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

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} + \dots . + {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}) + \dots \dots + 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 \dots . . i g n o r e t h e + \dots +) 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