设2^6a=3^3b=6^2c=k则6a=log(2,k)=1/log(k,2).3b=log(3,k）=1/log(k,3) 2c=log(6,k)=1/log(k,6)=1/[log(k,2)+log(k,3)]=1/[1/6a+1/3b] 即1/2c=1/6a+1/3b得3ab-2ac-bc=0