Single Variable Calculus, Chapter 7, 7.3-2, Section 7.3-2, Problem 22

Determine a.) the domain of $f(x) = \ln (2 + \ln x)$ and b.) $f^{-1}$ and its domain.

a.) The domain of the given function, should be


$\begin{equation} \begin{aligned} & x > 0 \text{ and } 2 + \ln x > 0 \\ \\ & \ln x < -2 \\ \\ & e^{\ln x} < e^{-2} \\ \\ & x < e^{-2} \\ \\ & x < \frac{1}{e^2} \end{aligned} \end{equation}$



Therefore, the domain of $f(x) = \ln (2 + \ln x)$ is $\displaystyle \left(0, \frac{1}{e^2} \right)$

b.) Solving for the inverse function


$\begin{equation} \begin{aligned} & y = \ln (2 + \ln x) \\ \\ & e^y = e^{\ln (2 + \ln x)} \\ \\ & e^y 2 + \ln x \\ \\ & \ln x = e^y - 2 \\ \\ & e^{\ln x} = e^{e^y} - e^2 \\ \\ & x = e^{e^y} - e^2 \\ \\ & \text{Interchange } x \text{ and } y \\ \\ & y = e^{e^x} - e^2 \\ \\ & y = e^{e^x - 2} \\ \\ & f^{-1} (x) = e^{e^x - 2} \end{aligned} \end{equation}$



The domain of $f^{-1} (x) = e^{e^x - 2}$ is $(- \infty, \infty)$