{\mathrm{2}}\right)\right]........\:\left({i}\right) \\ $$ $$ \\ $$ $$\frac{\mathrm{1}}{{ln}\mathrm{0}.\mathrm{5}}−\frac{\mathrm{1}}{{ln}\left({x}+\mathrm{5}\right)}<\mathrm{0}\Rightarrow\frac{{ln}\left(\mathrm{2}\left({x}+\mathrm{5}\right)\right)}{{ln}\left({x}+\mathrm{5}\right)}>\mathrm{0} \\ $$ $$\Rightarrow\left(\mathrm{1}\right)\:{ln}\left(\mathrm{2}\left({x}+\mathrm{5}\right)\right)>\mathrm{0}\:\&\:{ln}\left({x}+\mathrm{5}\right)>\mathrm{0} \\ $$ $${x}>−\mathrm{4}.\mathrm{5}\:\:\&\:{x}>−\mathrm{4}\Rightarrow{x}>−\mathrm{4} \\ $$ $$\left(\mathrm{2}\right){ln}\left(\mathrm{2}\left({x}+\mathrm{5}\right)\right)<\mathrm{0}\:\:\&\:{ln}\left({x}+\mathrm{5}\right)<\mathrm{0} \\ $$ $$\mathrm{0}<\mathrm{2}\left({x}+\mathrm{5}\right)<\mathrm{1}\:\:\&\mathrm{0}<{x}+\mathrm{5}<\mathrm{1} \\ $$ $$−\mathrm{5}<{x}<−\mathrm{4}.\mathrm{5}\:\:\&−\mathrm{5}<{x}<−\mathrm{4}\Rightarrow\:{x}\in\left(−\mathrm{5},−\mathrm{4}.\mathrm{5}\right) \\ $$ $$\therefore\:{x}\in\left[\left(−\mathrm{4},+\infty\right)\cup\left(−\mathrm{5},−\mathrm{4}.\mathrm{5}\right)\right]....\left({ii}\right) \\ $$ $${Region}\:{of}\:{overlap}\:{of}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{is} \\ $$ $${is}\:{x}\in\left[\left(−\frac{\mathrm{5}+\sqrt{\mathrm{7}}}{\mathrm{2}},−\frac{\mathrm{9}}{\mathrm{2}}\right)\cup\left(−\mathrm{3},\frac{\sqrt{\mathrm{5}}−\mathrm{7}}{\mathrm{2}}\right)\right]. \\ $$ $$−−−−−−−−−−−−−−−−−−−−−−−− \\ $$ $${Answer}:\:{x}\in\left[\left(\frac{−\mathrm{5}−\sqrt{\mathrm{7}}}{\mathrm{2}},−\frac{\mathrm{9}}{\mathrm{2}}\right)\cup\left(−\mathrm{3},\frac{\sqrt{\mathrm{5}}−\mathrm{7}}{\mathrm{2}}\right)\right] \\ $$ $$ \\ $$ | ||
Commented bylove math last updated on 07/Jun/16 | ||
$${Then}\:{what}\:{need}\:{to}\:{do}? \\ $$ | ||
Commented byprakash 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 byYozzii last updated on 07/Jun/16 | ||
$${What}\:{if}\:{x}=−\mathrm{3}?\:{Does}\:{the}\:{inequality}\: \\ $$ $${hold}\:{if}\:{we}\:{assume}\:{the}\:{logarithm}\:{is}\:{real}\:{valued}?\: \\ $$ | ||
Commented byprakash jain last updated on 08/Jun/16 | ||
$$\mathrm{Just}\:\mathrm{to}\:\mathrm{add}\:\mathrm{to}\:\mathrm{Yozzi}'\mathrm{s}\:\mathrm{comment}. \\ $$ $$\mathrm{Complex}\:\mathrm{logarithm}\:\mathrm{and}\:\mathrm{logs}\:\mathrm{of}\:−\mathrm{ve}\:\mathrm{numbers} \\ $$ $$\mathrm{are}\:\mathrm{complex}\:\mathrm{number}\:\mathrm{and}\:\mathrm{inequality}\:\mathrm{does} \\ $$ $$\mathrm{not}\:\mathrm{make}\:\mathrm{sense}.\:\mathrm{So}\:\mathrm{real}\:\mathrm{valued}\:\mathrm{logs}\:\mathrm{are} \\ $$ $$\mathrm{only}\:\mathrm{required}. \\ $$ $${x}\in\left[−\mathrm{3},−\mathrm{4}\right],{x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{12}\leqslant\mathrm{0}\:\mathrm{so}\:\mathrm{log}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{taken}. \\ $$ | ||