Simplifying $\log_4(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}})$

Question

Simplify$$\log_4(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}) .$$

Can we use the following formula to solve it?$$\sqrt{a+\sqrt{b}}= \sqrt{\frac{{a+\sqrt{a^2-b}}}{2}}$$

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.

Answer 1

Hint:

$(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}})=\sqrt k$

Your $k=6$, Now its just $\frac{1}{2} \log_46$

Answer 2

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.

Answer 3

$$\sqrt{2+\sqrt3}=\sqrt{\frac{4+2\sqrt3}2}=\sqrt{\frac{(\sqrt3)^2+1^2+2\cdot\sqrt3\cdot1}2}=\sqrt{\frac{(\sqrt3+1)^2}2}$$

$$\text{So,}\sqrt{2+\sqrt3}=\frac{\sqrt3+1}{\sqrt2}$$

$$\text{Similarly, }\sqrt{2-\sqrt3}=\frac{\sqrt3-1}{\sqrt2}$$