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Question Number 15097 by Tinkutara last updated on 07/Jun/17

If log_4  log_(1/2)  log_3  (x) > 0 then x belongs  to (1, a), then the value of a^2  is?

$$\mathrm{If}\:\mathrm{log}_{\mathrm{4}} \:\mathrm{log}_{\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{log}_{\mathrm{3}} \:\left({x}\right)\:>\:\mathrm{0}\:\mathrm{then}\:{x}\:\mathrm{belongs} \\ $$ $$\mathrm{to}\:\left(\mathrm{1},\:{a}\right),\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{a}^{\mathrm{2}} \:\mathrm{is}? \\ $$

Commented byprakash jain last updated on 07/Jun/17

log_(1/2) y>1⇒y<(1/2)  log_3 x<(1/2)⇒x<(√3)  x∈(1,(√3))⇒a=(√3)⇒a^2 =3

$$\mathrm{log}_{\frac{\mathrm{1}}{\mathrm{2}}} {y}>\mathrm{1}\Rightarrow{y}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$ $$\mathrm{log}_{\mathrm{3}} {x}<\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{x}<\sqrt{\mathrm{3}} \\ $$ $${x}\in\left(\mathrm{1},\sqrt{\mathrm{3}}\right)\Rightarrow{a}=\sqrt{\mathrm{3}}\Rightarrow{a}^{\mathrm{2}} =\mathrm{3} \\ $$

Commented byTinkutara last updated on 07/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Answered by mrW1 last updated on 07/Jun/17

log_4  log_(1/2)  log_3  (x) > 0  ⇒log_(1/2)  log_3  (x) > 1  ⇒log_3  (x) < 1/2  ⇒x < 3^(1/2) =(√3)  1<x<(√3)  ⇒a=(√3)  ⇒a^2 =3

$$\mathrm{log}_{\mathrm{4}} \:\mathrm{log}_{\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{log}_{\mathrm{3}} \:\left({x}\right)\:>\:\mathrm{0} \\ $$ $$\Rightarrow\mathrm{log}_{\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{log}_{\mathrm{3}} \:\left({x}\right)\:>\:\mathrm{1} \\ $$ $$\Rightarrow\mathrm{log}_{\mathrm{3}} \:\left({x}\right)\:<\:\mathrm{1}/\mathrm{2} \\ $$ $$\Rightarrow{x}\:<\:\mathrm{3}^{\mathrm{1}/\mathrm{2}} =\sqrt{\mathrm{3}} \\ $$ $$\mathrm{1}<\mathrm{x}<\sqrt{\mathrm{3}} \\ $$ $$\Rightarrow\mathrm{a}=\sqrt{\mathrm{3}} \\ $$ $$\Rightarrow\mathrm{a}^{\mathrm{2}} =\mathrm{3} \\ $$

Commented bymrW1 last updated on 07/Jun/17

the answer is 3

$$\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{3} \\ $$

Commented byTinkutara last updated on 07/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Answered by Tinkutara last updated on 08/Jun/17

Also finding the domain:  I: x > 0  II: log_3  x > 0 ⇒ x > 1  III: log_(1/2) log_3  x > 0  log_3  x < 1 ⇒ x < 3  x < 3 and x < (√3) ⇒ x < (√3)  ∴ x ∈ (1, (√3))  ⇒ a^2  = 3

$$\mathrm{Also}\:\mathrm{finding}\:\mathrm{the}\:\mathrm{domain}: \\ $$ $$\boldsymbol{\mathrm{I}}:\:{x}\:>\:\mathrm{0} \\ $$ $$\boldsymbol{\mathrm{II}}:\:\mathrm{log}_{\mathrm{3}} \:{x}\:>\:\mathrm{0}\:\Rightarrow\:{x}\:>\:\mathrm{1} \\ $$ $$\boldsymbol{\mathrm{III}}:\:\mathrm{log}_{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{log}_{\mathrm{3}} \:{x}\:>\:\mathrm{0} \\ $$ $$\mathrm{log}_{\mathrm{3}} \:{x}\:<\:\mathrm{1}\:\Rightarrow\:{x}\:<\:\mathrm{3} \\ $$ $${x}\:<\:\mathrm{3}\:\mathrm{and}\:{x}\:<\:\sqrt{\mathrm{3}}\:\Rightarrow\:{x}\:<\:\sqrt{\mathrm{3}} \\ $$ $$\therefore\:{x}\:\in\:\left(\mathrm{1},\:\sqrt{\mathrm{3}}\right) \\ $$ $$\Rightarrow\:{a}^{\mathrm{2}} \:=\:\mathrm{3} \\ $$

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