(1)log12 5
=lg5/lg12
=(1-lg2)/(2lg2+lg3)
=(1-a)/(2a+b)
(2)log14 56
=lg56/lg14
=(3lg2+lg7)/(lg2+lg7)
因为a=lg3/lg2 所以lg3=alg2
因为b=lg7/lg3 所以lg7=blg3=ablg2
所以原式=(3lg2+ablg2)/(lg2+ablg2)
=1+2/(1+ab)
换底:log12 5 =lg5/lg12=lg5/(2lg2+lg3)=lg5/(2a+b)
log14 56
=lg56/lg14=(2lg2+lg7)/(lg2+lg7)
log2 3=a,log3 7=b,log2 3*log3 7=lg3/lg2*(lg7/lg3)=lg7/lg2=ab,
lg7=ablg2
所以log14 56=(2lg2+lg7)/(lg2+lg7)=(2+ab)/(1+ab)