4. Given that $\left\{a_{n}\right\}$ is an arithmetic sequence with a non-zero common difference, $\left\{b_{n}\right\}$ is a geometric sequence, where $a_{1}=3, b_{1}=1, a_{2}=$ $b_{2}, 3 a_{5}=b_{3}$, and there exist constants $\alpha, \beta$ such that for every positive integer $n$ we have $a_{n}=\log _{\alpha} b_{n}+\beta$, then $\alpha+\beta=$ . $\qquad$
Let the common difference of the sequence $\left\{a_{n}\right\}$ be $d$, and the common ratio of the sequence $\left\{b_{n}\right\}$ be $q$, then
$$
\left\{\begin{array}{l}
3+d=q, \\
3(3+4 d)=q^{2}
\end{array} \Rightarrow(3+d)^{2}=9+12 d \Rightarrow d^{2}-6 d=0 \Rightarrow d=6\right. \text {. }
$$

Thus, $a_{n}=6 n-3, b_{n}=9^{n-1} \Rightarrow 6 n-3=\log _{\alpha} 9^{n-1}+\beta=n \log _{\alpha} 9+\beta-\log _{\alpha} 9$
$$
\Rightarrow\left\{\begin{array} { l } 
{ \operatorname { l o g } _ { \alpha } 9 = 6 , } \\
{ \beta - \operatorname { l o g } _ { \alpha } 9 = - 3 }
\end{array} \Rightarrow \left\{\begin{array}{l}
\alpha=\sqrt[3]{3}, \\
\beta=3
\end{array} \Rightarrow \alpha+\beta=\sqrt[3]{3}+3\right.\right.
$$