證明：
設(shè)2^6a=3^3b=6^2c=k
則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).不懂請解釋?log(a,b)*log(b,a)=1，這個顯然。那么取個倒數(shù)就行啦！log(a,b) 和llog(b,a)不等于啊6a=log(2,k)=1/log(k,2). 怎么才能得到這個 ？我比較笨求詳解log(a,b) 和log(b,a)乘積是1.log(a,b) =1/log(b,a)