Solve the logarithmic equation: 2 2 x - 2 2 (8 − 3x) = 1

$$\frac{2 \log_2 x - 2 \log_2 (8 - 3x)}{2} = \frac{1}{2}$$ $$\log_2 x - \log_2 (8 - 3x) = \frac{1}{2}$$

If $\log_b M = c$, then $b^c = M$.

$$\frac{x}{8 - 3x} = 2^{\frac{1}{2}}$$ $$\frac{x}{8 - 3x} = \sqrt{2}$$

Multiply both sides by $(8 - 3x)$.

$$x = \sqrt{2}(8 - 3x)$$ $$x = 8\sqrt{2} - 3\sqrt{2}x$$

Move all terms containing $x$ to one side.

$$x + 3\sqrt{2}x = 8\sqrt{2}$$

Factor out $x$.

$$x(1 + 3\sqrt{2}) = 8\sqrt{2}$$

Divide to isolate $x$.

$$x = \frac{8\sqrt{2}}{1 + 3\sqrt{2}}$$

Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $1 - 3\sqrt{2}$.

$$x = \frac{8\sqrt{2}}{1 + 3\sqrt{2}} \times \frac{1 - 3\sqrt{2}}{1 - 3\sqrt{2}}$$ $$x = \frac{8\sqrt{2}(1 - 3\sqrt{2})}{1^2 - (3\sqrt{2})^2}$$ $$x = \frac{8\sqrt{2} - 8\sqrt{2} \cdot 3\sqrt{2}}{1 - (9 \cdot 2)}$$ $$x = \frac{8\sqrt{2} - 24 \cdot 2}{1 - 18}$$ $$x = \frac{8\sqrt{2} - 48}{-17}$$ $$x = \frac{48 - 8\sqrt{2}}{17}$$

The arguments of the logarithms must be positive.

1. $x > 0$
2. $8 - 3x > 0 \implies 8 > 3x \implies x < \frac{8}{3}$

Let's approximate the value of $x$:
$x \approx \frac{48 - 8 \times 1.414}{17} \approx \frac{48 - 11.312}{17} \approx \frac{36.688}{17} \approx 2.158$
This value satisfies both conditions ($2.158 > 0$ and $2.158 < \frac{8}{3} \approx 2.667$).