Find the number of real solutions of the equation _ x+1 (2x^2 + 7x...

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

First, factor the quadratic expressions in the arguments:

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

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

Substitute these into the equation:

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

Using properties of logarithms:

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

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

Let $t = \log_{x+1}(2x + 5)$. Then $\log_{2x+5}(x + 1) = \dfrac{1}{t}$.

The equation becomes:

$t + 1 + \dfrac{2}{t} = 4$

$t - 3 + \dfrac{2}{t} = 0$

$t^2 - 3t + 2 = 0$

$(t - 1)(t - 2) = 0 \implies t = 1$ or $t = 2$

Case 1: $t = 1$

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

Case 2: $t = 2$

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

$2x + 5 = x^2 + 2x + 1 \implies x^2 = 4 \implies x = 2$ or $x = -2$

Now, we must check the domain conditions for the original logarithmic equation:

The bases must be strictly greater than $0$ and not equal to $1$:

$x + 1 > 0 \implies x > -1$

$x + 1 \neq 1 \implies x \neq 0$

$2x + 5 > 0 \implies x > -2.5$

$2x + 5 \neq 1 \implies x \neq -2$

Checking our candidate solutions against $x > -1$:

For $x = -4$, $x + 1 = -3 < 0$ (rejected)

For $x = -2$, $x + 1 = -1 < 0$ (rejected)

For $x = 2$, $x + 1 = 3 > 0$ and $2x + 5 = 9 > 0$. Both bases are positive and not equal to $1$. The arguments are also positive.

Thus, $x = 2$ is the only valid solution.

The number of real solutions is $1$.

Answer: $1$