设函数f(X)满足f(X)=1+f(二分之一)*log底数为2真数为x,则f(2)=?

f(1/2)=1+f(1/2)*log2 1/2
f(1/2)=1-f(1/2)
f(1/2)=1/2
f(2)=1+f(1/2)*log2 2
=1+f(1/2)
=3/2请问 怎么算的 f(1/2)=1-f(1/2)f(1/2)=1/2f(1/2)=1-f(1/2)移项2f(1/2)=1合并同类项f(1/2)=1/2那 f(1/2)=1-f(1/2)怎么求log2为底1/2的值是-1