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 $(- \infty, \infty)$