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Q1)
4 + log_4 x + log_x 64 = 0
4 + logx/log4 + logx/log64 = 0
4 + logx/2log2 + logx/6log2 = 0
24log2 + 3logx + logx = 0
24log2 = 4logx
6log2 = logx
x = 2^6 = 64

Q2)
g(x) = 6(3x-8)^(3/2)
g'(x) = 6(3/2)(3x-8)^(1/2)(3)
= 27(3x-8)^(1/2)

Q3)
For 5 boys and 3 girls,
Since no boys in both ends, therefore, we assume two girls are standing on two end,
In the middle, There are only remaining 6 place for 5boys and 1 girls,
Therefore,
If, the boys are the same,
No. of ways
(the girl can rearrange in six places, and the remaining boys must occupy the 5 places, and the boys are the same)
= 6
If the boys are not the same.
No. of ways
= 5! * 6
= 720

Q2 6times3times(3/2)times(3x-8)to the power (1/2)
=27(3x-8)^(1/2)

2011-04-14 12:53:36 補充：
Q3
6ways?

2011-04-14 13:03:43 補充：
is it 6 ways? for q 3