log應用題(f.4)

40.
(a)
設半衰期為 t­1/2 年。

W = Wo[3^(-kt)]

當 W = (1/2)Wo，t = t1/2 :
(1/2)Wo = Wo[3^(-kt1/2)]
3^(-kt1/2­) = 2^(-1)
3^(kt1/2­) = 2
log [3^(kt1/2)] = log 2
(kt1/2­) log 3 = log 2
kt1/2 = log 2 / log 3
kt1/2 = log3 2
t1/2 = (1/k) log3 2
半衰期 = (1/k) log3 2 年

(b)
W = Wo[3^(-kt)]

當 t = 4，W = 68%Wo：
68%Wo = Wo[3^(-kt)]
[3^(-4k)] = 0.68
log [3^(-4k)] = log 0.68
-4k log 3 = log 0.68
k = log 0.68 / (-4 log 3)
k = 0.0878

半衰期
= (1/0.8776) log2 / log3
= 7.19 年


=====
42.
α 和 β 是 3x² - 25x + 30 = 0 的根。
兩根之積：αβ = 30/3 = 10

log α + log β
= log (αβ)
= log 10
= 1


=====
43.
(a)
左式
= log4 xy
= log xy / log 4
= log xy / log 2^2
= (log x + log y) / (2 log 2)
= (1/2) [(log x / log 2) + (log y / log 2)]
= (1/2) (log2 x + log2 y)
= (1/2) * 4
= 2
= 右式

所以 log4 xy = 2

(b)
左式
= log8 xy
= log xy / log 8
= log xy / log 2^3
= (log x + log y) / (3 log 2)
= (1/3) [(log x / log 2) + (log y / log 2)]
= (1/3) (log2 x + log2 y)
= (1/3) * 4
= 4/3
= 右式

所以 log8 xy = 4/3