請教簡單數學 urgent!

Let ln ka = u, then

2au + a = -5

u = -(a + 5)/2a

Sub into the other equation:

-3 = a2 [-(a + 5)/2a] = -a(a + 5)/2

6 = a2 + 5a

a2 + 5a - 6 = 0

(a - 3)(a - 2) = 0

a = 2 or 3

2010-09-27 22:29:03 補充：
(a + 6)(a - 1) = 0

a = 1 or -6

Note: when a = 1, k should be positive and when a = -6, k should be negative.

2010-09-27 23:34:33 補充：
When a = 1:
(2a In ka)+ a =-5
2 ln k + 1 = -5
ln k = -3
k = e^(-3)

When a = -6:
(2a In ka)+ a =-5
-12 ln (-6k) - 6 = -5
ln (-6k) = -1/12
k = -[e^(-1/12)]/6

6尺將軍 :

你竟然連

a2 + 5a - 6 = 0

都....