若log2^3 log4^5 log5^6 =m

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

(log2(5)+log4(125))(log3(2)/log√3(5))=(log2(5)+3/2log2(5))(log3(2)/2log3(5))=5/2log2(5)*1/2log5(2)=5

(log2(5)+log4(125))(log3(2)/log√3(5))=(log2(5)+3/2log2(5))(2log3(2)/log3(5))=5/2log2(5)*2log5(2)=5lo

log4(5-x)=1/2log2(5-x)=log2(5-x)½=log2(x-2)因为log2x为单调函数,所以(5-x)½=(x-2)解得x=(3±√13)/2又因为x定义域

1、log2(4*5)-(2/2)log2(5)=log2(4)+log2(5)-log2(5)=2；2、换底公式：(lg3/lg2)*(lg5/lg2)*(lg5/lg3)=1；3、log2(5-l

log2(3)*log3(4)*log4(5)*log5(2)=log2(3)*log3(2^2)*log2^2(5)*log5(2)=log2(3)*2log3(2)*1/2*log2(5)*log

应运换底公式㏒aN=(㏒bN)/(㏒ba)将上述式子转换(Log43Log83)*(Log32Log92)=[log43(log43)/(log48)]×[log32(log32/log39)]=(l

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

log2(3)-log4(6)=log2(3)-log2^2((√6)^2)=log2(3)-log2(√6)=log2(3/√6)=log2(√3/√2)=log2(√3)-log2(√2)=1/2

原式=（log2(5)+(3/2)log2(5))×log3(2)/(2log3(5))=(5/2)log2(5)))×(1/2)log5(2)=5/4

log4(6)=1/2log2(6)=log2(√6)log4(6)

log43*log45=log4*43*5=log1615

log43=log2^23=1/2log23=p/2

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=

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(3-x)-log4(x+5)=1log2(3-x)=1+log4(x+5)log4(3-x)²=log44(x+5)(3-x)²=4(x+5)x²-6x+9=4

利用换底公式:log（a）（b）=log(n)(b)/log(n)（a）:log2(3)*log3(4)*log4(5)*.log(k+1)(k+2)=log2(3)*[log2(4)/log2(3)

我的对,看清楚log2^x>=log4^(3x+4)==>log2^x>=log2^(3x+4)/log2^4==>2log2^x>=log2^(3x+4)==>log2^(x^2)>=log2^(3

log76log47所以log23>log47>log76

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