1. The problem is to solve the equation $$\log_3(x + 4) = 2$$ and analyze the function $$f(x) = -\log_3(x + 2)$$. 2. To solve $$\log_3(x + 4) = 2$$, recall that $$\log_b(a) = c$$ means $$b^c = a$$. 3. Rewrite the equation as $$3^2 = x + 4$$. 4. Calculate $$3^2 = 9$$, so $$9 = x + 4$$. 5. Solve for $$x$$: $$x = 9 - 4 = 5$$. 6. The solution to the equation is $$x = 5$$. 7. Now consider the function $$f(x) = -\log_3(x + 2)$$. 8. The domain of $$f(x)$$ requires $$x + 2 > 0$$, so $$x > -2$$. 9. The function is the negative of the logarithm base 3 of $$x + 2$$, which reflects the graph of $$\log_3(x + 2)$$ across the x-axis. 10. The vertical asymptote is at $$x = -2$$, where the argument of the logarithm is zero. 11. The function decreases as $$x$$ increases because of the negative sign. Final answers: - Solution to $$\log_3(x + 4) = 2$$ is $$x = 5$$. - Function $$f(x) = -\log_3(x + 2)$$ has domain $$x > -2$$ and is a reflection of $$\log_3(x + 2)$$ across the x-axis.