✔ 最佳答案
1.
log(x-2) (5-x) 有意義。
(x - 2) > 0 及 (5 - x) > 0
x > 2 及 x < 5
x 的範圍: 2 < x < 5
2.
(1)
log0.001 10
= log10 10 / log10 0.001
= 1 / log10 10^(-3)
= 1 / (-3 log10 10)
= -1/3
(2)
根據定義:
若 a = b^N,則 N = logb a
設 u = 7^log7 5
則 log75 = log7u
u = 5
所以 7^log7 5 = 5
3.
loga 2 + loga 3 = 5
(log10 2 / log10 a) + (log10 3 / log10a) = 5
(log10 2 + log10 3) / log10 a = 5
log10 6 = 5 log10 a
log10 6 = log10 a^5
a^5 = 6
a = 6^(1/5)
4.
請檢查題目,log^[1/log_5(7)] 沒義意。
5.(1)
(√3)^(x+2) = 1/3
[3^(1/2)]^(x+2) = 1/3
3^[(x+2)/2] = 3^(-1)
(x+2)/2 = -1
x + 2 = -2
x = -4
6.
設 u = 2^x
4^x - 5•2^x - 24 = 0
(2^2)^x - 5•2^x - 24 = 0
(2^x)^2 - 5•2^x - 24 = 0
u^2 - 5u - 24 = 0
(u - 8)(u + 3) = 0
u = 8 或 u = -3
2^x = 8 或 2^x = -3 (不合)
2^x = 2^3
x = 3
7.
設 u = 2^x
4^x - 5•2^x - 24 > 0
(2^2)^x - 5•2^x - 24 > 0
(2^x)^2 - 5•2^x - 24 > 0
u^2 - 5u - 24 > 0
(u + 3)(u - 8) > 0
u < -3 或 u > 8
2^x < -3 (不合) 或 2^x > 8
2^x > 2^3
x > 3
2013-12-28 23:53:29 補充:
4.
題目是 7^log_3(7) + 7^[1/log_5(7)] 嗎?
設 u = 7^log_3(7) 及 v = 7^[1/log_5(7)]
log_u(7) = log_3(7) 及 log_v(7) = log_5(7)
u = 3 及 v = 5
7^log_3(7) + 7^[1/log_5(7)]
= u + v
= 3 + 5
= 8 ...... (答案)