domain

You need the expression under the square root sign to be non-negative 2^{2x} + 64^{\frac{x-2}{3}} - \frac{1}{2} \left(72 + 2^{2x} \right) = 2^{2x} + \frac{2^{2x}}{16} - \frac{2^{2x}}{2} -36 \\ \\ = \frac{9}{16} 2^{2x} - 36 \ge 0 \Rightarrow 2^{2x} \ge 64 \Rightarrow 2x \ge 8 \\ \\ \Rightarrow \boxed{x \ge 4}

edit: should be 2x≥6, so x≥3