Question

已知log12(7)=a,log12(3)=b,试用a,b表示log21(63)

Answer 1

log12(7)=a,log12(3)=b,

相除，得
lg7/lg3=a/b
log21(63)=lg63/lg21=
(lg21+lg3)/lg21=1+lg3/lg21=1+1/(lg21/lg3)
=1+1/(1+lg7/lg3)
=1+1/(1+a/b)

=1+b/(a+b)
=(a+2b)/(a+b)

Answer 2

log21(63)
=log12(63)/log12(21)
=log12(7*9)/log12(7*3)
=[log12(7)+log12(9)]/[log12(7)+log12(3)]
=[log12(7)+log12(3^2)]/[log12(7)+log12(3)]
=[log12(7)+2log12(3)]/[log12(7)+log12(3)]
=(a+2b)/(a+b)
=1+b/(a+b)

追问

log21(63)怎么到log12(63)/log12(21)

追答

用的是换底公式：loga(b)=logc(b)/logc(a)！