[tex]\left \{ {{x+y=34} \atop {\log_{2} x+ \log_{2} y=6}} \right. \: \: \: \left \{ {{x=34-y} \atop {\log_{2} (34-y)+\log_{2} y=6}} \right. \\\\\log_{2} (34-y)+\log_{2} y=6\\\log_{2} (y(34-y))=\log_{2} 64\\y(34-y)=64\\-y^{2} +34y-64=0\\y^{2} -34y+64=0\\y^{2} -2y-32y+64=0\\y(y-2)-32(y-2)=0\\(y-2)(y-32)=0\\y_{1} =2\\y_{2} =32\\\\x_{1} =34-2=32\\x_{2} = 34-32=2[/tex]

Ответ: (32; 2) и (2; 32)