Question Number 15097 by Tinkutara last updated on 07/Jun/17 | ||
$$\mathrm{If}\:\mathrm{log}_{\mathrm{4}} \:\mathrm{log}_{\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{log}_{\mathrm{3}} \:\left({x}\right)\:>\:\mathrm{0}\:\mathrm{then}\:{x}\:\mathrm{belongs} \\ $$ $$\mathrm{to}\:\left(\mathrm{1},\:{a}\right),\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{a}^{\mathrm{2}} \:\mathrm{is}? \\ $$ | ||
Commented byprakash jain last updated on 07/Jun/17 | ||
$$\mathrm{log}_{\frac{\mathrm{1}}{\mathrm{2}}} {y}>\mathrm{1}\Rightarrow{y}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$ $$\mathrm{log}_{\mathrm{3}} {x}<\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{x}<\sqrt{\mathrm{3}} \\ $$ $${x}\in\left(\mathrm{1},\sqrt{\mathrm{3}}\right)\Rightarrow{a}=\sqrt{\mathrm{3}}\Rightarrow{a}^{\mathrm{2}} =\mathrm{3} \\ $$ | ||
Commented byTinkutara last updated on 07/Jun/17 | ||
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$ | ||
Answered by mrW1 last updated on 07/Jun/17 | ||
$$\mathrm{log}_{\mathrm{4}} \:\mathrm{log}_{\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{log}_{\mathrm{3}} \:\left({x}\right)\:>\:\mathrm{0} \\ $$ $$\Rightarrow\mathrm{log}_{\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{log}_{\mathrm{3}} \:\left({x}\right)\:>\:\mathrm{1} \\ $$ $$\Rightarrow\mathrm{log}_{\mathrm{3}} \:\left({x}\right)\:<\:\mathrm{1}/\mathrm{2} \\ $$ $$\Rightarrow{x}\:<\:\mathrm{3}^{\mathrm{1}/\mathrm{2}} =\sqrt{\mathrm{3}} \\ $$ $$\mathrm{1}<\mathrm{x}<\sqrt{\mathrm{3}} \\ $$ $$\Rightarrow\mathrm{a}=\sqrt{\mathrm{3}} \\ $$ $$\Rightarrow\mathrm{a}^{\mathrm{2}} =\mathrm{3} \\ $$ | ||
Commented bymrW1 last updated on 07/Jun/17 | ||
$$\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{3} \\ $$ | ||
Commented byTinkutara last updated on 07/Jun/17 | ||
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$ | ||
Answered by Tinkutara last updated on 08/Jun/17 | ||
$$\mathrm{Also}\:\mathrm{finding}\:\mathrm{the}\:\mathrm{domain}: \\ $$ $$\boldsymbol{\mathrm{I}}:\:{x}\:>\:\mathrm{0} \\ $$ $$\boldsymbol{\mathrm{II}}:\:\mathrm{log}_{\mathrm{3}} \:{x}\:>\:\mathrm{0}\:\Rightarrow\:{x}\:>\:\mathrm{1} \\ $$ $$\boldsymbol{\mathrm{III}}:\:\mathrm{log}_{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{log}_{\mathrm{3}} \:{x}\:>\:\mathrm{0} \\ $$ $$\mathrm{log}_{\mathrm{3}} \:{x}\:<\:\mathrm{1}\:\Rightarrow\:{x}\:<\:\mathrm{3} \\ $$ $${x}\:<\:\mathrm{3}\:\mathrm{and}\:{x}\:<\:\sqrt{\mathrm{3}}\:\Rightarrow\:{x}\:<\:\sqrt{\mathrm{3}} \\ $$ $$\therefore\:{x}\:\in\:\left(\mathrm{1},\:\sqrt{\mathrm{3}}\right) \\ $$ $$\Rightarrow\:{a}^{\mathrm{2}} \:=\:\mathrm{3} \\ $$ | ||