Solving system of integer solution First I want to know what is the logarithm used in the following system of equations.

In the set-up protocol, $r_{0}^{\prime}, \cdots, r_{s}^{\prime}$ are found as a solution of the following system
$$
\left[\begin{array}{c}
R^{\prime} \\
\vdots \\
R^{\prime}
\end{array}\right]=\left[\begin{array}{cc}
r_{0}^{\prime} \cdot r_{1}^{\prime a_{1}} \cdot r_{2}^{\prime a_{1}^{2}} \cdots r_{s}^{a_{1}^{s}} & \bmod n^{2} \\
& \vdots & \\
r_{0}^{\prime} \cdot r_{1}^{\prime a_{s}} \cdot r_{2}^{\prime a_{s}^{2}} \cdots r_{s}^{a_{s}^{s}} & \bmod n^{2}
\end{array}\right]
$$
The above system has $s+1$ unknowns and $s$ equations. Therefore it has one degree of freedom. To avoid the trivial solution $r_{0}^{\prime}=R^{\prime}$ and $r_{1}^{\prime}=\cdots=r_{s}^{\prime}=1$, we choose a random $r_{0}^{\prime}$. Then we divide the system by $r_{0}^{\prime}$ and we take logarithms to get
$$
\left[\begin{array}{c}
\log \left(R^{\prime} / r_{0}^{\prime}\right) \\
\log \left(R^{\prime} / r_{0}^{\prime}\right) \\
\vdots \\
\log \left(R^{\prime} / r_{0}^{\prime}\right)
\end{array}\right] \quad \bmod n=\left[\begin{array}{cccc}
a_{1} & a_{1}^{2} & \cdots & a_{1}^{s} \\
\vdots & \vdots & \vdots & \\
a_{s} & a_{s}^{2} & \cdots & a_{s}^{s}
\end{array}\right] \cdot\left[\begin{array}{c}
\log r_{1}^{\prime} \\
\log r_{2}^{\prime} \\
\vdots \\
\log r_{s}^{\prime}
\end{array}\right] \quad \bmod n
$$
The matrix on the right-hand side of the above system is an $s \times s$ generalized Vandermonde matrix (not quite a Vandermonde matrix). Hence, using the techniques in $[7]$ it can be solved in $O\left(s^{2}\right)$ time for $\log r_{1}^{\prime}, \cdots, \log r_{s}^{\prime} .$ Then $s$ powers modulo $n^{2}$ need to be computed to turn $\log r_{i}^{\prime}$ into $r_{i}^{\prime}$ for $i=0, \cdots, s$.

Second I want to understand how one can obtain integer values $r^\prime_1, r^\prime_2, ..., r^\prime_s$ after using logarithm.
Note: All the other values ($R^\prime,r^\prime_0, a_i;\forall i$) in the system are integers.
This equation is from this article (page 16).
 A: It is a discrete logarithm with an unspecified base. Because it is a discrete logarithm it gives you integers. Some points follow.
First, you're following the math too linearly; you need to think from an implementation perspective. For example, you do not "obtain integer $(r_0', r_1', \dots r_n')$ after performing a logarithm" on them; instead you have a matrix equation which you are going to solve for integer $(\log r_0', \log r_1', \dots \log r_n').$ The logarithms in this expression are not logarithms that you actually have to take! Instead you simply have to invert them via modular exponentiation, which is much simpler. 
That leaves the $r_0', R'$ thing for the only reason you'd like to choose a specific logarithm. I do not know the exact reasoning of the authors of the paper but I can tell you that there is one logarithm in particular which will give these some nice properties, and it is base-$n+1,$ but it is not perfect.
In particular, choose a random $R'$ coprime to $n$ and then choose a random $ 0 < b < n$ so that you choose therefore $r_0' = R' + b n \mod n^2$  from a smaller set of possibilities. This smaller-set-of-possibilities is the "imperfection" I mentioned above. However in the paper they only seem to require that the $r_i'$ be nontrivial -- so probably this is OK?
Then $R' / r_0' \equiv 1 \mod n,$ which means that you can efficiently calculate its base-$n+1$ logarithm in the mod-$n^2$ ring, because $(1 + n)^p = 1 + p n + O(n^2).$ So we compute the inverse element for the $r_0'$ that we've chosen above, and then multiply by $R'$, reduce modulo $n^2$, subtract 1, and divide by $n$. The resulting number is not $0$ so the resulting expression is not trivial. 
Also, since multiplying by $(r_0')^{-1}$ was a permutation, I'm pretty sure that uniformly choosing $0 < b < n$ is the same as uniformly choosing $ 0 < p < n $ and working backwards from a known exponent, which is probably how they did this in practice.
