高一数学典型例题分析 对数VIP免费

2．7对数·例题解析【例1】计算：(1)(lg27lg8lg1000)lg1.2＋－÷(2)lg22＋lg4·lg50＋lg250(3)log(4)lg(3+5+35)(5)272237271949552332··loglogloglog解(1)=原式×32332323410322231223132lglglg(lglg)lglg(2)原式=lg22＋2lg2·(1＋lg5)＋(1＋lg5)2=(lg2＋1＋lg5)2=4(3)=12原式··loglogloglog25527332323798(4)=lg(6+252+6252)=lg(5+5)=lg原式．12122521012lg(5)=2727=(3)(3)2=3=9223log33233log32log323原式····233=98【例2】(1)已知10x＝2，10y=3，求1002x-y的值．(2)已知log89=a，log25＝b，用a、b表示lg3．解(1)∵10x＝2∴lg2＝x，∵10y=3∴lg3=y则1002x-y=100=100=10=1692lg2lg3lg43lg169．(2)log9=23lg3lg2=alog5=b82∵·①∵∴lglglg52211bb②用心爱心专心1由①得，把②代入上式得·．已知＝，求证：．lg3=3a2lg2lg3=3a2r=96xyz11321236babxyz()【例3】证设8x＝9y＝6z＝k(k＞0，且k≠1)则x=log8k，y＝log9k，z＝log6k，∴＋＋证毕．2323689xykkkzloglog=2log83log9=6(log2log3)=6log6=6logkkkkk6用心爱心专心2