2^x = 4x Solution 2^x = 4x This can be re write as (1+1)^x = 4x Using combination to epand from the identity. (1+x)^n = 1+nx+((n(n−1))/(2!))x^2 +((n(n−1)(n−2))/(3!))x^3 +....+nCrx^(r ) Therefore. (1+1)^x = 4x 1+x+((x(x−1))/(2!))(1^2 )+((x(x−1)(x−2))/(3!))(1^3 )+......+1 = 4x ignore the continuity (what rule is ..... ignore the +...+) is it linear approximation 1+x+((x^2 −x)/(2×1))+(((x^2 −x)(x−2))/(3×2×1))+1 = 4x 1+x+((x^2 −x)/2)+((x^3 −2x^2 −x^2 +2x)/6)+1 = 4x 1+x+((x^2 −x)/2)+((x^3 −3x^2 +2x)/6)+1 = 4x Multiply through by 6 6+6x+3(x^2 −x)+x^3 −3x^2 +2x+6 = 24x 12+6x+3x^2 −3x+x^3 −3x^2 +2x+6 = 24x 12+5x+x^3 = 24x 12+5x+x^3 −24x = 0 x^3 −19x+12 = 0 Factorize x^3 −4x^2 +4x^2 −16x−3x+12 = 0 (x^3 −4x^2 )+(4x^2 −16x)−(3x+12) = 0 x^2 (x−4)+4x(x−4)−3(x−4) = 0 Factor out (x−4) (x−4)(x^2 +4x−3) = 0 x−4 = 0 or x^( 2) +4x−3 = 0 x = 4 or x = 0.6458 or x = −4.6458 The only real solution is x = 4 Therefore x = 4 DONE! Please confirm the solution. is it correct or please corect it or show me alternative. This is my trial. Thanks.


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Question Number 5764 by sanusihammed last updated on 26/May/16

$2^{x} = 4 x$ $S o l u t i o n$ $2^{x} = 4 x$ $T h i s c a n b e r e w r i t e a s$ ${(1 + 1)}^{x} = 4 x$ $U s i n g c o m b i n a t i o n t o e p a n d$ $f r o m t h e i d e n t i t y .$ ${(1 + x)}^{n} = 1 + n x + \frac{n (n - 1)}{2!} x^{2} + \frac{n (n - 1) (n - 2)}{3!} x^{3} + . . . . + {n C r x}^{r}$ $T h e r e f o r e .$ ${(1 + 1)}^{x} = 4 x$ $1 + x + \frac{x (x - 1)}{2!} (1^{2}) + \frac{x (x - 1) (x - 2)}{3!} (1^{3}) + . . . . . . + 1 = 4 x$ $i g n o r e t h e c o n t i n u i t y (w h a t r u l e i s . . . . . i g n o r e t h e + . . . +) i s i t$ $l i n e a r a p p r o x i m a t i o n$ $1 + x + \frac{x^{2} - x}{2 \times 1} + \frac{(x^{2} - x) (x - 2)}{3 \times 2 \times 1} + 1 = 4 x$ $1 + x + \frac{x^{2} - x}{2} + \frac{x^{3} - 2 x^{2} - x^{2} + 2 x}{6} + 1 = 4 x$ $1 + x + \frac{x^{2} - x}{2} + \frac{x^{3} - 3 x^{2} + 2 x}{6} + 1 = 4 x$ $M u l t i p l y t h r o u g h b y 6$ $6 + 6 x + 3 (x^{2} - x) + x^{3} - 3 x^{2} + 2 x + 6 = 24 x$ $12 + 6 x + 3 x^{2} - 3 x + x^{3} - 3 x^{2} + 2 x + 6 = 24 x$ $12 + 5 x + x^{3} = 24 x$ $12 + 5 x + x^{3} - 24 x = 0$ $x^{3} - 19 x + 12 = 0$ $F a c t o r i z e$ $x^{3} - 4 x^{2} + 4 x^{2} - 16 x - 3 x + 12 = 0$ $(x^{3} - 4 x^{2}) + (4 x^{2} - 16 x) - (3 x + 12) = 0$ $x^{2} (x - 4) + 4 x (x - 4) - 3 (x - 4) = 0$ $F a c t o r o u t (x - 4)$ $(x - 4) (x^{2} + 4 x - 3) = 0$ $x - 4 = 0 o r x^{2} + 4 x - 3 = 0$ $x = 4 o r x = 0.6458 o r x = - 4.6458$ $T h e o n l y r e a l s o l u t i o n i s x = 4$ $T h e r e f o r e$ $x = 4$ $D O N E!$ $P l e a s e c o n f i r m t h e s o l u t i o n . i s i t c o r r e c t o r p l e a s e c o r e c t i t o r$ $s h o w m e a l t e r n a t i v e .$ $T h i s i s m y t r i a l .$ $T h a n k s .$
Commented by prakash jain last updated on 26/May/16

${(1 + x)}^{n} = \underset{i = 0}{\sum^{n}}^{n} C_{i} x^{i} only if n \in Z$
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