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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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