My attempt : $$\dfrac{1}{4}(\log_2x)^2+4\sqrt{\dfrac{1}{4}(\log_4x)^2-\log_2x-4}=\log_2x+16$$ $$(\log_2x)^2+16\sqrt{\dfrac{1}{4}(\log_2x)^2-\log_2x-4}=\log_2x+16$$ Let $a=\log_2x$ $$a^2+8\sqrt{a^2-4a-16}=a+16$$
Because this is fourth degree polynomials so I think this might not be the most efficient way to solve it.
It should be $$a^2+8\sqrt{a^2-4a-16}=4a+64.$$
Now, we can use a substitution $a^2-4a-16=t^2,$ where $t\geq0$.