(1)设{an}的公差为d,bn=e^an,则b(n+1)/bn=e^a(n+1)/e^an=e^(a(n+1)-an)=e^d,为不为零的常数,故{bn}是等比数列.(2)设{an}的公比为q,因为an>0,故q>0,设bn=loge(an),则b(n+1)-bn=loge(a(n+1))-loge(an)=loge(a(n+1)/an)...