1. **State the problem:** We are given a logarithmic function $$f(x) = -3 \log_{0.5}(x + 4) + 2$$ and a quadratic function $$g(x) = 5(x - 2)^2 + 7$$ which is said to be the inverse of the logarithmic function restricted to the domain $[2, \infty)$. We want to understand the relationship and verify the inverse. 2. **Recall the inverse function definition:** If $g$ is the inverse of $f$, then $$f(g(x)) = x \quad \text{and} \quad g(f(x)) = x$$ for all $x$ in the domain of $g$ and $f$ respectively. 3. **Rewrite the logarithmic function:** The base is $0.5$, so $$\log_{0.5}(y) = \frac{\log(y)}{\log(0.5)}$$ but we will work with the properties of logarithms directly. 4. **Find the inverse of $$f(x)$$:** Start with $$y = -3 \log_{0.5}(x + 4) + 2$$ Isolate the logarithm: $$y - 2 = -3 \log_{0.5}(x + 4)$$ Divide both sides by $-3$: $$\frac{y - 2}{-3} = \log_{0.5}(x + 4)$$ Intermediate step with cancellation: $$\frac{\cancel{y - 2}}{\cancel{-3}} = \log_{0.5}(x + 4)$$ (just showing division) Rewrite the logarithm in exponential form: $$x + 4 = (0.5)^{\frac{y - 2}{-3}}$$ Simplify the exponent: $$x + 4 = (0.5)^{-\frac{y - 2}{3}} = 2^{\frac{y - 2}{3}}$$ because $0.5 = 2^{-1}$. So, $$x = 2^{\frac{y - 2}{3}} - 4$$ 5. **Express inverse function:** Swap $x$ and $y$ to get inverse: $$f^{-1}(x) = 2^{\frac{x - 2}{3}} - 4$$ 6. **Compare with given quadratic inverse:** The given inverse is $$g(x) = 5(x - 2)^2 + 7$$ which is a quadratic, not an exponential. This suggests the quadratic is not the inverse of the logarithmic function. 7. **Conclusion:** The inverse of $$f(x) = -3 \log_{0.5}(x + 4) + 2$$ is $$f^{-1}(x) = 2^{\frac{x - 2}{3}} - 4$$, an exponential function, not the quadratic given. **Final answer:** $$f^{-1}(x) = 2^{\frac{x - 2}{3}} - 4$$