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Question Number 79263 by jagoll last updated on 23/Jan/20

4^(2x−1) +(1/4)^2 log^2 (2x)>^2 log(x) {^2 log((1/x))−2^(2x) }

$$\mathrm{4}^{\mathrm{2x}−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{4}}\:^{\mathrm{2}} \mathrm{log}^{\mathrm{2}} \left(\mathrm{2x}\right)>\:^{\mathrm{2}} \mathrm{log}\left(\mathrm{x}\right) \\ $$ $$\left\{^{\mathrm{2}} \mathrm{log}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)−\mathrm{2}^{\mathrm{2x}} \right\} \\ $$

Commented byjohn santu last updated on 24/Jan/20

(i) x>0 (ii)(4^(2x) /4)+(1/4){log_2 (2)+log_2 (x)}^2 >log_2 (x){−log_2 (x)−4^x } let 4^x = u , log_2 (x)= v (u^2 /4)+(1/4)(1+v)^2 >−4v^2 −4uv u^2 +4uv+4v^2 +(1+v)^2 >0 (u+2v)^2 +(1+v)^2 >0 ⇒ { ((u+2v=0⇒4^x +2.log_2 (x)=0)),((1+v=0⇒1+log_2 (x)=0)) :} x =(1/2) ∴ x∈R ∧x=(1/2)

$$\left({i}\right)\:{x}>\mathrm{0} \\ $$ $$\left({ii}\right)\frac{\mathrm{4}^{\mathrm{2}{x}} }{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\left\{\mathrm{log}_{\mathrm{2}} \:\left(\mathrm{2}\right)+\mathrm{log}_{\mathrm{2}} \left({x}\right)\right\}^{\mathrm{2}} \\ $$ $$>\mathrm{log}_{\mathrm{2}} \left({x}\right)\left\{−\mathrm{log}_{\mathrm{2}} \left({x}\right)−\mathrm{4}^{{x}} \right\} \\ $$ $${let}\:\mathrm{4}^{{x}} =\:{u}\:,\:\mathrm{log}_{\mathrm{2}} \left({x}\right)=\:{v} \\ $$ $$\frac{{u}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+{v}\right)^{\mathrm{2}} >−\mathrm{4}{v}^{\mathrm{2}} −\mathrm{4}{uv} \\ $$ $${u}^{\mathrm{2}} +\mathrm{4}{uv}+\mathrm{4}{v}^{\mathrm{2}} +\left(\mathrm{1}+{v}\right)^{\mathrm{2}} >\mathrm{0} \\ $$ $$\left({u}+\mathrm{2}{v}\right)^{\mathrm{2}} +\left(\mathrm{1}+{v}\right)^{\mathrm{2}} >\mathrm{0} \\ $$ $$\Rightarrow\begin{cases}{{u}+\mathrm{2}{v}=\mathrm{0}\Rightarrow\mathrm{4}^{{x}} +\mathrm{2}.\mathrm{log}_{\mathrm{2}} \left({x}\right)=\mathrm{0}}\\{\mathrm{1}+{v}=\mathrm{0}\Rightarrow\mathrm{1}+\mathrm{log}_{\mathrm{2}} \left({x}\right)=\mathrm{0}}\end{cases}\: \\ $$ $${x}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\therefore\:{x}\in\mathbb{R}\:\wedge{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented byjohn santu last updated on 24/Jan/20

check for x=(1/2) 4^(2((1/2))−1) +(1/4){log_2 (2.(1/2))}^2 > log_2 ((1/2)){log_2 (2)−4^(1/2) } 1+(1/4). 0 > (−1).(−1)

$${check}\:{for}\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$ $$\mathrm{4}^{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{4}}\left\{\mathrm{log}_{\mathrm{2}} \left(\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}\right)\right\}^{\mathrm{2}} > \\ $$ $$\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left\{\mathrm{log}_{\mathrm{2}} \left(\mathrm{2}\right)−\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{2}}} \right\} \\ $$ $$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}.\:\mathrm{0}\:>\:\left(−\mathrm{1}\right).\left(−\mathrm{1}\right) \\ $$

Commented byjohn santu last updated on 24/Jan/20

i think about this typing error. should 4^(2x−1) +(1/4).log_2 ^2 (2x)≥ log_2 (x){log_2 ((1/x))−2^(2x) }

$${i}\:{think}\:{about}\:{this}\:{typing}\:{error}. \\ $$ $${should}\:\mathrm{4}^{\mathrm{2}{x}−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{4}}.\mathrm{log}_{\mathrm{2}} ^{\mathrm{2}} \left(\mathrm{2}{x}\right)\geqslant \\ $$ $$\mathrm{log}_{\mathrm{2}} \left({x}\right)\left\{\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}}\right)−\mathrm{2}^{\mathrm{2}{x}} \right\} \\ $$

Commented byjagoll last updated on 24/Jan/20

thanks

$$\mathrm{thanks} \\ $$

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