problem similar to DLP I'm studying a protocol whose security is based on the following assumption: given $g_1, g_2, A, B \in \mathbb{F}_p ^*$ with $g_1, g_2, A \neq 1$ and $g_1 \neq g_2$ , it is difficult to find exponents $r_1,r_2, d$ s.t. the following equation holds:
$$
g_1^{r_1}g_2^{r_2} \equiv A^dB \ \ (\mbox{mod} \ p)
$$
The paper I'm reading states "this is a type of discrete logarithm problem, hence it is assumed hard to solve". I'm not sure why this last sentence is true: initially I thought I could show this problem was harder than the DLP by saying "if I could solve this problem for given $g_1, g_2, A, B$, in particular I could solve it for $g_2 = A = 1$, so I could solve $g_1^{r_1}=B$, which is a DLP". But my argument is fallacious as $g_2$ and $A$ are different from $1$ by hypothesis. I would really like to have some proof like the one I was trying to show, so that I can bring this problem back to the DLP hardness assumpion, and don't assume anything more.
If someone know about similar-discrete-log-problem or see any way to show this problem is harder than the DLP, please help me! Thank you in advance
The standard way to prove this problem is at least as hard as the DLP problem is to show that, with an Oracle that can solve this problem, you can solve the DLP problem.

Here's how to do that (and we'll assume that the Oracle will solve random instances with nontrivial probability):

Given the DH problem $G^x = H$, find $x$, we pick random values $s_1, s_2, s_3, s_4$, and form the problem:

$$(G^{s_1})^{r_1} \cdot (G^{s_2})^{r_2} = (G^{s_3})^d \cdot (H^{s_4})$$

This is a random instance, and so the Oracle gives us, with nontrivial probability, values $r_1, r_2, d$, and so we know that:

$$s_1r_1 + s_2r_2 \equiv s_3d + xs_4 \pmod{p-1}$$

We know everything other than $x$ in the above equation, we solve it, and our work is done