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.
The only real solution is given by $x=0$, since $\sqrt{2-\sqrt{3}}$ and $\sqrt{2+\sqrt{3}}$ are positive real numbers with product one, hence your equation is equivalent to: $$ e^{cx}+e^{-cx}=2 $$ or: $$ \cosh(cx) = 1 $$ for $c=\log\sqrt{2+\sqrt{3}}$.
HINT :
$$(2-\sqrt 3)^{\frac x2}+(2+\sqrt 3)^{\frac x2}-2=0\iff \left((2-\sqrt 3)^{\frac x4}-(2+\sqrt 3)^{\frac x4}\right)^2=0$$
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 $$x=0$$
since $\sqrt { 2-\sqrt { 3 } } =\frac { 1 }{ \sqrt { 2+\sqrt { 3 } } } $ $t=\left( \sqrt { 2-\sqrt { 3 } } \right) ^{ x }$
you should solve only this eqv. $t+\frac { 1 }{ t } =2$
HINT:
As $(2+\sqrt3)(2-\sqrt3)=1$ set $(2+\sqrt3)^{x/2}=u$ to find $u=1=(2+\sqrt3)^0$
Now as $(2+\sqrt3)^{x/2}=(2+\sqrt3)^0,$
$\implies\dfrac x2=0$ as $2+\sqrt3\ne0,\pm1$
See also: Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$
