(1+logn^m)logmn^x=logn^x

Assuming that log (mn) = log m + logn prove that log x^(n) = n log x, n in N

Given that lim_(n to oo)sum_(r=1)^(n)(log(n^(2)+r^(2))-2logn)/(n)=log2+(pi)/(2)-2 , then lim_(n to oo) (1)/(n^(2m))[(n^(2)+1^(2))^(m)(n^(2)+r^(2))^(m)......(n^(2))^(m)]^(1//n) is equal to

Prove that: log_a x=log_bx xx log_c b xx…xx log_n m xx log_a n

lim_(n->oo)[log_(n-1)(n)log_n(n+1)*log_(n+1)(n+2).....log_(n^k-1) (n^k)] is equal to :

If n=1999! then sum_(x=1)^(1999) log_n x=

Prove that log_n(n+1)>log_(n+1)(n+2) for any natural number n > 1.

Let g(x)=((x-1)^(n))/(logcos^(m)(x-1)),0ltxlt2 m and n integers, m ne0, n gt0 and. If lim_(xrarr1+) g(x)=-1 , then

If m =! n and (m + n)^(-1) (m^(-1) + n^(-1)) = m^(x) n^(y) , show that : x + y + 2 = 0.

Consider the equation log _(2)^(2) x -4 log _(2)x-m^(2) -2m -13=0, m in R. Let the real roots of the equation be x _(1), x _(2) such that x _(1)lt x _(2). The sum of maximum value of, x _(1) and minimum value of x _(2) is :

int(nx^(n-1)+m^(x)logm)/(x^(n)+m^(x))dx equals