1. The problem is to solve for $x$ in the equation $\log_x \left( \frac{4}{9} \right) = -2$. 2. Recall the definition of logarithm: $\log_a b = c$ means $a^c = b$. 3. Using this definition, rewrite the equation: $$x^{-2} = \frac{4}{9}$$ 4. To solve for $x$, take the reciprocal of the exponent: $$x^{-2} = \frac{4}{9} \implies \frac{1}{x^2} = \frac{4}{9}$$ 5. Cross-multiply to isolate $x^2$: $$9 = 4x^2$$ 6. Divide both sides by 4: $$\cancel{\frac{9}{4}} = \cancel{\frac{4x^2}{4}} \implies \frac{9}{4} = x^2$$ 7. Take the square root of both sides: $$x = \pm \sqrt{\frac{9}{4}} = \pm \frac{3}{2}$$ 8. Since the base of a logarithm must be positive and not equal to 1, $x = \frac{3}{2}$ is the valid solution. **Final answer:** $$x = \frac{3}{2}$$