Answer: [tex]\Large\boxed{x=log_2\frac{12\sqrt{2}+4}{17}}[/tex]
Step-by-step explanation: [tex]\displaystyle\ \bigg{2^x-3(2^{x+\frac{1}{2} }})+2^2=0 \qquad ; \ \ \boxed{t=2^x} \\\\t-3t\sqrt{2} +4=0 \\\\ t(1-3\sqrt{2} )=-4 \\\\ t=\frac{4}{3\sqrt{2} -1} \cdot \frac{3\sqrt{2}+1 }{3\sqrt{2}+1 } =\frac{12\sqrt{2}+4}{17} \\\\\\2^x=\frac{12\sqrt{2}+4}{17} \\\\x=log_2 \ \ \frac{12\sqrt{2}+4}{17}[/tex]
