已知f(x)={2^x,(x≥4);f(x+1)(x

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已知f(x)={2^x,(x≥4);f(x+1)(x已知f(x)={2^x,(x≥4);f(x+1)(x已知f(x)={2^x,(x≥4);f(x+1)(x∵log2(3)∈(1,2)∴f(log2(

已知f(x)={2^x,(x≥4);f(x+1)(x
已知f(x)={2^x,(x≥4);f(x+1)(x

已知f(x)={2^x,(x≥4);f(x+1)(x
∵ log2(3)∈(1,2)
∴ f(log2(3))
=f(log2(3)+1)
=f(log2(6))
∵ log2(6)∈(2,3)
=f(log2(6)+1)
=f(log2(12))
∵ log2(12)∈(3,4)
=f(log2(12)+1)
=f(log2(24)
∵ log2(24)∈(4,5)
=2^(log2(24))
=24

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解log2(3)＜log2（16）=4
故f(log2(3))=f(log2(3)+1)=f(log2(3)+log2（2）)=f(log2(6))
而log2(6)＜log2（16）=4
f(log2(3))=f(log2(6))=f(log2(6)+1)=f(log2(6)+log2（2）)=f(log2(12))
而log2(12)＜log2（16）=4
故f(log2(3))=f(log2(6))=f(log2(12))=f(log2(12)+1)=f(log2(24))
而log2(24)＞log2（16）=4
故f(log2(3))=f(log2(6))=f(log2(12))=f(log2(24))
=2^(log2(24))
=24.

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