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已知log12(7)=a,log12(3)=b,试用a,b表示log28(63)
人气:434 ℃ 时间:2020-06-24 01:32:16
解答
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)
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