北师大版 换底公式
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2.2 换底公式必备知识基础练知识点一 利用换底公式求值1.若log a x =2,log b x =3,log c x =6,则log abc x =( )A .1B .2C .3D .52.若log 34·log 48·log 8m =log 416,则m =________.3.设3x =4y =36,求2x +1y的值.知识点二 利用换底公式计算4.(log 134)·(log 227)=( )A .23B .32C .6D .-6 5.计算:(1)log 927;(2)log 21125 ×log 3132 ×log 513; (3)(log 43+log 83)(log 32+log 92).知识点三 利用换底公式证明6.证明:log a a b m =m n log a b (a >0,且a ≠1,n ≠0).7.已知2x =3y =6z ≠1,求证:1x +1y =1z.关键能力综合练1.log 29log 23=( )A .12B .2C .32D .922.已知log 23=a ,log 37=b ,则log 27=( )A .a +bB .a -bC .abD .a b3.设2a =5b =m ,且1a +1b =2,则m =( )A .10B .10C .20D .1004.1log 1419 +1log 1513=( )A .lg 3B .-lg 3C .1lg 3D .-1lg 35.(多选题)已知2x =3y =a ,且(x -1)(y -1)=1,则a 的值可能为() A .1 B .2 C .3 D .66.(探究题)设a ,b ,c 都是正数,且4a =6b =9c ,那么( )A .ab +bc =2acB .ab +bc =acC .2c =2a +1bD .1c =2b -1a7.已知log 32=m ,则log 3218=________.(用m 表示)8.(易错题)计算:(log 2125+log 425+log 85)(log 52+log 254+log 1258).9.计算:5log 53-log 311·log 1127+log 82+log 48.核心素养升级练1.(多选题)已知正数x ,y ,z 满足等式2x =3y =6z ,下列说法正确的是( )A .x >y >zB .3x =2yC .1x +1y -1z =0D .1x -1y +1z=0 2.(学科素养—逻辑推理)已知a ,b ,c 是不等于1的正数,且a x =b y =c z ,1x +1y +1z=0,求abc 的值.2.2 换底公式必备知识基础练1.答案:A解析:∵log a x =1log x a =2,∴log x a =12. 同理log x c =16 ,log x b =13.∴log abc x =1log x abc =1log x a +log x b +log x c=1. 2.答案:9解析:由换底公式,得lg 4lg 3 ×lg 8lg 4 ×lg m lg 8 =lg m lg 3=log 416=2,∴lg m =2lg 3=lg 9,∴m =9.3.解析:∵3x =36,4y=36,∴x =log 336,y =log 436,由换底公式,得 x =log 3636log 363 =1log 363 ,y =log 3636log 364 =1log 364, ∴1x=log 363,1y =log 364, ∴2x +1y=2log 363+log 364=log 36(32×4) =log 3636=1.4.答案:D解析:(log 13 4)·(log 227)=(log 13 22)·(log 2(13 )-3)=(2log 132)·(-3log 213 )=-6·lg 2lg 13·lg 13lg 2 =-6. 5.解析:(1)log 927=log 327log 39 =log 333log 332 =3log 332log 33 =32. (2)log 21125 ×log 3132 ×log 513=log 25-3×log 32-5×log 53-1=-3log 25×(-5log 32)×(-log 53)=-15×lg 5lg 2 ×lg 2lg 3 ×lg 3lg 5=-15. (3)原式=(lg 3lg 4 +lg 3lg 8 )(lg 2lg 3 +lg 2lg 9) =(lg 32lg 2 +lg 33lg 2 )(lg 2lg 3 +lg 22lg 3) =12 +14 +13 +16 =54. 6.证明: log a a b m =lg b m lg a n =m lg b n lg a =m n log a b .7.证明:设2x =3y =6z =k (k ≠1),∴x =log 2k ,y =log 3k ,z =log 6k ,∴1x=log k 2,1y =log k 3,1z =log k 6=log k 2+log k 3, ∴1z =1x +1y. 关键能力综合练1.答案:B解析:由换底公式得log 39=log 29log 23 ,又∵log 39=2,∴log 29log 23 =2. 2.答案:C解析:log 27=log 23×log 37=ab .3.答案:A解析:∵2a =5b =m ,∴a =log 2m ,b =log 5m .1a +1b=log m 2+log m 5=log m 10=2,∴m 2=10.又m >0,∴m =10 ,选A.4.答案:C解析:原式=log 19 14 +log 13 15 =log 13 12 +log 13 15 =log 13110 =log 310=1lg 3 .选C. 5.答案:AD解析:由(x -1)(y -1)=1,可得xy =x +y .当xy =0时,x =y =0,此时a =1满足;当xy ≠0时,由1x +1y=1. 又2x =3y =a ,所以x =log 2a ,y =log 3a ,则1x =1log 2a =log a 2,1y =1log 3a=log a 3. 所以有1x +1y=log a 2+log a 3=log a 6=1,解得a =6. 综上所述,a =1或a =6.故选AD.6.答案:AD解析:由a ,b ,c 都是正数,可设4a =6b =9c =M ,∴a =log 4M ,b =log 6M ,c =log 9M ,则1a =log M 4,1b =log M 6,1c=log M 9,∵log M 4+log M 9=2log M 6,∴1c +1a =2b ,即1c =2b -1a,去分母整理得ab +bc =2ac .故选AD. 7.答案:m +25m解析:log 23=1log 32 =1m ,log 3218=lg 18lg 32 =lg 2+2lg 35lg 2 =15 +25 log 23=15 +25m=m +25m. 8.解析:解法一:原式=(log 253+log 225log 24 +log 25log 28 )(log 52+log 54log 525 +log 58log 5125)=(3log 25+2log 252log 22 +log 253log 22 )(log 52+2log 522log 55 +3log 523log 55 )=(3+1+13)log 25·(3log 52)=13log 25·log 22log 25=13. 解法二:原式=(lg 125lg 2 +lg 25lg 4 +lg 5lg 8 )(lg 2lg 5 +lg 4lg 25 +lg 8lg 125 )=(3lg 5lg 2 +2lg 52lg 2 +lg 53lg 2 )(lg 2lg 5 +2lg 22lg 5 +3lg 23lg 5 )=(13lg 53lg 2 )·(3lg 2lg 5)=13. 解法三:原式=(log 2 53+log 2252+log 235)(log 52+log 5222+log 5323)=(3log 2 5+log 2 5+13 log 2 5)(log 5 2+log 5 2+log 5 2)=(3+1+13 )log 2 5·3log 5 2=3×133=13. 9.解析:原式=3-log 311×3log 113+13 log 22+32log 22 =3-3+13 +32 =116 . 核心素养升级练1.答案:AC解析:设2x =3y =6z=k (k >1),则x =log 2k ,y =log 3k ,z =log 6k .因为x =log 2k =1log k 2 ,y =log 3k =1log k 3 ,z =log 6k =1log k 6 ,且0<log k 2<log k 3<log k 6, 所以1log k 2 >1log k 3 >1log k 6,即x >y >z ,故A 正确; 3x =3ln k ln 2 ,2y =2ln k ln 3 ,则3x 2y =3ln 32ln 2>1,故B 错误; 1x +1y =log k 2+log k 3=log k 6=1z,故C 正确;1x -1y +1z=log k 2-log k 3+log k 6=log k 4≠0,故D 错误.故选AC. 2.解析:解法一:设a x =b y =c z =t ,则x =log a t ,y =log b t ,z =log c t , ∴1x +1y +1z =1log a t +1log b t +1log c t=log t a +log t b +log t c =log t (abc )=0, ∴abc =t 0=1,即abc =1.解法二:设a x =b y =c z =t ,∵a ,b ,c 是不等于1的正数,∴t >0且t ≠1,∴x =lg t lg a ,y =lg t lg b ,z =lg t lg c, ∴1x +1y +1z =lg a lg t +lg b lg t +lg c lg t =lg a +lg b +lg c lg t, ∵1x +1y +1z=0,且lg t ≠0, ∴lg a +lg b +lg c =lg (abc )=0,∴abc =1.。
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对数换底公式 一、换底公式:)0,1,0,1,0(log log log >≠>≠>=b c c a a a b b c c a 二、常用关系:1、自然对数与常用对数之间关系:e N N ln ln lg =2、)0,1,0(lg lg log >≠>=b a a ab b a 3、)1,0,1,0(log 1log ≠>≠>=b b a a ab b a 4、 )0,0,1,0(log log ≠>≠>=m b a a b b m a a m5、)1,0,1,0(log log ≠>≠>=n b a a b n m ba m a n 三、例题:例1、求证:1log log =⋅a b b a例2、求下列各式的值。
(1)、log 98•log 3227(2)、(log 43+log 83)•(log 32+log 92)(3)、log 49•log 32(4)、log 48•log 39(5)、(log 2125+log 425+log 85)•(log 52+log 254+log 1258)例3、若log 1227=a,试用a 表示log 616.解:法一、换成以2为底的对数。
法二、换成以3为底的对数。
法三、换成以10为底的对数。
练习:已知log 189=a, 18b =5,求log 3645。
例4、已知12x =3,12y =2,求y x x+--1218的值。
练习:已知7log log ,5log log 248248=+=+a b b a ,求a •b 的值;例5、有一片树林,现有木材22000方,如果每年比上一年增长2.5%,求15年后约有多少方木材?解:设15年后约有木材A 方,则A=22000(1+2.5%)15=22000×1.02515lgA=lg22000+15×lg1.025=4.3424+15×0.0107=4.5029∴ A=131840教学无忧/专注中小学 教学事业! 客服唯一联系qq 1119139686 欢迎跟我们联系答:15年后约有木材131840方。