[アルゴリズム x 数学] p.78 の L <= log1.5 (S) の導出

解いたのでメモ。

(2/3)^L * S >= 1
(2/3)^L >= 1/S (S = A + B なので > 0)
(2/3)^L >= S^-1
log(3/2) (2/3)^L >= log(3/2) S^-1
L * log(3/2) (2/3) >= -log(3/2) S
L * log(3/2) (3/2)^-1 >= -log(3/2) S
-L * log(3/2) (3/2) >= -log(3/2) S
-L >= -log(3/2) S
L <= log(3/2) S
L <= log1.5 (S)