Question

(sqrt(3 + sqrt(8))) ^ x + (sqrt(3 - sqrt(8))) ^ x = 6​

Answer 1

$\Big(\sqrt{3+\sqrt{8} } \Big)^x+\Big(\sqrt{3-\sqrt{8} }\Big)^x=6\\\\ \boxed{\sqrt{a\pm \sqrt{b}} =\sqrt{\cfrac{a+\sqrt{a^2-b} }{2} } \pm \sqrt{\cfrac{a-\sqrt{a^2-b} }{2} } }$

$\sqrt{3\pm\sqrt{8}} =\sqrt{\cfrac{3+\sqrt{3^2-8} }{2} }\pm \sqrt{\cfrac{3-\sqrt{3^2-8} }{2} }=\\\\=\sqrt{\cfrac{3+\sqrt{9-8} }{2} } \pm\sqrt{\cfrac{3-\sqrt{9-8} }{2}}=\sqrt{\cfrac{3+\sqrt{1} }{2}}\pm\sqrt{\cfrac{3-\sqrt{1} }{2}}=\\\\=\sqrt{\cfrac{3+1}{2}}\pm\sqrt{\cfrac{3-1}{2} }=\sqrt{\cfrac{4}{2} } \pm \sqrt{\cfrac{2}{2} }=\sqrt{2} \pm 1$

$\sqrt{3+\sqrt{8} } =\sqrt{2} +1;~\sqrt{3-\sqrt{8} } =\sqrt{2} -1\\\\ \Big(\sqrt{3+\sqrt{8} }\Big)^x+\Big(\sqrt{3-\sqrt{8} }\Big)^x=6\\\\ \Big(\sqrt{2} +1\Big)^x+\Big(\sqrt{2}-1\Big)^x=6\\\\ \Big(\sqrt{2} +1\Big)^x+\Big(\Big(\sqrt{2}+1\Big)^x\Big)^{-1} =6\\\\ \Big(\sqrt{2} +1\Big)^x=t\\\\ t+t^{-1}=6 \Rightarrow t+\cfrac{1}{t} =6 \Rightarrow t^2+1=6t \Rightarrow t^2-6t+1=0$

$\displaystyle \Delta=(-6)^2-4 \cdot 1 \cdot 1=36-4=32 > 0\\\\ t_1=\frac{-(-6)-\sqrt{32} }{2 \cdot 1}=\frac{6-4\sqrt{2} }{2} =\frac{\not2\Big(3-2\sqrt{2} \Big)}{\not 2} =3-2\sqrt{2} \\\\ t_2=\frac{-(-6)+\sqrt{32} }{2 \cdot 1}=\frac{6+4\sqrt{2} }{2} =\frac{\not2\Big(3+2\sqrt{2} \Big)}{\not 2} =3+2\sqrt{2}$

$\displaystyle \Big(\sqrt{2} +1\Big)^x=t_1 \Rightarrow \Big(\sqrt{2} +1\Big)^x=3-2\sqrt{2} \Rightarrow x=log_{\sqrt{2} +1}\Big(3-2\sqrt{2} \Big) \Rightarrow \\\\ \Rightarrow x=log_{\sqrt{2} +1}\Big(\sqrt{2} +1\Big)^{-2} \Rightarrow x=-2log_{\sqrt{2}+1 }\Big(\sqrt{2} +1\Big)\Rightarrow \boxed{x=-2}$

$\displaystyle \Big(\sqrt{2} +1\Big)^x=t_2 \Rightarrow \Big(\sqrt{2} +1\Big)^x=3+2\sqrt{2} \Rightarrow x=log_{\sqrt{2} +1}\Big(3+2\sqrt{2} \Big) \Rightarrow \\\\ \Rightarrow x=log_{\sqrt{2} +1}\Big(\sqrt{2} +1\Big)^2 \Rightarrow x=2log_{\sqrt{2}+1 }\Big(\sqrt{2} +1\Big)\Rightarrow \boxed{x=2}$