已知log12（7）=a,log12（3）=b,试用a,b表示log28（63）
log12(7)=a,
lg7/lg12=a
lg7=alg12
log12(3)=b
lg3/lg12=b
lg3=blg12
lg3=b(lg3+lg4)
lg3=blg3+blg4
(1-b)lg3=blg4
lg4=(1-b)lg3/b
lg4=(1-b)blg12/b
lg4=(1-b)lg12
log28(63)
=lg63/lg28
=(lg7+lg9)/(lg7+lg4)
=(alg12+2blg12)/(alg12+(1-b)lg12)
=(a+2b)/(a+1-b)