:已知 3  a, 21  b,试用a.b 的代数式表示 0.28

lg5=lg(10/2)=lg10-lg2=1-a

log(2)3=lg3/lg2=alog(3)7=lg7/lg3=bab=lg7/lg2log(14)56=lg56/lg14=(lg7+lg8)/(lg2+lg7)=(3lg2+lg7)/(lg2+

log(14)56=[log3(56)]/[log3(14)]=[3log3(2)＋log3(7)]/[log3(2)＋log3(7)]=[(3/a)＋b]/[(1/a)＋b]=[ab＋3]/[ab＋

根据换底公式：log125=lg5/lg12lg5=lg(10/2)=lg10-lg2=1-alg12=lg(2²×3)=2lg2+lg3=2a+b原式=(1-a)/（2a+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)

lg18/25=lg18-lg25=lg2+lg3+lg3-lg5-lg5=a+2b-2lg5(没有lg5的c?)

即a=lg2b=lg3log12(5)=lg5/lg12=lg(10/2)/(lg4+lg3)=(lg10-lg2)/(2lg2+lg3)=(1-a)/(2a+b)

log23=lg3/lg2=a,所以lg3=alg2,同理：log37=lg7/lg3=b,所以lg7=blg3即lg7=ab*lg2log1456=lg56/lg14其中lg56=lg7+lg8=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

(1)log125=lg5/lg12=(1-lg2)/(2lg2+lg3)=(1-a)/(2a+b)(2)log1456=lg56/lg14=(3lg2+lg7)/(lg2+lg7)因为a=lg3/l

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

log12(5)=lg5/lg12=lg(10/2)/lg(3*2*2)=(1-lg2)/(lg3+2lg2)=(1-a)/(b+2a)

lg56/lg14＝lg(2*2*2*7)/lg(2*7)＝(3lg2+lg7)/(lg2+lg7)分子分母同除以lg3＝(3/a+b)/(1/a+b)接下来你可以化简一下＝(3+ab)/(1+ab)

解题思路：利用换底公式解题过程：

log(21)35=log(21)5+log(21)7=[1/(log(5)3+log(5)7)]+[1/(log(7)3+log(7)7)]因为log(9)5=a所以log(5)3=1/2alog(

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

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=log18(3)b=log18(5)log36(45)=log18(45)/log18(36)=log18(3X3X5)/log18(18X2)=[2log18(3)+log18(5)]/[1+l