由loga(b)=1/logb(a)
又loga(x)=2,logb(x)=3,logc(x)=6,
∴logx(a)=1/2,logx(b)=1/3,logx(c)=1/6
logabc(x)
=1/logx(abc)
=1/[logx(a)+logx(b)+logx(c)]
=1/(1/2+1/3+1/6)
=1
由loga(b)=1/logb(a)
又loga(x)=2,logb(x)=3,logc(x)=6,
∴logx(a)=1/2,logx(b)=1/3,logx(c)=1/6
logabc(x)
=1/logx(abc)
=1/[logx(a)+logx(b)+logx(c)]
=1/(1/2+1/3+1/6)
=1