difference between the math of RSA problem and DH problem
If you could take discrete logarithms modulo a composite number of unknown factorization, you could recover the decryption exponent.

Compute $c = m^e \bmod N$ for some $m &lt; N$. We know that $c^d \equiv m \bmod N$. 

You have $m, c, N$, and knowledge that $c^d \equiv m \bmod N$. Recovery of $d$ is the discrete logarithm problem.

The factorization of $N$ being unknown will probably work against you when attempting to compute the discrete logarithm. Depending on the hypothetical algorithm that you employ to compute discrete logarithms, knowledge of the factors of $N$ may be required, which implies that you still have to factor $N$ to compute the discrete log, which means breaking RSA would still be as hard as factoring.

Diffie-Hellman usually works with a prime modulus, so the factors of the modulus (which are just $1, P$) are public information.