指數函數 & 對數函數 MC (20點)

37.
原式
= {x*[x^(1/m)]}^(1/n)
= {x^[1 + (1/m)]}^(1/n)
= x^[(1/n) + (1/mn)]
= x^[(m + 1)/mn] ...... 答案選 C


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35.
令 a = log x < 0
則 x = 10^a< 1
而負數的對數無定義，所以 x > 0
因此 0 < x < 1 ...... 答案選 C


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39.
I. f(x)g(x) = (2^x)*(1/2)^(-x) = (2^x)*(2^x) = 2^(2x) ≠ 1
II. f(-x)g(-x) = 2^(-x)*(1/2)^(x) = (1/2)^x*(1/2)^x = (1/2)^(2x) ≠ 1
III. f(-x)g(x) = 2^(-x)*(1/2)^(-x) = [2*(1/2)]^(-n) = 1^(-n) = 1
答案選 B


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33.
a^x1 = b^x2
log a^x1 = log b^x2
x1 log a = x2 log b
x1 / x2 = log b / log a

由於 b < a﹐故此 log b < log a
log b / log a < 1
所以 x1/ x2 = log b / log a < 1
x1 < x2

但當 x1 = 0 及 x2 = 0， x1 = x2

綜合以上，x1­ ≤ x2。
因此，III. x1 > x2必定不正確 ...... 答案選 C


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40.
log2x
= log2(8y)²
= log2(8) + log2(y)²
= log2(2)³ + 2log2y
= 2log2y + 3log22
= 2log2y + 3 ...... 答案選 A


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2.
2^2n * 9^n / 3^n
= 2^2n * (3^2)^n / 3^n
= (2^2)^n * 3^2n / 3^n
= 4^n * 3^(2n - n)
= 4^n * 3^n
= (4 * 3)^n
= 12^n ...... 答案選 C

2011-06-18 23:43:55 補充：
37.
x^(1/n + 1/mn)
= x^(m/mn+1/mn)
= x^[(m+ 1)/mn]


35.
設 a = log x
根據 log 的定義 x = 10^a
因 log x < 0，而 a = log x，故 a < 0
當 a < 0 時，10^a < 1
因 x = 10^a，所以 x < 1 ... (1)

若 x ≤ 0，log x 無定義，所以 x > 0 ... (2)

根據 (1) 和 (2): 0 < x < 1

2011-06-18 23:44:20 補充：
33.
x1 log a = x2 log b
(x1 log a)/(x2 loga) = (x2 log b)/(x2 log a)
x1 / x2 = log b / log a

2011-06-20 13:16:50 補充：
有甚麼不明白呢？一大段只一句不明白，怎樣幫你？