4-2x-1-1-4-2-log-2-2x-gt-2-log-x-2-log-1-x-2-2x-

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

4^{2 x - 1} + \frac{1}{4}^{2} \log^{2} (2 x) >^{2} \log (x)

{^{2} \log (\frac{1}{x}) - 2^{2 x}}

Commented by john santu last updated on 24/Jan/20

(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 by john santu last updated on 24/Jan/20

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 by john santu last updated on 24/Jan/20

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 by jagoll last updated on 24/Jan/20

thanks

Facebook Tweet Pin