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指數函數 & 對數函數 MC (20點)
圖片參考：http://imgcld.yimg.com/8/n/HA00230702/o/701106180079413873460705.jpg 教我做...唔該哂 33,35題請解釋 答案 : 37. C 35. C 39.B 33. C 40. A 2. C 更新: 37. 我唔明點解 x^(1/n + 1/mn) = x^(m+1/mn) 35題唔明你嘅解釋 33. a^x1 = b^x2 log a^x1 = log b^x2 x1 log a = x2 log b x1 / x2 = log b / log a <--- 我唔明點解係log b / log a, 點解唔係x1 / x2 = log a / log b 更新 2: 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 <---呢度開始唔明解釋 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
最佳解答:
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 ===== 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 ===== 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 ===== 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 ===== 40. log2x = log2(8y)2 = log2(8) + log2(y)2 = log2(2)3 + 2log2y = 2log2y + 3log22 = 2log2y + 3 ...... 答案選 A ===== 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？
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指數函數 & 對數函數 MC (20點)
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發問:圖片參考：http://imgcld.yimg.com/8/n/HA00230702/o/701106180079413873460705.jpg 教我做...唔該哂 33,35題請解釋 答案 : 37. C 35. C 39.B 33. C 40. A 2. C 更新: 37. 我唔明點解 x^(1/n + 1/mn) = x^(m+1/mn) 35題唔明你嘅解釋 33. a^x1 = b^x2 log a^x1 = log b^x2 x1 log a = x2 log b x1 / x2 = log b / log a <--- 我唔明點解係log b / log a, 點解唔係x1 / x2 = log a / log b 更新 2: 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 <---呢度開始唔明解釋 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
最佳解答:
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 ===== 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 ===== 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 ===== 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 ===== 40. log2x = log2(8y)2 = log2(8) + log2(y)2 = log2(2)3 + 2log2y = 2log2y + 3log22 = 2log2y + 3 ...... 答案選 A ===== 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？
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