证明：
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=06a=log(2,k)=1/log(k,2).不懂请解释?log(a,b)*log(b,a)=1，这个显然。那么取个倒数就行啦！log(a,b) 和llog(b,a)不等于啊6a=log(2,k)=1/log(k,2). 怎么才能得到这个 ？我比较笨求详解log(a,b) 和log(b,a)乘积是1.log(a,b) =1/log(b,a)