1. The problem is to find the inverse function $f^{-1}(x)$ of the function $f(x) = e^{3x + 5}$. 2. To find the inverse, we start by setting $y = f(x)$, so: $$y = e^{3x + 5}$$ 3. The goal is to solve for $x$ in terms of $y$. First, take the natural logarithm (ln) of both sides to undo the exponential: $$\ln y = \ln \left(e^{3x + 5}\right)$$ 4. Using the property $\ln(e^a) = a$, we get: $$\ln y = 3x + 5$$ 5. Now solve for $x$: $$\ln y - 5 = 3x$$ 6. Divide both sides by 3: $$x = \frac{\cancel{\ln y - 5}}{\cancel{3}}$$ 7. Replace $y$ with $x$ to write the inverse function: $$f^{-1}(x) = \frac{\ln x - 5}{3}$$ 8. Therefore, the correct inverse function is: $$f^{-1}(x) = \frac{\ln x - 5}{3}$$ This matches the first option given. The other options are incorrect because: - $\frac{e^x - 5}{3}$ is not the inverse of an exponential function. - $\frac{\ln x + 5}{3}$ has the wrong sign inside the numerator. Hence, the answer is $f^{-1}(x) = \frac{\ln x - 5}{3}$.