已知log2=a,log3=b,试用a,b 表示log12

运用换底log14(56)=log3(56)/log3(14)=log3(7*8)/log3(2*7)=〔log3(7)＋log3(8)〕/〔log3(7)＋log3(2)〕log3(2)=1/log

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)

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

lg12=2lg2+lg3=2a+blg12^5=(2a+b)5

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

题目错了

log5=log2*log3=a

由lg2=a,lg3=b得log2（3）=b/a,log3(2)=a/b,所以log3（4）=2log3（2）=2a/blog2(12)=log2(4X3)=2+log2(3)=2+b/a再问：不太懂

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)=

等于2/a利用换底公式,log3^2=log2^2/log2^3=a,即log2^3=1/a;log2^9=log2^(3^2)=2log2^3=2*(1/a)=2/a

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

a=lg3/lg2lg3=alg2b=lg7/lg3lg7=blg3=ablg2log42(56)=lg56/lg42=lg(2^3*7)/lg(2*3*7)=(3lg2+lg7)/(lg2+lg3+