jmc

Converting $131_a$ to base 10 and setting it equal to 55, we find that $$\begin{array}{rl}1(a^2)+3(a^1)+1(a^0)& =55\\ a^2+3a+1& =55\\ \Rightarrow a^2+3a-54& =0\\ \Rightarrow (a+9)(a-6)& =0\end{array}$$This tells us that $a$ is either $-9$ or $6$. Since $a$ must be greater than 0, $a=\boxed{6}$.