又一道对数题

已知log5 35=m，试用m表示

log7 1.4

若log12 27=a，求证log6 16=4（3-a）/3+a

∵log5(35)=log5(5)+log5(7)=1+log5(7)

1+log5(7)=m

∴log5(7)=m-1

∴log7(1.4)=log5(1.4)/log5(7)=[log5(7)-log5(5)]/log5(7)=[log5(7)-1]/log5(7)=(m-2)/(m-1)

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∵log12(27)=log3(27)/log3(12)=3/[log3(3)+log3(4)]=3/(1+2log3(2)]

log12(27)=a

∴1+2log3(2)=3/a

∴log3(2)=(3-a)/(2a)

∵log6(16)=log3(16)/log3(6)=4log3(2)/[log3(2)+1]

4log3(2)=2(3-a)/a

log3(2)+1=(3+a)/(2a)

∴log6(16)=4(3-a)/(3+a)