Solve Complex Logarithmic Equation: 1/(log_x3) × x²log_(1/x)27 + 4x + 6 = 0

To solve the given equation, we need to simplify the logarithmic expressions and then solve for $x$. Let's proceed with the given equation:

$\frac{1}{\log_x 3} \times x^2 \log_{1/x} 27 + 4x + 6 = 0$

Step 1: Simplify the logarithmic terms.

Apply the change of base formula to the logarithms:

$\log_x 3 = \frac{\ln 3}{\ln x}$

Thus, $\frac{1}{\log_x 3} = \frac{\ln x}{\ln 3}$.

For the second logarithmic term: $\log_{1/x} 27 = -\log_x 27 = -\frac{\ln 27}{\ln x}$.

Step 2: Substitute these simplifications back into the equation.

We have:

$\frac{\ln x}{\ln 3} \times x^2 \times -\frac{\ln 27}{\ln x} + 4x + 6 = 0$

Simplify this expression:

The $\ln x$ terms cancel each other out in the expression $\frac{\ln x}{\ln 3} \times x^2 \times -\frac{\ln 27}{\ln x}$.

Thus, it becomes:

$-\frac{\ln 27}{\ln 3} x^2 + 4x + 6 = 0$

The value of $-\frac{\ln 27}{\ln 3}$ is actually $-\log_3 27 = -3$ because $27 = 3^3$.

Therefore, the simplified equation is:

$-3x^2 + 4x + 6 = 0$

Step 3: Solve the quadratic equation.

Rearrange it to $3x^2 - 4x - 6 = 0$.

Apply the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Here, $a = 3$, $b = -4$, $c = -6$.

So, the solution becomes:

$x = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 3 \times (-6)}}{2 \times 3}$

This simplifies to:

$x = \frac{4 \pm \sqrt{16 + 72}}{6}$

$x = \frac{4 \pm \sqrt{88}}{6}$

Simplify $\sqrt{88} = \sqrt{4 \times 22} = 2\sqrt{22}$.

Thus,

$x = \frac{4 \pm 2\sqrt{22}}{6}$

Simplifying further gives us:

$x = \frac{2 \pm \sqrt{22}}{3}$

The valid positive solution (since logarithms are not satisfied with negative bases) is:

$x = \frac{2}{3} + \frac{\sqrt{22}}{3}$

Therefore, the correct answer is choice $3$: $\frac{2}{3}+\frac{\sqrt{22}}{3}$.