Let $$x = x(y)$$ be the solution of the differential equation $$2(y+2)\log_e(y+2)dx + (x + 4 - 2\log_e(y+2))dy = 0$$, $$y > -1$$ with $$x(e^4 - 2) = 1$$. Then $$x(e^9 - 2)$$ is equal to

Rearranging the equation:

$$2(y+2)\ln(y+2) \frac{dx}{dy} + x = 2\ln(y+2) - 4$$

$$\frac{dx}{dy} + \frac{x}{2(y+2)\ln(y+2)} = \frac{1}{y+2} - \frac{2}{(y+2)\ln(y+2)}$$

Let $$t = \ln(y+2)$$, then $$dt = \frac{1}{y+2} dy$$. The equation becomes:

$$\frac{dx}{dt} + \frac{x}{2t} = 1 - \frac{2}{t}$$

The integrating factor (IF): $$IF = e^{\int \frac{1}{2t} dt} = e^{\frac{1}{2}\ln t} = \sqrt{t}$$

Multiplying by IF:

$$\frac{d}{dt}(x\sqrt{t}) = \sqrt{t} - \frac{2}{\sqrt{t}}$$

$$x\sqrt{t} = \int (t^{1/2} - 2t^{-1/2}) dt = \frac{2}{3}t^{3/2} - 4t^{1/2} + C$$

$$x = \frac{2}{3}t - 4 + \frac{C}{\sqrt{t}}$$

Using $$x(e^4-2) = 1$$ (where $$t = \ln(e^4) = 4$$):

$$1 = \frac{2}{3}(4) - 4 + \frac{C}{2} \implies 1 = \frac{8}{3} - 4 + \frac{C}{2}$$

$$1 = -\frac{4}{3} + \frac{C}{2} \implies \frac{C}{2} = \frac{7}{3} \implies C = \frac{14}{3}$$

Finding $$x(e^9-2)$$ (where $$t = \ln(e^9) = 9$$):

$$x = \frac{2}{3}(9) - 4 + \frac{14/3}{3}$$

$$x = 6 - 4 + \frac{14}{9} = 2 + \frac{14}{9} = \frac{32}{9}$$