Find the value of x 2^x = 4x workings is needed please.


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Question Number 5712 by sanusihammed last updated on 24/May/16

$F i n d t h e v a l u e o f x$ $2^{x} = 4 x$ $w o r k i n g s i s n e e d e d p l e a s e .$
Commented by Yozzii last updated on 24/May/16

$2^{x} = 4 x$ $\Rightarrow 2^{x - 2} = x$ $T r y x = 4 . \Rightarrow 2^{4 - 2} = 4 = 2^{2}$ $2^{2} = 2^{2} ∴ x = 4 i s o n e s o l u t i o n .$ $2^{x - 2} > 0 \forall x \in R \Rightarrow i f 2^{x - 2} = x \Rightarrow x > 0 i f$ $r e a l s o l u t i o n s a r e t o e x i s t .$ $I f x \in N \Rightarrow x i s e v e n s i n c e x = 2^{x - 2}$ $a n d 2^{x - 2} i s e v e n f o r x - 2 ⩾ 1 \Rightarrow x ⩾ 3 .$ $F o r x ⩾ 6 2^{x - 2} ⩾ 2^{6 - 2} = 16 a n d 2^{x - 2} d i v e r g e s$ $r a p i d l y (n o t e t h a t \frac{d}{d x} (2^{x - 2}) = 2^{x - 2} l n 2) w h e r e a s x r i s e s a t$ $a l i n e a r r a t e (\frac{d}{d x} (x) = 1 < 2^{x - 2} l n 2 f o r x ⩾ 3) .$ $T h i s i n d i c a t e s t h a t n o f u r t h e r i n t e g e r$ $s o l u t i o n s c o u l d p o s s i b l y e x i s t f o r 2^{x} = 4 x .$ $G r a p h i n g t h e t w o f u n c t i o n s y = 4 x$ $a n d y = 2^{x} (x \in R) s h o w s t h a t s o m e o t h e r$ $r o o t l i e s i n t h e i n t e r v a l [0, 1] .$ $A n y m e t h o d o f d e t e r m i n i n g t h e r o o t$ $a p p r o x i m a t e l y c o u l d b e u s e d : I n t e r v a l$ $b i s e c t i o n, N e w t o n - R a p h s o n m e t h o d, L i n e a r$ $i n t e r p o l a t i o n, e t c . . .$
Commented by sanusihammed last updated on 24/May/16

$T h a k s . n o a l g e r a i c s o l u t i o n . ?$
Commented by Yozzii last updated on 25/May/16

$M y l e v e l o f k n o w l e d g e o f m a t h$ $c a n n o t l e t m e s a y t h a t a n a l g e b r a i c$ $s o l u t i o n e x i s t s o r n o t .$
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