If $$\int_{1/3}^{3} |\log_e x| dx = \frac{m}{n} \log_e\left(\frac{n^2}{e}\right)$$, where $$m$$ and $$n$$ are coprime natural numbers, then $$m^2 + n^2 - 5$$ is equal to _____.

We need to evaluate $$\displaystyle\int_{1/3}^{3} |\ln x|\,dx$$. Since $$\ln x < 0$$ for $$x \in [1/3, 1)$$ and $$\ln x \ge 0$$ for $$x \in [1, 3]$$, we split the integral as $$\displaystyle\int_{1/3}^{1} (-\ln x)\,dx + \int_{1}^{3} \ln x\,dx$$.

Using $$\int \ln x\,dx = x\ln x - x + C$$, the first part evaluates to $$\bigl[-x\ln x + x\bigr]_{1/3}^{1} = (0 + 1) - \bigl(-\tfrac{1}{3}\ln\tfrac{1}{3} + \tfrac{1}{3}\bigr) = 1 - \tfrac{1}{3}\ln 3 - \tfrac{1}{3} = \tfrac{2}{3} - \tfrac{\ln 3}{3}$$. The second part gives $$\bigl[x\ln x - x\bigr]_{1}^{3} = (3\ln 3 - 3) - (0 - 1) = 3\ln 3 - 2$$.

Adding these: $$\tfrac{2}{3} - \tfrac{\ln 3}{3} + 3\ln 3 - 2 = \tfrac{8\ln 3 - 4}{3} = \tfrac{4}{3}(2\ln 3 - 1) = \tfrac{4}{3}\ln\!\bigl(\tfrac{3^2}{e}\bigr)$$.

Comparing with $$\tfrac{m}{n}\ln\!\bigl(\tfrac{n^2}{e}\bigr)$$, we get $$m = 4$$ and $$n = 3$$, which are coprime. Therefore $$m^2 + n^2 - 5 = 16 + 9 - 5 = \boxed{20}$$.