Earlier quoted context omitted.
I don't think so. Computing the discrete log of g^x assumes g to be known. In RSA the ciphertext is c = m^e and by definition, m is unknown. The reverse operation of RSA is the modular e-th root, not the discrete log.
Pick a base k. log_k(c) = e*log_k(m) log_k(c)*e^-1 = log_k(m) Raising k by both sides yields k^(log_k(c)*e^-1) => m
Take for instance N = 15, you can't find a log base 12 of 7, for instance (12 -> 9 -> 3 -> 6 -> 12). So you'd first need to find a fitting base, not sure how hard this is.
It has been proven [0] that factoring is as hard as solving DLP, but RSA also relies on the modular e-th root hypothesis, which we don't know to be as hard or easier than factoring.
In the end, because of [0] is you solve DLP you solve factoring yes, but we usually don't talk about DLP for RSA, even if as I said I was nitpicking.
[0] https://www2.eecs.berkeley.edu/Pubs/TechRpts/1984/5973.html