x-log-2-x-gt-2-

Question Number 4614 by love math last updated on 14/Feb/16

\sqrt{x^{{l o g}_{2} \sqrt{x}}} > 2

Commented by Yozzii last updated on 14/Feb/16

S q u a r i n g b o t h s i d e s o f t h e g i v e n

i n e q u a l i t y y i e l d s t h e f o l l o w i n g .

x^{{l o g}_{2} \sqrt{x}} > 4 (*)

T a k i n g l o g s t o b a s e 2 o n b o t h s i d e s

o f (*) w e o b t a i n

{l o g}_{2} x^{{l o g}_{2} \sqrt{x}} > {l o g}_{2} 2^{2}

B y t h e p o w e r r u l e w e h a v e

({l o g}_{2} \sqrt{x}) ({l o g}_{2} x) > 2 {l o g}_{2} 2 = 2

\frac{1}{2} ({l o g}_{2} x) ({l o g}_{2} x) > 2

{({l o g}_{2} x)}^{2} > 4

({l o g}_{2} x - 2) ({l o g}_{2} x + 2) > 0

\Rightarrow (1) {l o g}_{2} x - 2 > 0 a n d {l o g}_{2} x + 2 > 0

\Rightarrow x > 4 a n d x > 2^{- 2} \Rightarrow x > 4

o r \Rightarrow (2) {l o g}_{2} x - 2 < 0 a n d {l o g}_{2} x + 2 < 0

\Rightarrow x < 4 a n d x < 2^{- 2} \Rightarrow x < 1 / 4

x m u s t b e s t r i c t l y p o s i t i v e h o w e v e r

f o r \sqrt{x} \in R . S o, t r u l y 0 < x < 1 / 4 .

H e n c e, t h e s o l u t i o n s e t Υ f o r t h e

i n e q u a l i t y i s Υ = {x \in R ∣ 0 < x < 1 / 4 o r x > 4} .

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