由题设条件知f(x12)+f(x22)+…+f(x20092)=logax12+logax22+…+logax20092=loga(x1x2…x2009)2,由此能够求出f(x1x2…x2009)=16,则f(x12)+f(x22)+…+f(x20092)的值.
【解析】
∵f(x)=logax(a>0,a≠1),且f(x1x2…x2009)=16,
∴f(x12)+f(x22)+…+f(x20092)
=logax12+logax22+…+logax20092
=loga(x1x2…x2009)2
=2×f(x1x2…x2009)=2×16=32.
故答案:32.