對(duì)數(shù)換底公式證明及相關(guān)推論證明

1、對(duì)數(shù)換底公式對(duì)數(shù)換底公式)0; 1, 0; 1, 0(logloglogbccaaacbcba且且loglogloglogloglogloglog321MaaNaMaNMaNaMaMNanMn質(zhì)一、對(duì)數(shù)的三條運(yùn)算性復(fù)習(xí)：), 10(2)0, 10(log1logRxaaxaNaaNxNaaa且且二、兩個(gè)對(duì)數(shù)的恒等式問題的值。求已知loglglg3232,4771.0,3010.0loglglglglglglg2log3223323322103xxxxx由對(duì)數(shù)的運(yùn)算性質(zhì)有有為底的對(duì)數(shù)等式的兩邊同時(shí)取以由對(duì)數(shù)的定義有設(shè)解：解：?)10(loglogloglogloglglglnlnlglglog3

2、22323acbcbabaabbaccc即為底的對(duì)數(shù)且為以是否可以表示進(jìn)一步猜想們發(fā)現(xiàn)又通過計(jì)算器計(jì)算，我的形式？表示為都可以是不是任意的對(duì)數(shù)猜想：換底公式證明換底公式證明logloglogloglogloglogloglog,acbcbaacbcaccbcxbaxxacbxxa即從而得：為底的對(duì)數(shù)，等式兩邊取以則證法一：設(shè)證法二：證法二：成立故即則logloglogloglogloglogloglog,acbcbabamcacbcacbcmbammam成立即又則設(shè)loglogloglogloglogloglog,1,acbcbaacbcbaMNNNNMbcacMNMNbMbaNMaaccc證

3、法三：證法三：換底公式推論換底公式推論loglogloglog1NbMbNaMa、推論比與底數(shù)無關(guān)真數(shù)相同時(shí)，兩對(duì)數(shù)之所以證明：因?yàn)閘oglogloglogloglogloglogloglog;NbMbNaMaMNNbMbMNNaMa之比與真數(shù)無關(guān)）（底數(shù)相同時(shí)，兩對(duì)數(shù)變形二：互換）兩對(duì)數(shù)之積，真數(shù)可以變形一：loglogloglogloglogloglog(NbNaMbMaMbNaNbMa）和對(duì)數(shù)的真數(shù)可以互換數(shù)以對(duì)數(shù)為指數(shù)的冪，底由此有得變形三：由以上幾式可(loglogloglogloglogMNNaMaNaMaMNbbloglogloglogloglog1:) 1, 10, 012ababbbbaabbababa證明且（、推論1logloglogaccbba推廣1loglogbaab等價(jià)等于原對(duì)數(shù)值的倒數(shù)）得到的對(duì)數(shù)值（真數(shù)與底數(shù)互換，所 logloglogloglogloglogloglogloglog,),0(,:)0, 10, 0(3babaxnnxbxnnabbanababannnnnnnnnnbanxnnxbababaaabababaa即所以即所以則設(shè)證法二證法一：且、推論logloglogloglogloglogloglogloglogloglogloglogloglog1)0, 10,0(51)0, 1,0(14baaababaaabaNaaNaNNaNnm