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

Question Number 103914 by bobhans last updated on 18/Jul/20

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

Commented by som(math1967) last updated on 18/Jul/20

let {(2 + \sqrt{3})}^{x^{2}} = a

∴ {(2 - \sqrt{3})}^{x^{2}} = \frac{1}{a}

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

\Rightarrow a = (2 \pm \sqrt{3})

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

\Rightarrow x = \pm 1

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

x = i

Commented by bemath last updated on 18/Jul/20

t h a n k y o u b o t h

Answered by OlafThorendsen last updated on 18/Jul/20

X = {(2 + \sqrt{3})}^{x^{2}}

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

then X + \frac{1}{X} = 4

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

Δ = {(- 4)}^{2} - 4 (1) (1) = 12

X = \frac{- (- 4) \pm \sqrt{12}}{2 (1)} = 2 \pm \sqrt{3}

1) X = 2 + \sqrt{3}

x^{2} = 1, x = \pm 1

2) X = 2 - \sqrt{3}

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

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

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

x^{2} + 1 = 0

x = \pm i

S = {\pm 1; \pm i}

Answered by Rauny last updated on 18/Jul/20

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

t := 2 + \sqrt{3}

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

∴ 2 - \sqrt{3} = \frac{1}{t}

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

u := t^{x^{2}} (\forall u > 0)

u - 4 + \frac{1}{u} = 0

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

u = 2 \pm \sqrt{3}

t^{x^{2}} = 2 \pm \sqrt{3}

x^{2} = \log_{2 + \sqrt{3}} (2 \pm \sqrt{3})

if u = 2 + \sqrt{3},

x^{2} = 1

x = \pm 1

if u = 2 - \sqrt{3},

x^{2} = \log_{2 + \sqrt{3}} \frac{1}{2 + \sqrt{3}} = - 1 (∵ 2 - \sqrt{3} = \frac{1}{2 + \sqrt{3}})

x = \pm i

∴ x = \pm 1 or \pm i

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