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Question Number 187731 by Rupesh123 last updated on 21/Feb/23

	Answered by Frix last updated on 21/Feb/23

	$Very obviously x = 0$
	Answered by Rasheed.Sindhi last updated on 21/Feb/23

	$2^{x} + 2^{- x} = 2$ $2^{x} = y \Rightarrow 2^{- x} = \frac{1}{y}$ $y + \frac{1}{y} = 2$ $y^{2} - 2 y + 1 = 0$ ${(y - 1)}^{2} = 0$ $y = 1$ $2^{x} = 1$ $2^{x} = 2^{0}$ $x = 0$
	Answered by CElcedricjunior last updated on 21/Feb/23

	$2^{2 x} - 2^{x} + 1 = 0 => {(2^{x} - 1)}^{2} = 0 ◼ M o i v r e$ $=> 2^{x} = 1 => x l n 2 = l n 1 => x = \frac{l n 1}{l n 2} = \frac{0}{l n 2} = 0$ $=> x = 0 ★ C e d r i c j u n i o r$
	Answered by manxsol last updated on 21/Feb/23

	$a + \frac{1}{a} ≫ 2 \Rightarrow a = 1 1 + 1 = 2$ $2^{z} = 1 \Rightarrow x = 0$
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