Question

The sum of squares of all the real solutions of the equation \(\log_{(x+1)}(2x^2 + 5x + 3) = 4 - \log_{(2x+3)}(x^2 + 2x + 1)\) is equal to ________.

Answer 1

Correct Answer: 2

Step 1: Understanding the Concept:
This problem involves logarithmic identities and base change properties. We must also consider the constraints for the existence of logarithms: the base must be positive and not equal to 1, and the argument must be positive. 
Step 2: Key Formula or Approach: 
1. Factorize the arguments: $2x^2 + 5x + 3 = (2x+3)(x+1)$ and $x^2 + 2x + 1 = (x+1)^2$. 
2. Use $\log_a(bc) = \log_a b + \log_a c$ and $\log_a(b^k) = k \log_a b$. 
Step 3: Detailed Explanation: 
The equation is: \[ \log_{(x+1)}((2x+3)(x+1)) = 4 - \log_{(2x+3)}((x+1)^2) \] Using log properties: \[ \log_{(x+1)}(2x+3) + \log_{(x+1)}(x+1) = 4 - 2\log_{(2x+3)}(x+1) \] Since $\log_a a = 1$: \[ \log_{(x+1)}(2x+3) + 1 = 4 - \frac{2}{\log_{(x+1)}(2x+3)} \] Let $t = \log_{(x+1)}(2x+3)$: \[ t + 1 = 4 - \frac{2}{t} \implies t - 3 + \frac{2}{t} = 0 \] Multiplying by $t$: \[ t^2 - 3t + 2 = 0 \implies (t-1)(t-2) = 0 \] Case 1: $t = 1 \implies 2x + 3 = x + 1 \implies x = -2$. Check Constraints: Base $x+1 = -2+1 = -1$. Logarithm base must be $>0$. So, $x=-2$ is rejected. Case 2: $t = 2 \implies 2x + 3 = (x + 1)^2$ \[ 2x + 3 = x^2 + 2x + 1 \implies x^2 = 2 \implies x = \pm\sqrt{2} \] Check Constraints: For $x = \sqrt{2}$, base $x+1>0$. Valid. For $x = -\sqrt{2} \approx -1.414$, base $x+1 = -0.414 < 0$. Rejected. The only real solution is $x = \sqrt{2}$. 
Step 4: Final Answer: 
Sum of squares $= (\sqrt{2})^2 = 2$.