Question

F[X]是周期为三的R上的奇函数当X大于0小于等于2时,F[X]=2的X次方减1；求当X=以0.5为底24的对数F[X]

Answer 1

f（x）=f(x-3)

log(0.5)24=-log(2)24

f(log(0.5)24)=f(-log(2)24)=-f(log(2)24)

4＜log(2)24＜5

-f(log(2)24)=-f（log(2)24-3）=-f（log(2)24-log(2)8）=-f（log(2)3）=-（3-1）=-2