log以3为底的4乘以log以4为底的8乘以log以8为底的M

log2（log3(log4x)=0（log3(log4x)=2^0=1log4x=3^1=3x=4^3=64log3（log4(log2y)=0log4(log2y)=3^0=1log2y=4^1=

D首先化减:log8(M)*log3(4)=log4(16)=2log8(M)=2/log3(4)=log3(9)/log3(4)运用换底公式log3(9)/log3(4)=log4(9)所以log8

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

是64log以2为底3的对数乘以log以81为底64的对数=lg3/lg2*(lg64/lg81)=lg3/lg2*(6/4lg2/lg3)=3/2

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/(

=2log23×3log32=6log23×log32=6

log以4为底(x+12)的对数乘以log以x为底2的对数=【ln（x+12）/ln4】×【ln2/lnx】=ln（x+12）/lnx=ln（x+12-x）其中x+12-x＞0,即-3＜x＜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

解；可以利用换底公式,变为相同的底数来计算原式=(log以2为底3为真数÷log以2为底4为真数)×(log以2为底2为真数÷log以2为底9为真数)×(log以2为底4√32为真数÷log以2为底1

log26*log36-(log23+log32)=log26*log36-log23-log32=log2(3x2)*log3(3x2)-log23-log32=(log23+log22)(log3

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

根据换底公式和对数运算法则,通式为:log(2^n)3^n=n/n*log₂3=log₂3∴[log₂3+log(4)9+log(8)27+……+log(2^n)3

log以2为底25乘以log以3为底2√2乘以log以5为底9=2log25×(3/2)log32×2log53=6log25×log32×(log23/log25)=6(log25×1/log52)

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

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：

首先必须满足3^x-1>0原式为：(lg(3^x-1)/lg4)*(lg(3^x-1)-lg16)/lg0.25=0△=4(lg4)^2-3(lg4)^2恒大于零解为lg(3^x-1)>=1.5lg4

log_3[log_4(log_5(a))]=0(1)log_4[log_3(log_5(b))]=0(2)(1)=>log_4(log_5(a))=1=>log_5(a)=4=>a=5^4(2)=>