1. Given equation

We need to solve, for $x>0$,

$$\log_{x+1}(2x^2+7x+5)+\log_{2x+5}(x+1)^2-4=0.$$

So,

$$\log_{x+1}(2x^2+7x+5)+\log_{2x+5}(x+1)^2=4.$$

2. Factor the quadratic

Observe that

$$2x^2+7x+5=(2x+5)(x+1).$$

Hence,

$$\log_{x+1}\big((2x+5)(x+1)\big)+\log_{2x+5}(x+1)^2=4.$$

3. Check domain

For logarithms to exist:

• base $x+1>0$ and $x+1\neq 1$
• base $2x+5>0$ and $2x+5\neq 1$
• arguments positive

Given $x>0$, we have:

• $x+1>1$
• $2x+5>5>1$
• $(2x+5)(x+1)>0$
• $(x+1)^2>0$

So the domain is simply $x>0$.

4. Use logarithm properties

First term:

$$\log_{x+1}\big((2x+5)(x+1)\big) =\log_{x+1}(2x+5)+\log_{x+1}(x+1) =\log_{x+1}(2x+5)+1.$$

Second term:

$$\log_{2x+5}(x+1)^2=2\log_{2x+5}(x+1).$$

Therefore the equation becomes

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

so

$$\log_{x+1}(2x+5)+2\log_{2x+5}(x+1)=3.$$

5. Introduce a variable

Let

$$a=\log_{x+1}(2x+5).$$

Then by reciprocal property,

$$\log_{2x+5}(x+1)=\frac{1}{a}.$$

So the equation reduces to

$$a+\frac{2}{a}=3.$$

Multiply by $a$:

$$a^2-3a+2=0.$$

Thus,

$$(a-1)(a-2)=0.$$

Hence,

$$a=1 \quad \text{or} \quad a=2.$$

6. Solve each case

Case 1: $a=1$

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

which means

$$2x+5=x+1.$$

So,

$$x=-4,$$

which does not satisfy $x>0$.

So this case gives no valid solution.

Case 2: $a=2$

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

which means

$$2x+5=(x+1)^2.$$

Expanding,

$$2x+5=x^2+2x+1.$$

Therefore,

$$x^2-4=0,$$

so

$$x=\pm 2.$$

Given $x>0$, only

$$x=2$$

is valid.

7. Count the number of solutions

There is exactly one valid solution.

Therefore, the number of solutions is

$$\boxed{1}.$$

8. Comparison with stored answer

Stored correct answer: $1$

Our derived answer also gives $1$, so they agree.