determinant-2-3-x-1-2-2-3-x-x-

Question Number 151205 by EDWIN88 last updated on 19/Aug/21

\underset{⏟}{} \begin{array}{cc} {(2 + \sqrt{3})}^{x} + 1 = {(2 \sqrt{2 + \sqrt{3}})}^{x} \\ x = ? \end{array}

Answered by bramlexs22 last updated on 19/Aug/21

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

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

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

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

x = 2, check \Rightarrow LHS \equiv {(2 + \sqrt{3})}^{2} + {(2 - \sqrt{3})}^{2} = 7 + 7 = 14

RHS \equiv 4^{2} - 2 = 14