How to solve this equation for $x$?$\left(\sqrt{2-\sqrt{3}}\right)^x + \left(\sqrt{2+\sqrt{3}}\right)^x = 2$ 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.
 A: 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 
$$x=0$$
A: 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.
$t+\frac { 1 }{ t } =2$
A: 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 $c=\log\sqrt{2+\sqrt{3}}$.
A: HINT :
$$(2-\sqrt 3)^{\frac x2}+(2+\sqrt 3)^{\frac x2}-2=0\iff \left((2-\sqrt 3)^{\frac x4}-(2+\sqrt 3)^{\frac x4}\right)^2=0$$
A: 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 $\frac{8^x+27^x}{12^x+18^x}=\frac76$
