填空题设函数f(x)=logax(a>0,且a≠1),若f(x1x2…x2007)=8

发布时间:2020-07-09 04:06:20

填空题设函数f(x)=logax(a>0,且a≠1),若f(x1x2…x2007)=8,则f(x12)+f(x22)+…+f(x20072)=________.

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16解析分析:由题设条件知f(x12)+f(x22)+…+f(x20072)=logax12+logax22+…+logax20072=loga(x1x2…x2007)2,由已知能够求出f(x1x2…x2007)=8,则f(x12)+f(x22)+…+f(x20092)的值可求.解答:f(x12)+f(x22)+…+f(x20072)=logax12+logax22+…+logax20072=loga(x1x2…x2007)2=2loga(x1x2…x2007)=2f(x1x2…x2007)=2×8=16故
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