化簡式子↓ (1)5x的負1次方*y，分之3a的負2次方 = ? 另外一題: 設67的x次方 = 27 ，603的y次方=81 ， 則x分之3 - y分之4 = ? (1)的答案是5a的平方*y分之3x 另外一提的答案是 -2 我需要祥解謝謝 高中 回應 jason12715 Re: 指數 2008-06-10 11:30 (1) 定義a^(-b) = 1/(a^b) 由此可得[3a^(-2)] / [5x^(-1)] = (3/5)*[a^(-2) / x^(-1)] = (3/5)*(x / a^2) = (3x) / (5a^2) (2) 67^x = 27 → x*log67 = 3*log3 → 3/x = log67 / log3.....(第一式) 603^y = 81 → y*log603 = 4*log3 → y*[(log67)+(2*log3)] = 4*log3 → 4/y = (log67 / log3) + 2.....(第二式) 將第一式和第二式代入 3/x - 4/y = (log67 / log3) - [(log67 / log3) + 2] = -2
jason12715 Re: 指數 2008-06-10 11:30 (1) 定義a^(-b) = 1/(a^b) 由此可得[3a^(-2)] / [5x^(-1)] = (3/5)*[a^(-2) / x^(-1)] = (3/5)*(x / a^2) = (3x) / (5a^2) (2) 67^x = 27 → x*log67 = 3*log3 → 3/x = log67 / log3.....(第一式) 603^y = 81 → y*log603 = 4*log3 → y*[(log67)+(2*log3)] = 4*log3 → 4/y = (log67 / log3) + 2.....(第二式) 將第一式和第二式代入 3/x - 4/y = (log67 / log3) - [(log67 / log3) + 2] = -2
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jason12715
Re: 指數
2008-06-10 11:30定義a^(-b) = 1/(a^b)
由此可得[3a^(-2)] / [5x^(-1)] = (3/5)*[a^(-2) / x^(-1)] = (3/5)*(x / a^2) = (3x) / (5a^2)
(2)
67^x = 27 → x*log67 = 3*log3 → 3/x = log67 / log3.....(第一式)
603^y = 81 → y*log603 = 4*log3
→ y*[(log67)+(2*log3)] = 4*log3
→ 4/y = (log67 / log3) + 2.....(第二式)
將第一式和第二式代入 3/x - 4/y = (log67 / log3) - [(log67 / log3) + 2] = -2