<s>[INST] 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..
 [/INST] $$\sqrt{2+\sqrt3}=\sqrt{\frac{4+2\sqrt3}2}=\sqrt{\frac{(\sqrt3)^2+1^2+2\cdot\sqrt3\cdot1}2}=\sqrt{\frac{(\sqrt3+1)^2}2}$$
$$\text{So,}\sqrt{2+\sqrt3}=\frac{\sqrt3+1}{\sqrt2}$$
$$\text{Similarly, }\sqrt{2-\sqrt3}=\frac{\sqrt3-1}{\sqrt2}$$
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