difference between the math of RSA problem and DH problem I'm trying to understand why the RSA problem is not related to the Discrete Log problem, and how breaking the Discrete Log problem does not directly compromise RSA.

I will explain my reasoning of why it seems to me that they are the same in hope that someone will find the flaw in my reasoning and correct me.

If you don't take into account how you generate the public and private keys and just assume they exist, we could minimize the RSA problem to just modular exponentiation in this way:


$M$: Plain text
$C$: cipher text
$E$: some number
$D$: some other number
This is easy (encrypt): $M^E \bmod N = {?}$
This is hard (decrypt): ${?}^E \bmod N = C$


Now, according to this khan academy video, ${?}^E \bmod N = C$ is equivalent to $C^{D} \bmod N = {?}$ for some unknown $D$, this will act as our private key.

Now my question is: If I can identify if I have decrypted $M$, by some file header for example. isn't finding $D$ for $C^{D} \bmod N = M$ the Discrete Log problem? Given this khan academy video about the discrete log problem, it would appear it is.
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.