Solve for n: 4^n + 2^n − 6 = (2^n − 4)^3 + (4^n − 2)^3 ....


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Question Number 47139 by Tawa1 last updated on 05/Nov/18

$Solve for n : 4^{n} + 2^{n} - 6 = {(2^{n} - 4)}^{3} + {(4^{n} - 2)}^{3} . . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 05/Nov/18
	$2^n =x x^2 +x−6=(x−4)^3 +(x^2 −2)^3 (x^2 +x−6)=(x^2 +x−6){(x−4)^2 −(x−4)(x^2 −2)+(x^2 −2)^2 } (x^2 +x−6)[{x^2 −8x+16−x^3 +2x+4x^2 −8+x^4 −4x^2 +4}−1]=0 either x^2 +x−6=0 x^2 +3x−2x−6=0 x(x+3)−2(x+3)=0 (x+3)(x−2)=0 x=−3 x=2 so 2^n =2 n=1← n=1 2^n =−3 n=ln_2 (−3)←not feasible solution again... [{x^2 −8x+16−x^3 +2x+4x^2 −8+x^4 −4x^2 +4−1]=0 x^4 −x^3 +x^2 −6x−11=0 soljtion are 2.4077and −1.12227 2^n =2.4077 n=ln_2 (2.4077) 2^n =−1.2227 n=ln_2 (−1.1227) ←nog feasible soluution so n=(1 and ln_2 2.4077)$
	$2^{n} = x$ $x^{2} + x - 6 = {(x - 4)}^{3} + {(x^{2} - 2)}^{3}$ $(x^{2} + x - 6) = (x^{2} + x - 6) {{(x - 4)}^{2} - (x - 4) (x^{2} - 2) + {(x^{2} - 2)}^{2}}$ $(x^{2} + x - 6) [{x^{2} - 8 x + 16 - x^{3} + 2 x + 4 x^{2} - 8 + x^{4} - 4 x^{2} + 4} - 1] = 0$ $e i t h e r$ $x^{2} + x - 6 = 0$ $x^{2} + 3 x - 2 x - 6 = 0$ $x (x + 3) - 2 (x + 3) = 0$ $(x + 3) (x - 2) = 0$ $x = - 3 x = 2$ $s o 2^{n} = 2 n = 1 \leftarrow n = 1$ $2^{n} = - 3$ $n = {l n}_{2} (- 3) \leftarrow n o t f e a s i b l e s o l u t i o n$ $a g a i n . . .$ $[{x^{2} - 8 x + 16 - x^{3} + 2 x + 4 x^{2} - 8 + x^{4} - 4 x^{2} + 4 - 1] = 0$ $x^{4} - x^{3} + x^{2} - 6 x - 11 = 0$ $s o l j t i o n a r e 2.4077 a n d - 1.12227$ $2^{n} = 2.4077$ $n = {l n}_{2} (2.4077)$ $2^{n} = - 1.2227$ $n = {l n}_{2} (- 1.1227) \leftarrow n o g f e a s i b l e s o l u u t i o n$ $s o n = (1 a n d {l n}_{2} 2.4077)$
	Commented by Tawa1 last updated on 05/Nov/18

	$God bless you sir$
	Commented by tanmay.chaudhury50@gmail.com last updated on 05/Nov/18

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