题目内容

函数f(x)=logax(a>0,a≠1),若f(x1x2…x2009)=16,则f(x12)+f(x22)+…+f(x20092)=______.
∵f(x)=logax(a>0,a≠1),且f(x1x2…x2009)=16,
∴f(x12)+f(x22)+…+f(x20092
=logax12+logax22+…+logax20092
=loga(x1x2…x20092
=2×f(x1x2…x2009)=2×16=32.
故答案:32.
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