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


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Question Number 79263 by jagoll last updated on 23/Jan/20
$4^(2x−1) +(1/4)^2 log^2 (2x)>^2 log(x) {^2 log((1/x))−2^(2x) }$
$4^{2 x - 1} + \frac{1}{4}^{2} \log^{2} (2 x) >^{2} \log (x)$ ${^{2} \log (\frac{1}{x}) - 2^{2 x}}$
Commented byjohn santu last updated on 24/Jan/20
$(i) x>0 (ii)(4^(2x) /4)+(1/4){log_2 (2)+log_2 (x)}^2 >log_2 (x){−log_2 (x)−4^x } let 4^x = u , log_2 (x)= v (u^2 /4)+(1/4)(1+v)^2 >−4v^2 −4uv u^2 +4uv+4v^2 +(1+v)^2 >0 (u+2v)^2 +(1+v)^2 >0 ⇒ { ((u+2v=0⇒4^x +2.log_2 (x)=0)),((1+v=0⇒1+log_2 (x)=0)) :} x =(1/2) ∴ x∈R ∧x=(1/2)$
$(i) x > 0$ $(i i) \frac{4^{2 x}}{4} + \frac{1}{4} {\log_{2} (2) + \log_{2} (x)}^{2}$ $> \log_{2} (x) {- \log_{2} (x) - 4^{x}}$ $l e t 4^{x} = u, \log_{2} (x) = v$ $\frac{u^{2}}{4} + \frac{1}{4} {(1 + v)}^{2} > - 4 v^{2} - 4 u v$ $u^{2} + 4 u v + 4 v^{2} + {(1 + v)}^{2} > 0$ ${(u + 2 v)}^{2} + {(1 + v)}^{2} > 0$ $\Rightarrow {\begin{cases} u + 2 v = 0 \Rightarrow 4^{x} + 2 . \log_{2} (x) = 0 \\ 1 + v = 0 \Rightarrow 1 + \log_{2} (x) = 0 \end{cases}$ $x = \frac{1}{2} ∴ x \in R \land x = \frac{1}{2}$
Commented byjohn santu last updated on 24/Jan/20
$check for x=(1/2) 4^(2((1/2))−1) +(1/4){log_2 (2.(1/2))}^2 > log_2 ((1/2)){log_2 (2)−4^(1/2) } 1+(1/4). 0 > (−1).(−1)$
$c h e c k f o r x = \frac{1}{2}$ $4^{2 (\frac{1}{2}) - 1} + \frac{1}{4} {\log_{2} (2 . \frac{1}{2})}^{2} >$ $\log_{2} (\frac{1}{2}) {\log_{2} (2) - 4^{\frac{1}{2}}}$ $1 + \frac{1}{4} . 0 > (- 1) . (- 1)$
Commented byjohn santu last updated on 24/Jan/20
$i think about this typing error. should 4^(2x−1) +(1/4).log_2 ^2 (2x)≥ log_2 (x){log_2 ((1/x))−2^(2x) }$
$i t h i n k a b o u t t h i s t y p i n g e r r o r .$ $s h o u l d 4^{2 x - 1} + \frac{1}{4} . \log_{2}^{2} (2 x) ⩾$ $\log_{2} (x) {\log_{2} (\frac{1}{x}) - 2^{2 x}}$
Commented byjagoll last updated on 24/Jan/20

$thanks$
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