解對數方程

圖片參考：http://hk.geocities.com/abc2004y2k/MATHS/maths1.jpg

Guttvennnnnnnnnn 始終做不到題目給的答案
(loga)^2>= -32 是一個好弱的結果 (因為x^2>=0)
[log(ax)][log(ax^2)] = 4
[loga + logx][loga + logx^2] = 4
[loga + logx][loga + 2logx] = 4
(loga)^2 + 3logalogx + 2(logx)^2 = 4
Let u = logx, the equation becomes
((loga)^2 - 4) + (3loga)u + 2u^2 = 0
u=1/4[-3loga-√[9(loga)^2-8((loga)^2 - 4)]
u=1/4[-3loga-√[(loga)^2 +32)]
Now u>0 [since x>1 , logx>0]
1/4[-3loga-√[(loga)^2 +32)]>0
[-3loga-√[(loga)^2 +32)]>0
√[(loga)^2 +32)]<3loga
(loga)^2 +32<9(loga)^2
8(loga)^2>32
(loga)^2>4
loga>2, loga<-2
loga>2 means a>100 but this is not true
loga<-2 meana a<1/100
ALSO a>0 (otherwise ax will be negative)
so
0<a<1/100

2007-02-10 14:31:38 補充：
u=1/4[-3loga+√[9(loga)^2-8((loga)^2 - 4)] 是沒有用的因為出到-2&lt;2a&lt;100 and a&gt;1/100而a&gt;1/100與題設不合

2007-02-10 14:53:51 補充：
-2&lt; loga &lt;2