若log2=a,log3=b

log23=a;log37=b即lg3/lg2=a;lg7/lg3=b;所以lg7/lg2=a*b所以log72=1/a*b同理log142=1/1+ab;所以结果为（3+ab）/(1+ab)

运用换底:log14(56)=log3(56)/log3(14)=〔log3(7)＋log3(8)〕/〔log3(7)＋log3(2)〕log3(2)=1/log2(3)=1/alog3(8)=3lo

答：ab=log2^7ab+1=log2^7+1=log2^14ab+3=log2^7+3=log2^56log14^56=(ab+3)/(ab+1)

log14^56=log3^56/log3^14=(log3^7+3log3^2)/(log3^7+log3^2)=(b+3/a)/(b+1/a)=(3+ab)/(1+a)

若log23=a,log37=b则log23*log37=ab即log27=ablog1456=log214/(log256)=(log22*7)/(log27*8)=(log22+log27)/(l

从大到小：abcd因为a大于1,b介于0和1之间,c在负1和0之间,d在负1和负2之间

题目错了

log2(log0.5(log2a))=0log0.5(log2a)=1log2a=0.5a=平方根下2,log3(log1/3(log3b))=0log1/3(log3b)=1log3b=1/3b=

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

运用换底:log14(56)=log3(56)/log3(14)=〔log3(7)＋log3(8)〕/〔log3(7)＋log3(2)〕log3(2)=1/log2(3)=1/alog3(8)=3lo

log1256=lg56/lg12=(3lg2+lg7)/(lg3+lg4)因为a=lg3/lg2所以lg3=alg2因为b=lg7/lg3所以lg7=blg3=ablg2所以原式=(3lg2+abl

a=lg3/lg2b=lg7/lg3log4256=lg56/lg42=(lg7+3lg2)/(lg2+lg3+lg7)=(lg7/lg3+3lg2/lg3)/(lg2/lg3+1+lg7/lg3)=

log(2)3=a==>lg3/lg2=a==>lg3=alg2log(3)5=b==>lg5/lg3=b==>lg5=blg3=ablg2log(15)20=lg20/lg15=(lg2+lg2+l

已知log210=a,换底公式1/lg2=alg2=1/alog310=b1/lg3=blg3=1/blog3(4)=lg4/lg3=2lg2/lg3=2b/a请参考……

log2(3)=a,则log3(2)=1/a.则log36(45)=[log3(36)]/[log3(45)]=[log3(4)＋log3(9)]/[log3(9)＋log3(5)]=[2log3(2

a=lg3/lg2,lg2=lg3/ab=lg5/lg3,lg5=blg3log15(20)=lg20/lg15=lg(2²*5)/lg(3*5)=(2lg2+lg5)/(lg3+lg5)=

log3(2)=lg2/lg3=a/blog3(4)=2log3(2)=2a/

c=log42(56)=lg56/lg42=(3lg2+lg7)/(lg2+lg3+lg7)ab=lg7/lg2,lg7=ablg2.a=lg3/lg2,lg3=alg2.代入：c=(3lg2+abl

a=lg3/lg2lg2=lg3/ab=lg7/lg3lg7=blg3log1456=lg56/lg14=(lg7+lg8)/(lg7+lg2)=(lg7+3lg2)/(lg7+lg2)=(blg3+