1. **State the problem:** Solve for $x$ in the equation $\log_4(-2x + 4) = 2$. 2. **Recall the definition of logarithm:** If $\log_b(a) = c$, then $a = b^c$. 3. **Apply the definition:** From $\log_4(-2x + 4) = 2$, we get $$-2x + 4 = 4^2$$ 4. **Calculate the right side:** $$4^2 = 16$$ So, $$-2x + 4 = 16$$ 5. **Isolate $x$:** $$-2x = 16 - 4$$ $$-2x = 12$$ 6. **Divide both sides by $-2$:** $$x = \frac{12}{-2}$$ Show cancellation: $$x = \frac{\cancel{12}}{\cancel{-2}} = -6$$ 7. **Check the domain:** The argument of the logarithm must be positive: $$-2x + 4 > 0$$ $$-2x > -4$$ $$x < 2$$ Since $x = -6$ satisfies $x < 2$, it is valid. **Final answer:** $$x = -6$$