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Question-13151




Question Number 13151 by Tinkutara last updated on 15/May/17
Answered by mrW1 last updated on 16/May/17
x>9  log (2x−1)(x−9)>2  (2x−1)(x−9)>100  2x^2 −19x−91>0  x_(1,2) =((19±(√(19^2 +4×2×91)))/4)=((19±33)/4)=−3.5 or 13    ⇒x>13
$${x}>\mathrm{9} \\ $$$$\mathrm{log}\:\left(\mathrm{2}{x}−\mathrm{1}\right)\left({x}−\mathrm{9}\right)>\mathrm{2} \\ $$$$\left(\mathrm{2}{x}−\mathrm{1}\right)\left({x}−\mathrm{9}\right)>\mathrm{100} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{19}{x}−\mathrm{91}>\mathrm{0} \\ $$$${x}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{19}\pm\sqrt{\mathrm{19}^{\mathrm{2}} +\mathrm{4}×\mathrm{2}×\mathrm{91}}}{\mathrm{4}}=\frac{\mathrm{19}\pm\mathrm{33}}{\mathrm{4}}=−\mathrm{3}.\mathrm{5}\:{or}\:\mathrm{13} \\ $$$$ \\ $$$$\Rightarrow{x}>\mathrm{13} \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 17/May/17
mrW1 is right.
$${mrW}\mathrm{1}\:{is}\:{right}. \\ $$
Commented by Tinkutara last updated on 16/May/17
Answer seems correct but in my book it is   given 9 < x < 13.
$$\mathrm{Answer}\:\mathrm{seems}\:\mathrm{correct}\:\mathrm{but}\:\mathrm{in}\:\mathrm{my}\:\mathrm{book}\:\mathrm{it}\:\mathrm{is}\: \\ $$$$\mathrm{given}\:\mathrm{9}\:<\:{x}\:<\:\mathrm{13}. \\ $$
Commented by Tinkutara last updated on 16/May/17
Commented by Tinkutara last updated on 16/May/17
Please help me to choose the correct  answer.
$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{choose}\:\mathrm{the}\:\mathrm{correct} \\ $$$$\mathrm{answer}. \\ $$
Commented by mrW1 last updated on 16/May/17
The answer in your book is wrong.  It were correct if the inequality is  (1/2)log (2x−1)+log (√(x−9))<1
$${The}\:{answer}\:{in}\:{your}\:{book}\:{is}\:{wrong}. \\ $$$${It}\:{were}\:{correct}\:{if}\:{the}\:{inequality}\:{is} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\left(\mathrm{2}{x}−\mathrm{1}\right)+\mathrm{log}\:\sqrt{{x}−\mathrm{9}}<\mathrm{1} \\ $$
Commented by Tinkutara last updated on 16/May/17
Thanks. I understood.
$$\mathrm{Thanks}.\:\mathrm{I}\:\mathrm{understood}. \\ $$
Commented by mrW1 last updated on 16/May/17
You can check with x=10,  (1/2)log (2×10−1)+log (√(10−9))  =(1/2)log (19)+log 1  =(1/2)log (19)<(1/2)log 100=1    with x=20,  (1/2)log (2×20−1)+log (√(20−9))  =(1/2)log (39)+log (√(11))  >(1/2)log (10)+(1/2)log 10=1
$${You}\:{can}\:{check}\:{with}\:{x}=\mathrm{10}, \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\left(\mathrm{2}×\mathrm{10}−\mathrm{1}\right)+\mathrm{log}\:\sqrt{\mathrm{10}−\mathrm{9}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\left(\mathrm{19}\right)+\mathrm{log}\:\mathrm{1} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\left(\mathrm{19}\right)<\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\mathrm{100}=\mathrm{1} \\ $$$$ \\ $$$${with}\:{x}=\mathrm{20}, \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\left(\mathrm{2}×\mathrm{20}−\mathrm{1}\right)+\mathrm{log}\:\sqrt{\mathrm{20}−\mathrm{9}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\left(\mathrm{39}\right)+\mathrm{log}\:\sqrt{\mathrm{11}} \\ $$$$>\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\left(\mathrm{10}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\mathrm{10}=\mathrm{1} \\ $$
Commented by Tinkutara last updated on 16/May/17
Thanks very much!
$$\mathrm{Thanks}\:\mathrm{very}\:\mathrm{much}! \\ $$

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