If the sum of all the roots of the equation $$e^{2x} - 11e^x - 45e^{-x} + \frac{81}{2} = 0$$ is $$\log_e P$$, then $$P$$ is equal to ______

Find $$P$$ such that the sum of all roots of $$e^{2x} - 11e^x - 45e^{-x} + \frac{81}{2} = 0$$ equals $$\log_e P$$.

Substituting $$t = e^x$$ (where $$t > 0$$) and then multiplying the entire equation by $$2e^x$$ gives

$$2e^{3x} - 22e^{2x} + 81e^x - 90 = 0$$

In terms of $$t$$ this becomes

$$2t^3 - 22t^2 + 81t - 90 = 0$$

Now testing $$t = 2$$ yields $$2(8) - 22(4) + 81(2) - 90 = 16 - 88 + 162 - 90 = 0$$ ✔

Therefore one factor is $$(t - 2)$$ and we can write

$$2t^3 - 22t^2 + 81t - 90 = (t - 2)(2t^2 - 18t + 45)$$

Verifying, we have

$$ (t-2)(2t^2 - 18t + 45) = 2t^3 - 18t^2 + 45t - 4t^2 + 36t - 90 = 2t^3 - 22t^2 + 81t - 90$$ ✔

Solving $$2t^2 - 18t + 45 = 0$$ gives

$$t = \frac{18 \pm \sqrt{324 - 360}}{4} = \frac{18 \pm \sqrt{-36}}{4} = \frac{18 \pm 6i}{4} = \frac{9 \pm 3i}{2}$$

Thus the complex roots are $$t_2 = \frac{9 + 3i}{2}$$ and $$t_3 = \frac{9 - 3i}{2}$$.

Since $$e^{x_1} = 2$$ for the real root, it follows that $$x_1 = \ln 2$$. For the complex roots we have $$e^{x_2} = \frac{9+3i}{2}$$ and $$e^{x_3} = \frac{9-3i}{2}$$, so

$$x_2 + x_3 = \ln\left(\frac{9+3i}{2}\right) + \ln\left(\frac{9-3i}{2}\right) = \ln\left(\frac{9+3i}{2} \cdot \frac{9-3i}{2}\right)$$

$$= \ln\left(\frac{81 + 9}{4}\right) = \ln\left(\frac{90}{4}\right) = \ln\left(\frac{45}{2}\right)$$

Therefore the sum of all roots is

$$x_1 + x_2 + x_3 = \ln 2 + \ln\frac{45}{2} = \ln\left(2 \cdot \frac{45}{2}\right) = \ln 45$$

Alternatively, by Vieta's formulas for the cubic $$2t^3 - 22t^2 + 81t - 90 = 0$$, the product of the roots is $$t_1 t_2 t_3 = \frac{90}{2} = 45$$. Since $$t_i = e^{x_i}$$ we obtain $$e^{x_1 + x_2 + x_3} = t_1 t_2 t_3 = 45$$, which implies $$x_1 + x_2 + x_3 = \ln 45 = \log_e P$$.

So $$P = 45$$.

The answer is $$\boxed{45}$$.