Logarithm simplification Simplify: $\log_4(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}})$
Can we use the formula to solve this: $\sqrt{a+\sqrt{b}}= \sqrt{\frac{{a+\sqrt{a^2-b}}}{2}}$
Therefore first term will become: $\sqrt{\frac{3}{2}}$ + $\sqrt{\frac{1}{2}}$
$\log_4$ can be written as $\frac{1}{2}\log_2$
Please guide further..

Let $x=\sqrt{2+\sqrt 3}+\sqrt{2-\sqrt 3}$
$x^2=(2+\sqrt 3)+(2-\sqrt 3)+2\sqrt{(2+\sqrt 3)(2-\sqrt 3)}=6$
I think you get it from here.
