已知ab=M(a>0,b>0,M≠1),且logM(b)=2,則logM(a)=

已知ab=M(a>0,b>0,M≠1),且logM(b)=2,則logM(a)=

logM(b)+logM(a)
=logM(ab)
=logM(M)
=1
所以logM(a)
=1-logM(b)
=-1