log以3为低9的对数 lg1 1000

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

=lg2*lg25=lg50=1+lg5再问：那，5倍log以10为低2的对数，加2倍的log以10为低2的对数，等于什么？再答：=5lg2+2lg2=7lg2=lg128这种化简没什么意义一开始就已

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

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/(

=(lg3/lg4+;g3/lg8)(lg2/lg3+lg2/lg9)=(lg3/2lg2+;g3/3lg2)(lg2/lg3+lg2/2lg3)=(1/2+1/3)*lg3/lg2*(1+1/2)*

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

换底公式我喜欢ln所以log2(9)=ln9/ln2=ln(3^2)/ln2=2*ln3/ln2log3(4)=ln4/ln3=ln(2^2)/ln3=2*ln2/ln3log2(9)*log3(4)

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/

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

换底公式原式=(lg3/lg4+lg3/lg8)(lg2/lg3+lg2/lg9)=(lg3/2lg2+lg3/3lg2)(lg2/lg3+lg2/2lg3)=(lg3/lg2)(1/2+1/3)*(

log以3为底的4的对数/log以9为底的8的对数=log以3为底的2^2的对数/log以3^2为底的2^3的对数=log3(2^2)/log3^2(2^3)=4/3再问：最后一步在详细点再答：log

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

=log3（1/27)=log3（1）-log3（27）=0-log3（27）=-log3（27）=-3

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

利用：log(a^n)[b^m]=(m/n)log(a)[b]则：log(3)[2]=log(3²)[2²]=log(9)[4]