f(x)中x为任意常数，f(xy)=f(x)+f(y) 已知 f(8)=3 求f(32)=???????
\u51fd\u6570f(x)\u7684\u5b9a\u4e49\u57df\u4e3aR+,\u4e14f(xy)=f(x)+f(y),\u4e14f(8)=3,\u5219f(32)=\u7b54\uff1a
f(x)\u7684\u5b9a\u4e49\u57df\u4e3aR+\uff0cf(xy)=f(x)+f(y)
f(8)=3
f(8)=f(2\u00d74)=f(2)+f(4)=f(2)+f(2\u00d72)=f(2)+f(2)+f(2)=3
\u6240\u4ee5\uff1af(2)=1
f(32)=f(4\u00d78)=f(4)+f(8)=f(4)+3=f(2)+f(2)+3=1+1+3=5
\u6240\u4ee5\uff1af(32)=5
\u5df2\u77e5\u51fd\u6570\u5bf9\u4efb\u610fx,y\u2208R\uff0c\u90fd\u6709f(xy)=f(x)+f(y)\uff0c\u4e14f(2)=3,
\u90a3\u4e48
f(4)=f(2\u00d72)=f(2)+f(2)=2f(2)=6
\u2234
f(8)=f(2\u00d74)=f(2)+f(4)=3+6=9
得f(2)=1
f(4)=f(2)+f(2)=2
f(32)=f(4*8)=f(4)+f(8)=2+3=5
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