若log2[log3(log4x)]=0,log3[log4(log2y]=0

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

(log25+log4125)log32/log根号35=(log25+log4125)log32/log325=(log25+log4125)log252=(log25+log25根号5)log25

log3(log4x)=2^0log3(log4x)=1log4x=3x=4³x=64

原式=lg25/lg4-2lg10/lg4+lg9/lg2×lg√5/lg3×lg2/lg5=(2lg5)/(2lg2)-2/(2lg2)+(2lg3)/lg2×(½lg5)/lg3×lg2

alog23＜b=log32＜c=log46再问：为什么c＞b再答：哦，开始以为同底呢，应该这样：b=log32＜1a=log23＞1c=log46＞1a=log23=lg3/lg2c=log46=l

因为log2[log3(log4x)]=0所以：log3(log4x)=1进一步：log4x=3所以x=4^3=64,.log3[log4(log2y)]=0则有：log4(log2y)=1进一步：l

log4log3log2（x）=0=log4（1）所以log3log2（x）=1所以log3log2（x）=log3（3）所以log2（x）=3所以x=2^3=8

log2^3×log3^4×log4^x=log9^3(lg3/lg2)×(lg4/lg3)×(lgx/lg4)=log9^(9^1/2)所以lgx/lg2=1/2lgx=1/2*lg2=lg√2x=

log4(9)=2*1/2log2(3)=log2(3)log3(4)=2log3(2)log9(2)=1/2log3(2)[log2(3)+log4(9)][log3(4)+log9(2)]=2lo

∵√log2512=9log3√9=log33=1√log4√1=√log41=0∴原式=0

换底公式原式=(lg3/lg4+lg3/lg8)(lg2/lg3+lg4/lg3)-log2(2^6)=(lg3/2lg2+lg3/3lg2)(lg2/lg3+2lg2/lg3)-log2(2^6)=

1、log23*log34*log45*log52=lg2/lg3xlg3/lg4xlg4/lg5xlg5/lg2（换底公式）=1（约分）2.(log43+log83)*(log32+log92)中间

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

从外到里计算log4底[log3底(log2底x)]=0log3底(log2底x)]=1log2底x)]=3x=8

换底公式全部换位以10为底的

利用换底公式,将上式换成以10为底的常用对数,即：[(lg7)/(lg3)]·[(lg9)/(lg2)]·[(lga)/(lg49)]=[lg(1/2)]/(lg4)公式loga(m)^n=n·log

log2为底0.4,log3为底0.4,log4为底0.4分别等于lg0.4/lg2,lg0.4/lg3,lg0.4/lg4,因为lg0.4小于0,lg2小于lg3小于lg4,lg2大于0,所以log

3再问：怎么算的呢······重点是2log3(2)×log4(3)，前面我会算再答：运用换底公式即可

/>希望能够帮到你好像弄错了,log2^(log3^x)中2是底数吧?如果是这样应该这么算：log2^(log3^x)=0,则log3^x=1,x=3log3^(log4^y)=0,则log4^y=1