Discrete Logarithm Problem.
Fix a cyclic group $\mathbb{G}$ of order $q$, and generator $g\in\mathbb{G}$. We know that ${g^0,g^1,…,g^{q-1}}=\mathbb{G}$.
Equilently, for every $h\in\mathbb{G}$, there is a unique $x\in \mathbb{Z}_q$, such that $g^x=h$. We define $\log_g{h}$ to be this $x$. This is the discrete logarithm of $h$ with respect to $g$ in the group $\mathbb{G}$.
Discrete Logarithm problem in $\mathbb{G}$: Given $g\in \mathbb{G}$ and uniform $h\in \mathbb{G}$, compute $\log_g{h}$.
Discrete-Logarithm Experiment
Similarly, we have the discrete-logarithm experiment $\mathtt{DLog}_{\mathscr{A}, \mathscr{G}} (n)$ where $\mathscr{G}$ is an efficient group-generation algorithm that, on input $1^n$, outputs a cyclic group $\mathbb{G}$, its order $q$, and a generator $g$.
Adversary $\mathscr{A}$ with $\mathbb{G},q,g$ and a uniform $h\in\mathbb{G}$ needs to efficiently compute correponding $x$.
Definition. RSA problem is hard relative to $\mathscr{G}$ if for all probabilistic polynomial time algorithm $\mathscr{A}$, there exists a negligible $\varepsilon(n)$ that $\Pr[\mathtt{DLog}_{\mathscr{A}, \mathscr{G}}(n)=1]\leq \varepsilon(n)$.
The Discrete Logarithm Assumption is the assumption that there exists a group-generation algorithm $\mathscr{G}$ relative to which the Discrete Logarithm problem is hard.
离散对数问题本质上是模意义下的对数运算。另外还需要注意一点，无论是 RSA 实验还是离散对数实验，对敌手的挑战 (challenge) 都是均匀随机 (uniform) 的。
其实，RSA 问题 (模意义下的开次方运算) 与离散问题 (模意义下的对数运算) 均可以视为幂运算 $h=g^x$ 的逆运算：RSA 问题给出指数 $x$，要求由 $h$ 计算底数 $g=h^{1/x}$；而离散对数问题给出底数 $g$，要求由 $h$ 计算指数 $x=\log_g h$。