解析：∵{an}是等比数列,令首项=a1,公比=q∴an=a1*q^(n-1),an＞0,n∈N+∵log[an]-log[a(n-1)]=log[a1*q^(n-1)]-log[a1*q^(n-2)]=log{[a1*q^(n-1)]/[a1*q^(n-2)]}=logq=定值∴{logan}是首项loga1,公差=logq的等差数列...