2-3-x-2-3-x-4-x-

Question Number 179529 by mathlove last updated on 30/Oct/22

{(2 - \sqrt{3})}^{x} + {(2 + \sqrt{3})}^{x} = 4 x = ?

Answered by mr W last updated on 30/Oct/22

(2 - \sqrt{3}) (2 + \sqrt{3}) = 2^{2} - {(\sqrt{3})}^{2} = 4 - 3 = 1

\Rightarrow 2 - \sqrt{3} = \frac{1}{2 + \sqrt{3}}

\frac{1}{{(2 + \sqrt{3})}^{x}} + {(2 + \sqrt{3})}^{x} = 4

l e t t = {(2 + \sqrt{3})}^{x}

\frac{1}{t} + t = 4

t^{2} - 4 t + 1 = 0

t = 2 \pm \sqrt{3}

{(2 + \sqrt{3})}^{x} = 2 \pm \sqrt{3} = 2 + \sqrt{3}, \frac{1}{2 + \sqrt{3}}

\Rightarrow x = \pm 1

Commented by mathlove last updated on 30/Oct/22

t h a n k s m r

Answered by cortano1 last updated on 30/Oct/22

let {(2 - \sqrt{3})}^{x} = y \Rightarrow {(2 + \sqrt{3})}^{x} = \frac{1}{y}

\Rightarrow y + \frac{1}{y} = 4 \Rightarrow y^{2} - 4 y + 1 = 0

\Rightarrow {(y - 2)}^{2} = 3

\Rightarrow y = 2 + \sqrt{3} or y = 2 - \sqrt{3}

{\begin{cases} y = 2 - \sqrt{3} \Rightarrow x = 1 \\ y = 2 + \sqrt{3} \Rightarrow x = - 1 \end{cases}

Answered by Frix last updated on 30/Oct/22

b^{x} + b^{- x} = y; b > 0 \land y ⩾ 2

2 \cosh (x \ln b) = y

x = \frac{\cosh^{- 1} \frac{y}{2}}{\ln b}

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