Exponential equation with double radical I'm trying ti solve this exponential equation:
$(\sqrt{2+\sqrt{3}})^x+(\sqrt{2-\sqrt{3}})^x=2^x$.
Here my try:
$\sqrt{2+\sqrt{3}}=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}$ and
$\sqrt{2-\sqrt{3}}=\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}$
So i get this relation:
$\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}}$
Using the substitution $t=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}$ the equation can be written as 
$(t)^x+(\frac{1}{t})^x=2^x$.
And from this i wrote:
$t^{2x}-(2t)^x-1=0$.
Now i don't how to proceed. Any suggestions?
Thanks!

Hint: 
Write $s= \sqrt{2+\sqrt{3}}$, then $\sqrt{2-\sqrt{3}} = {1\over s}$ 
Then $$s^x+({1\over s})^x\geq 2 \implies x\geq 1$$
