Solve the Logarithmic Equation: Base-8 and Base-4 Logs with Quadratic Terms

To solve the equation: $\log_8 9 - \log_8 3 + \log_4 x^2 = \log_8 1.5 + \log_8 2 + \log_4 (-x^2 - 11x - 9)$, we proceed as follows:

Step 1: Simplify Both Sides
On the left-hand side (LHS), apply logarithmic subtraction: $\log_8 \left(\frac{9}{3}\right) + \log_4 x^2 = \log_8 3 + \log_4 x^2$.
Note $\log_8 3$ remains and convert $\log_4 x^2$ using the base switch to $8$:
$\log_4 x^2 = 2\log_4 x = 2 \times \frac{\log_8 x}{\log_8 2^2} = \frac{\log_8 x}{\log_8 2}$.
Thus, the LHS combines into:
$\log_8 3 + \frac{2\log_8 x}{\log_8 4}$ (because $\log_4 x^2 = 2 \log_4 x$).

On the right-hand side (RHS):
Combine: $\log_8 (1.5 \times 2) = \log_8 3$.
Also apply for $\log_4$ term:
$\log_4 (-x^2 - 11x - 9) = \frac{\log_8 (-x^2 - 11x - 9)}{\log_8 4}$.

Step 2: Equalize Both Sides
Equate LHS and RHS logarithmic expressions:
$\log_8 3 + \frac{2\log_8 x}{\log_8 4} = \log_8 3 + \frac{\log_8 (-x^2 - 11x - 9)}{\log_8 4}$.
The $\log_8 3$ cancels out on both sides, leaving:
$\frac{2\log_8 x}{\log_8 4} = \frac{\log_8 (-x^2 - 11x - 9)}{\log_8 4}$.

Step 3: Solve for $x$
Since the denominators are equal, set the numerators equal:
$2\log_8 x = \log_8 (-x^2 - 11x - 9)$.
Translate this into an exponential equation:
$(x^2)^2 = -x^2 - 11x - 9$ or
$8^{2\log_8 x} = -x^2 - 11x - 9$.
Let $y = x$, solve the resulting quadratic equation:
$x^2 = -x^2 - 11x - 9$.
Then, finding valid $x$ by allowing roots of polynomial calculations should yield laws consistency:
$-x^2 - 11x - 9 = 0$ or rather substituting potential values. After appropriate checks:

The valid $x$ that satisfies the problem is thus $x = -4.5$.