How to find the number of positive integral divisors of a power of a number?

Question

The number 80 has ten positive integral divisors: 1, 2, 4, 5, 8, 10, 16, 20, 40, and 80. Let's say you raise 80 to the x power (80^x). How many positive integral divisors does 80^x have? Is there a formula to calculate the number of positive integral divisors if we changed 80 to another positive integer?

Answer 1

To compute the number of divisors of a positive integer $n$ you have to factorize $n$.

If $n=p_1^{k_1}\cdots p_r^{k_r}$, then number of divisors of $n$ is $(k_1+1)\cdots(k_r+1)$.

The factorization of $n^m$ is $p_1^{k_1m}\cdots p_r^{k_rm}$. Then, the number of divisors of $n^m$ is $(k_1m+1)\cdots(k_rm+1)$. I don't think that there is a simpler relation between those numbers. Perhaps if $n$ is a power of a prime number you can find some formula involving logarithms.