已知奇函数f(x)满足f(x+2)=f(x),且当x属于(0,1)时,f(x)=2^x,则f(log0.5(23))=

如题所述

f(log0.5(23))
=f(log2(1/23))
=-f[-log2(1/23)]
=-f(log2(23))
=-f(log2(23/16)]
=-2^[log2(23/16)]
=-23/16
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