求[log(2)9+log(4)9+log(8)27+……+log(2^n)3^n]*log(9)n次根号32(n∈N*

计算(log以2为底的3+log以4为底的9+log以8为底的7+……+log以2^n为底的3^n)*log以9为底^n (log2 3+log4 9.+log8 27+.+log(2^n底)(3^n))log9 n次根号32 log(n+2）n+1和log(n+1)n (n大于1),比较大小 log(2)(25)*log(3)(根号2)*log(5)(9) a^[log(a)b×log(b)c×log(c)N] 用换底公式简化[log(4)3+log(8)3][log(3)2+log(9)2] (log(4)3+log(8)3)(log(3)2+log(9)2) 怎么个思路 (log(4)3+log(8)3)(log(3)2+log(9)8) 怎么个思路 求log（2）(3+n^2)-2=log(2)3解题过程. 设n∈N,n>1.求证：logn (n+1)>log(n+1) (n+2) log(2)9×log(3)4 log(2)9log(3)4等于