The number of solutions of the equation $$\log_{(x+1)}(2x^2 + 7x + 5) + \log_{(2x+5)}(x+1)^2 - 4 = 0$$, $$x > 0$$, is ___.

We need to solve $$\log_{(x+1)}(2x^2 + 7x + 5) + \log_{(2x+5)}(x+1)^2 - 4 = 0$$ for $$x > 0$$.

First, note that $$2x^2 + 7x + 5 = (2x+5)(x+1)$$. So the equation becomes: $$\log_{(x+1)}[(2x+5)(x+1)] + 2\log_{(2x+5)}(x+1) - 4 = 0$$

$$\log_{(x+1)}(2x+5) + \log_{(x+1)}(x+1) + 2\log_{(2x+5)}(x+1) - 4 = 0$$

$$\log_{(x+1)}(2x+5) + 1 + 2\log_{(2x+5)}(x+1) - 4 = 0$$

Let $$t = \log_{(x+1)}(2x+5)$$, so $$\log_{(2x+5)}(x+1) = \frac{1}{t}$$. The equation becomes: $$t + \frac{2}{t} - 3 = 0$$ $$t^2 - 3t + 2 = 0$$ $$(t-1)(t-2) = 0$$ $$t = 1 \text{ or } t = 2$$

Case 1: $$t = 1$$, i.e., $$\log_{(x+1)}(2x+5) = 1 \Rightarrow 2x+5 = x+1 \Rightarrow x = -4$$. This is excluded since $$x > 0$$.

Case 2: $$t = 2$$, i.e., $$\log_{(x+1)}(2x+5) = 2 \Rightarrow (x+1)^2 = 2x+5 \Rightarrow x^2 + 2x + 1 = 2x + 5 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$$. Since $$x > 0$$, we take $$x = 2$$.

We verify the domain conditions for $$x = 2$$: the bases are $$x+1 = 3 > 0, \neq 1$$ and $$2x+5 = 9 > 0, \neq 1$$, and the argument $$2x^2+7x+5 = 8+14+5 = 27 > 0$$. All conditions are satisfied.

Therefore, the number of solutions is $$\boxed{1}$$.