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Question Number 120284 by Algoritm last updated on 30/Oct/20

Commented by Algoritm last updated on 30/Oct/20

$x = ?$
	Answered by TITA last updated on 30/Oct/20
	$(x+(1/x))^(x+1) =(1+(1/(2017)))^(2018) ⇒(x+(1/x))^(x+1) =x^(x+1) (1+(1/x^2 ))^(x+1) by equating coefficient x^(x+1) =1 ⇒(x+1)ln x=ln 1 ⇒ln x=0 hence x=1 x^2 =2017⇒x= +_− (√(2017)) x+1=2018 ⇒ x=2017 hence x={1,2017, +_− (√(2017 ))}$
	${(x + \frac{1}{x})}^{x + 1} = {(1 + \frac{1}{2017})}^{2018} \Rightarrow {(x + \frac{1}{x})}^{x + 1} = x^{x + 1} {(1 + \frac{1}{x^{2}})}^{x + 1}$ $b y e q u a t i n g c o e f f i c i e n t$ $x^{x + 1} = 1 \Rightarrow (x + 1) \ln x = \ln 1$ $\Rightarrow \ln x = 0 h e n c e x = 1$ $x^{2} = 2017 \Rightarrow x = \underline{+} \sqrt{2017}$ $x + 1 = 2018 \Rightarrow x = 2017$ $h e n c e x = {1, 2017, \underline{+} \sqrt{2017}}$
	Commented by JDamian last updated on 30/Oct/20

	$W o u l d y o u m i n d e x p l a i n i n g h o w$ ${(1 + \frac{1}{1})}^{1 + 1} = {(1 + \frac{1}{2017})}^{2018} ?$
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