Алгоритм решения показательного уравнения с иррациональностью в основании. [latex] ( \sqrt{5+ \sqrt{24} } )^x + ( \sqrt{5- \sqrt{24} } )^x = 10[/latex]

[latex]\star \quad \sqrt{5+\sqrt{24}}\cdot \sqrt{5-\sqrt{24}}=\sqrt{25-24}=1\; \; \Rightarrow \\\\\sqrt{5-\sqrt{24}}=\frac{1}{\sqrt{5+\sqrt{24}}}\\\\a=\sqrt{5+\sqrt{24}}\; \; \Rightarrow \; \; \sqrt{5-\sqrt{24}}=\frac{1}{a}\; \; \star \\\\\\(\sqrt{5+\sqrt{24}})^{x}+(\sqrt{5-\sqrt{24}})^{x}=10\\\\a^{x}+\frac{1}{a^{x}}-10=0\\\\ \frac{(a^{x})^2-10\cdot a^{x}+1}{a^{x}}=0 \; ,\\\\ t=a^{x}\ \textgreater \ 0\; ,\; \; \; \frac{t^2-10t+1}{t}=0 \; \to \; t^2-10t+1=0\; ,\; t\ne 0\\\\D/4=25-1=24\; \\\\ t_1=5-\sqrt{24}=5-2\sqrt{6}>0\; ,[/latex] [latex]t_2=5+2\sqrt6\ \textgreater \ 0[/latex] [latex]1)\; \; a^{x}=(\sqrt{5+\sqrt{24}})^{x}=5-2\sqrt6\; \to \; x_1=log_{\sqrt{5+\sqrt{24}}}(5-2\sqrt6) ,[/latex] [latex]x_1=log_{\sqrt{5+\sqrt{24}}}\left ( \frac{1}{\sqrt{5+\sqrt{24}}} \right )^2=log_{\sqrt{5+\sqrt{24}}}(\sqrt{5+\sqrt{24}})^{-2}=-2[/latex] [latex]2)\; \; a^{x}=(\sqrt{5+\sqrt{24}})^{x}=5+2\sqrt6\; \to \; x_2=log_{\sqrt{5+\sqrt{24}}}(5+2\sqrt6)\\\\x_2=2\cdot log_{5+\sqrt{24}}(5+\sqrt{24})=5\cdot 1=2[/latex] [latex]Otvet;\; \; x_1=-2,\; x_2=2\; .[/latex]