Given that the power function satisfies: X ≥ 4, then f (x) = (1 / 2) ^ X; when x < 4, f (x) = f (x + 1), then f (2 + L, logarithm of 3 with 2 as the base) is,

Given that the power function satisfies: X ≥ 4, then f (x) = (1 / 2) ^ X; when x < 4, f (x) = f (x + 1), then f (2 + L, logarithm of 3 with 2 as the base) is,

Since 2 + log234, then f (3 + log23) = (1 / 2) ^ (3 + log23)
=(1/8)*(1/2)^(log23)
=(1/8)*2^(-log23)
=(1/8)*2^[log2(1/3)]=(1/8)(1/3)=1/24