log以9为底72的对数是多少-搜答案-神马作文网

log2(25)*log3(1/16)*log5(1/9)=[2log2(5)]*[-4log3(2)]*[-2log5(3)=[2*lg5/lg2]*[-4*lg2/lg3]*[-2*lg3/lg5

[log9(4)+log3(8)]/log1/3(16)=[lg4/lg9+lg8/lg3]/[lg16/lg1/3]=[2lg2/2lg3+3lg2/lg3]/[4lg2/(-lg3)]=4lg2/

log1020=lg20=lg20/lg10=lg2再问：我要一个具体的数再答：就是等于lg2你需要具体的数字？那就是小数了

log以4为底8的对数-log以9分之1为底3的对数-log以根号2为底4的对数＝lg8/lg4-lg3/lg(1/9)-lg4/lg(√2)＝3lg2/2lg2-lg3/(-2)lg3-2lg2/(

用对数换底公式及商和幂的对数运算公式得：log3^(32/9)=lg(32/9)/lg3=(5lg2-2lg3)/lg3=（5*0.3010-2*0.4771）/0.4771=0.5508/0.477

用换底公式log(8)9/log(2)3=(lg9/lg8)/(lg3/lg2)=(2lg3/3lg2)/(lg3/lg2)=(2lg3*lg2)/(3lg2*lg3)=2/3

换底变成常用对视：lg2/lg3,中学生是要求背lg2和lg3的,自己计算一下就行了.

9乘以log以3为底5的对数=9×[(log5)÷(log3)]=9×[0.69897000433601880478626110527551÷0.477121254719662437295027903

log2(25)*log3(4)*log5(9)=lg25/lg2*lg4/lg3*lg9/lg5(换底公式)=lg5^2/lg2*lg2^2/lg3*lg3^2/lg5=2lg5/lg2*2lg2/

8再问：是不是换成分数形式可以互相约掉再答：log2(25)*log3(4)*log5(9)=lg25*lg4*lg9/lg2*lg3*lg4=log2(4)*log3(9)*log5(25)=2*2

log2(25)*log3(4)*log5(9)=lg25/lg2*lg4/lg3*lg9/lg5(换底公式)=lg5^2/lg2*lg2^2/lg3*lg3^2/lg5=2lg5/lg2*2lg2/

就是lg2,估计在0.30左右

log64(32)=log(2^6)(2^5)=(5/6)log2(2)=5/6再问：看不懂再答：根据公式：log(a^n)(b^m)=(m/n)loga(b)即以a^n为底，b^m的对数等于(m/n

楼上写错了[log(3,2)+log(9,2)]*[log(4,3)+log(8.3)]=[log(3,2)+1/2log(3,2)]*[1/2log(2,3)+1/3log(2,3)]=3/2log

9的多少次方等于27答案是3/2

换底公式：log以3为底12+log以9为底36-log以27为底512的对数=lg12/lg3+lg36/lg9-lg512/lg27=（2lg2+lg3)/lg3+(lg2+lg3)/lg3-3l

对数的加法就是真数的乘积,前提是底数相同,这里显然的,就是说5/4*4/5=1,既然真数是1,那值就是0咯.

解题思路：本题柱考察学生对于对数的运算的理解和应用。解题过程：

2log3(2)-log3(32/9)+log3(8)-5*2*log5(3)=log3(4)-log3(32/9)+log3(8)-5*2*log5(3)=log3(4/(32/9))+log3(8