MATHS (指數與對數函數)

3. 答案是 b。

(4√a3) x (a2)3 / (√a)
= (a)3/4 x (a)6 / (a)1/2
= (a)3/4 + 6 - 1/2
= (a)3/4 + 24/4 - 2/4
= (a)23/4


6. 答案是 b。

3x - y
= 3log6 - log8
= 3log6 - log23
= 3log6 - 3log2
= 3(log6 - log2)
= 3log(6/2)
= 3log3


7.
設 y = logx

(logx)2 + logx2 = 3
(logx)2 + 2logx - 3 = 0
y2 + 2y - 3 = 0
(y - 1)(y + 3) = 0
y = 1 或 y = -3
logx = 1 或 log x = -3
x = (10)1 或 x = (10)-3
x = 10 或 x = 0.001


9.
y = logx

當 x = xo, y = yo:
yo = logxo

當 x = 10xo, y = y1:
y1 = log(10xo)
y1 = log10 + logxo
y1 = 1 + yo

y 的值會增加 1。

2009-07-18 19:38:17 補充：
3. 最尾一步錯了。答案應是 c。

(4√a^3) x (a^2)^3 / (√a)
= (a)^(3/4) x (a)^6 / (a)^(1/2)
= (a)^(3/4 + 6 - 1/2)
= (a)^(3/4 + 24/4 - 2/4)
= (a)^ 25/4

3. Ans. is c.

4√a^3 * (a^2)^3 除 √a
= a^(3/4) x a^(2x3) / a^(1/2)
= a^(3/4) x a^6 / a^(1/2)
= a^(3/4 + 6 – 1/2)
= a^(25/4)