We know that `(a^m)^n = a^(mn)` Let `a^m = x`, then `m = log_ax x^n = a^(mn)`, then `log_ax^n= mn = n log_ax` (why?)

Using log_ax^n = n log_a x , expand the following : log_2^7^25 .

log ( x )^(n) = n .log x is true for n .

If I_(m,n)=int x^(m)(logx)^(n)dx then I_(m.n)=

If I_(n)= int (log x)^(n)dx then I_(n)+nI_(n-1)=

If I_(m.n) = int sin^(m) x cos^(n) xdx then I_(5,4) =

Obtain reduction formula for If I_(n)=int (log x) ^(n)dx, then show that I_(n) =x (log x)^(n) -nI_(n-1), and hence field int (log x) ^(4) dx.

If y = 2 ^(ax ) and (dy)/(dx) =log 256 at x =1, then a =