Sum of Squares of Solutions to Logarithmic Equation

To find the sum of squares of all real solutions to the given logarithmic equation:

$\log_{(x+1)}\left(2x^2 + 5x + 3\right) = 4 - \log_{(2x+3)}\left(x^2 + 2x + 1\right)$

Step 1: Determine the Domain of the Equation

For the logarithmic expressions to be well-defined:

1. Bases must be positive and not equal to $1$:

   • $x + 1 > 0 \text{ and } x + 1 \neq 1 \implies x > -1 \text{ and } x \neq 0$
   • $2x + 3 > 0 \text{ and } 2x + 3 \neq 1 \implies x > -\frac{3}{2} \text{ and } x \neq -1$

2. Arguments must be strictly positive:

   • $2x^2 + 5x + 3 = (x + 1)(2x + 3) > 0$
   • $x^2 + 2x + 1 = (x + 1)^2 > 0 \implies x \neq -1$

Combining these constraints, the domain for $x$ is: $x \in (-1, 0) \cup (0, \infty)$

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Step 2: Simplify the Logarithmic Terms

We can factor the quadratic expressions inside the logarithms: $2x^2 + 5x + 3 = (x + 1)(2x + 3)$ $x^2 + 2x + 1 = (x + 1)^2$

Substitute these back into the original equation: $\log_{(x+1)}\left((x + 1)(2x + 3)\right) = 4 - \log_{(2x+3)}\left((x + 1)^2\right)$

Using the properties of logarithms ($\log_b(a \cdot c) = \log_b a + \log_b c$ and $\log_b(a^k) = k \log_b a$): $\log_{(x+1)}(x + 1) + \log_{(x+1)}(2x + 3) = 4 - 2 \log_{(2x+3)}(x + 1)$

Since $\log_{(x+1)}(x + 1) = 1$, we get: $1 + \log_{(x+1)}(2x + 3) = 4 - 2 \log_{(2x+3)}(x + 1)$ $\log_{(x+1)}(2x + 3) + 2 \log_{(2x+3)}(x + 1) = 3$

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Step 3: Solve the Algebraic Equation

Recall the change of base property: $\log_{(2x+3)}(x + 1) = \frac{1}{\log_{(x+1)}(2x + 3)}$.

Let $y = \log_{(x+1)}(2x + 3)$. The equation transforms into: $y + \frac{2}{y} = 3$

Multiplying by $y$ (where $y \neq 0$): $y^2 - 3y + 2 = 0$ $(y - 1)(y - 2) = 0$

This yields two cases: $y = 1$ or $y = 2$.

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Step 4: Check Solutions for $x$

• Case 1: $y = 1$ $\log_{(x+1)}(2x + 3) = 1 \implies 2x + 3 = x + 1 \implies x = -2$ Checking against the domain $x \in (-1, 0) \cup (0, \infty)$, $x = -2$ is invalid (extraneous).

• Case 2: $y = 2$ $\log_{(x+1)}(2x + 3) = 2 \implies 2x + 3 = (x + 1)^2$ $2x + 3 = x^2 + 2x + 1$ $x^2 = 2 \implies x = \sqrt{2} \quad \text{or} \quad x = -\sqrt{2}$

  Checking against the domain $x \in (-1, 0) \cup (0, \infty)$:

  • $x = -\sqrt{2} \approx -1.414 < -1$ is invalid.
  • $x = \sqrt{2} \approx 1.414 > 0$ is valid.

Thus, the only real solution to the equation is $x = \sqrt{2}$.

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Step 5: Calculate the Sum of Squares

The sum of squares of all real solutions is: $(\sqrt{2})^2 = 2$