Let $n$ be the product of all positive factors of $10^4$. Find $log_{10} n$ I can't see any relationship between the products of the positive factors of powers of 10.
The answer given here is 50, by the way, but I simply don't know where to start.
 A: As @NL1992 pointed out, the key is write $10^4 = 2^4 \times 5^4$. 
Consider the factors which have no contribution from the "2". These are $5^0, 5^1, 5^2, 5^3$, and $5^4$. Their product is $5^{10}$.
Now, consider the factors which exactly one "2". These are $2 \times 5^0$, $2 \times 5^1$, $2 \times 5^2$, $2 \times 5^3$, and $2 \times 5^4$. Their product is $2^5 \times 5^{10}$.
Generalizing, you can see that if you multiply all factors which have a $2^k$ term, you get $2^{5k} \times 5^{10}$. 
So the final answer is $n = \prod_{k=0}^4 2^{5k}5^{10} = 2^{50} 5^{50}$, implying $\log_{10}(n) = 50$.
A: Hint:
We can see, either through investigation or just trying a few examples if you want, that, if a number $n$ has $d(n)$ divisors (sometimes denoted $\sigma_0(n)$ instead), then we can give the product of $n$'s divisors by
$$\prod_{d|n} d = n^{d(n) / 2}$$
This is discussed in more length at Product of Divisors of some $n$ proof
Personally I've seen this occasionally denoted $\pi(n)$ and used such a notation myself when I was looking into the product of a number's divisors. Either way, not important. But with that in mind, you are seeking
$$\log_{10} \left( \pi(10^4) \right) \;\;\; \text{or equivalently} \;\;\; \log_{10} \left( \prod_{d|10^4} d \right)$$
With the previous fact in mind, this problem is essentially reduced to simply finding the number of divisors of $10^4$ after using a few basic properties of logarithms and exponents. 
You should be able to easily continue from here.
