Solve the Mixed-Base Logarithm Equation: log₂(x²+3x+3)·log₃(1/4) = -2log₃((4x+2)/-2)

To solve the given logarithmic equation, follow these steps:

• Step 1: Simplify the logarithmic expressions.
• Step 2: Solve the resulting equation for $x$.
• Step 3: Verify that the solutions are within the domain of the original logarithm expressions.

Let's work through each step:

Step 1. Simplify the expression
The given equation is:

$\log_2(x^2+3x+3)\cdot\log_3\frac{1}{4}=-2\log_3\left(\frac{4x+2}{-2}\right)$

Recognizing that $\log_3\frac{1}{4} = \log_3(4^{-1}) = -\log_3(4)$, and $-2\log_3\left(\frac{4x+2}{-2}\right) = 2\log_3\left(\frac{-1(4x+2)}{-2}\right) = 2\log_3(x+1)$.

This simplifies to:

$\log_2(x^2+3x+3)\cdot (-\log_3(4)) = 2\log_3(x+1)$

Step 2. Simplify further
Rewriting it with all terms in base 3 logarithm by using change of base:

$\frac{\log_3(x^2+3x+3)}{\log_3(2)} \cdot (-\log_3(4)) = 2\log_3(x+1)$

This results in:

$-\frac{\log_3(x^2+3x+3)\log_3(4)}{\log_3(2)} = 2\log_3(x+1)$

Let $a = \log_3(x^2+3x+3)$ temporarily for easier manipulation:

$-a \frac{\log_3(4)}{\log_3(2)} = 2\log_3(x+1)$

Using change base for $\frac{\log_3(4)}{\log_3(2)} = \log_2(4) = 2$:

$-2a = 2\log_3(x+1)$

Which means:

$a = -\log_3((x+1)^2)$

Therefore returning to original substitution:

$\log_3(x^2 + 3x + 3) = -\log_3((x+1)^2)$

Since $-\log_3((x+1)^2)$ is equivalent to $\log_3\left(\frac{1}{(x+1)^2}\right)$

$\log_3(x^2 + 3x + 3) = \log_3\left(\frac{1}{(x+1)^2}\right)$

Equating inside terms gives:

$x^2 + 3x + 3 = \frac{1}{(x+1)^2}$

Step 3. Solving the quadratic equation

Clear the fraction:

$(x^2 + 3x + 3) \cdot (x+1)^2 = 1$

Expanding and simplifying results in the quadratic equation:

$x^4+2x^3+9x^2+8x+2 -1 = 0$

This reduces to solving the known quadratic terms:

$(x + 1)(x + 4) = 0$

Therefore, the potential solutions are $x = -1$ and $x = -4$.

Step 4. Validating solutions

Both solutions must satisfy domain conditions:

For $x = -1$ → Argument of all logs remain positive.

For $x = -4$ → Argument of all logs remain positive.

Therefore, both solutions are valid.

Thus, the correct answer is $\mathbf{-1, -4}$.