Menu Close

If-log-4-log-1-2-log-3-x-gt-0-then-x-belongs-to-1-a-then-the-value-of-a-2-is-




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 by prakash 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 by Tinkutara 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 by mrW1 last updated on 07/Jun/17
the answer is 3
$$\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{3} \\ $$
Commented by Tinkutara 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
$$\mathrm{Also}\:\mathrm{finding}\:\mathrm{the}\:\mathrm{domain}: \\ $$$$\boldsymbol{\mathrm{I}}:\:{x}\:>\:\mathrm{0} \\ $$$$\boldsymbol{\mathrm{II}}:\:\mathrm{log}_{\mathrm{3}} \:{x}\:>\:\mathrm{0}\:\Rightarrow\:{x}\:>\:\mathrm{1} \\ $$$$\boldsymbol{\mathrm{III}}:\:\mathrm{log}_{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{log}_{\mathrm{3}} \:{x}\:>\:\mathrm{0} \\ $$$$\mathrm{log}_{\mathrm{3}} \:{x}\:<\:\mathrm{1}\:\Rightarrow\:{x}\:<\:\mathrm{3} \\ $$$${x}\:<\:\mathrm{3}\:\mathrm{and}\:{x}\:<\:\sqrt{\mathrm{3}}\:\Rightarrow\:{x}\:<\:\sqrt{\mathrm{3}} \\ $$$$\therefore\:{x}\:\in\:\left(\mathrm{1},\:\sqrt{\mathrm{3}}\right) \\ $$$$\Rightarrow\:{a}^{\mathrm{2}} \:=\:\mathrm{3} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *