TITLE: How to solve this equation for $x$?$\left(\sqrt{2-\sqrt{3}}\right)^x + \left(\sqrt{2+\sqrt{3}}\right)^x = 2$
QUESTION [8 upvotes]: This is probably such a beginner question (and it's not homework). I've stumbled upon this:
$$\left(\sqrt{2-\sqrt{3}}\right)^x + \left(\sqrt{2+\sqrt{3}}\right)^x = 2$$
How to solve this equation for $x$? I've tried applying the $\log$ and also
raising the expression to the power of 2, but couldn't go much further. I am probably missing something obvious.

REPLY [8 votes]: multiply the equation by $(\sqrt{2-\sqrt{3}})^x$
$$(2-\sqrt{3})^x+(\sqrt{4-3})^x=2(\sqrt{2-\sqrt{3}})^x$$
$$(2-\sqrt{3})^x-2(\sqrt{2-\sqrt{3}})^x+1=0$$
$$((2-\sqrt{3})^{x/2})^2-2(2-\sqrt{3})^{x/2}+1=0$$
$$((2-\sqrt{3})^{x/2}-1)^2=0$$
$$(2-\sqrt{3})^{x/2}-1=0$$
$$(2-\sqrt{3})^{x/2}=1$$
take the $\log$
$$x/2\log(2-\sqrt{3})=\log 1$$
hence 
$$x=0$$