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高中數學*對數函數*

這是從http://www.csu.edu.tw/csitshow/Csitcomd/0166/抓下來的但是沒有解答而且有不會的 希望大大能幫我回答 謝謝^^作圖題3不用
1.(1)log(2) 64－log(1/3) 81＝log(2) 2^6－log[3^(-1)] 3^4＝log(2) 2^6＋log(3) 3^4＝6×log(2) 2＋4×log(3) 3＝6＋4＝10(2) log(1/2) (32√2)＋100^[log(10) √10]＝log[2^(-1)] [2^(5＋1/2)]＋100^[log(10) 10^(1/2)]＝-(5＋1/2)×log(2) 2＋100^(1/2)＝-(5＋1/2)＋10＝4＋1/2(3) log(10) 8＋log(10) 50＋log(10) 25＝log(10) (8×50×25)＝log(10) (10^4)＝4×log(10) 10＝4(4) log(3) 25．log(5) 49．log(7) 9＝log(3) 5^2．log(5) 7^2．log(7) 3^2＝2×log(3) 5．2×log(5) 7．2×log(7) 3＝8×[log(3) 5．log(5) 7．log(7) 3]＝8×1＝82.(1) log(5) x＝-1/2x＝5^(-1/2)＝(1/5)^(1/2)＝√(1/5)(2) x^[log(2) 1/8]＝125x^[log(2) 2^(-3)]＝5^3x^[(-3)×log(2) 2]＝5^3x^(-3)＝5^3x^(-1)＝5^1x＝1/53. 跳過4. log(15) 72　　換個底＝log(10) 72 / log(10) 15＝log(10) (2^3×3^2) / log(10) (3×10/2)＝[log(10) (2^3)＋log(10) (3^2)] / [log(10) 3＋log(10) 10－log(10) 2]＝[3×log(10) 2＋2×log(10) 3] / (0.4771＋1－0.3010)＝(3×0.3010＋2×0.4771) / (0.4771＋1－0.3010)＝1.5795. log(231) 99＝log(3×7×11) (3×3×11)　換個底＝log(3) (3^2×11) / log(3) (3×7×11)＝[log(3) (3^2)＋log(3) 11] / [log(3) 3＋log(3) 7＋log(3) 11]＝[2＋log(3) 11] / [1＋log(3) 7＋log(3) 11]＝[2＋log(3) 7×log(7) 11] / [1＋log(3) 7＋log(3) 7×log(7) 11]＝(2＋ab)/(1＋a＋ab)6.(1) [log(5) x]^2＋4 log(5) x－5＝0　（A^2 4A-5=0）[log(5) x＋5][log(5) x－1]＝0log(5) x＋5＝0　-