log2=a,log3=b试用a2b表示

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)

倒数第二个式子分子分母同时提出来一个lg2然后约掉了啊

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

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

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=

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

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

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

log3(2)=A3^B=5log3(5)=Blog3(根号30)=log3[30^(1/2)]=1/2*log3(30)=1/2*log3(3*10)=1/2*[log3(3)+log3(10)]=

方法名称我忘了.就是利用l0gab=lgb/lga(化成底数为10的常用对数）这样就好做了.a=log95=lg5/lg9=lg5/lg(3^2)=(1/2)lg5/lg3所以lg5=2alg3b=l

哦.由已知可得：log8(3)=lg3/lg8=lg3/(3lg2)=a,log3(5)=lg5/lg3=b所以：[lg3/(3lg2)]×(lg5/lg3)=ab即lg5/(3lg2)=ablg5=

log23=a----->log32=1/alog37=blog3(56)=log3(7*2^3)=log3(7)+3log3(2)=b+3/a=(ab+3)/alog3(14)=log3(7*2)=