log以4为底8的对数-搜答案-神马作文网

即lg4/lg3*lg8/lg4*lgm/lg8=lg16/lg4lgm/lg3=2lg4/lg4=2lgm=2lg3=lg3²所以m=9

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

为了书写方便,不妨记以a为底b的对数为：log【a】b(log【2】5+log【4】125)×[(log【3】2)/(log【√3】5)]=[(lg5)/(lg2)+(lg125)/(lg4)]×{[

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)*

9.

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/

换底公式原式=(lg3/lg4+lg3/lg8)×lg2/lg3=(lg3/2lg2+lg3/3lg2)×lg2/lg3=(1/2+1/3)×lg3/lg2×lg2/lg3=5/6

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

具体过程都在这儿了,自己看吧!

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

1/2log4(8)-log4(√2)=log4(√8)-log4(√2)=log4(√8/√2)=log4(√4)=1/2

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

根号3