College Algebra, Chapter 5, Review Exercise, Section Review Exercise, Problem 88

Determine the inverse of the function $f(x) = 2^{3^x}$ and state its domain and range.
To find the inverse, we set $y = f(x)$

$
\begin{equation}
\begin{aligned}
y &= 2^{3^x} && \text{Solve for $x$; take ln of both sides}\\
\\
\ln y &= 3^x (\ln 2) && \text{Divide both sides by ln 2}\\
\\
\frac{\ln y }{\ln 2} &= 3^x && \text{Take ln of both sides}\\
\\
\ln \left( \frac{\ln y}{\ln 2} \right) &= x (\ln 3) && \text{Divide both sides by ln 3}\\
\\
\frac{\ln \left( \frac{\ln y}{\ln 2} \right)}{\ln 3} &= x \\
\\
\frac{\ln \left( \frac{\ln x}{\ln 2} \right)}{\ln 3} &= y
\end{aligned}
\end{equation}
$

Thus, the inverse of $\displaystyle f(x) = f^{-1}(x) = \frac{\ln \left( \frac{\ln x}{\ln 2} \right)}{\ln 3}$
To find the domain, we want

$
\begin{equation}
\begin{aligned}
\frac{\ln x}{\ln 2} &> 0 && \text{Multiply ln 2 both sides}\\
\\
\ln x &> 0 && \text{Raise to $e$ both sides}\\
\\
x &> e^0 && \text{Evaluate}\\
\\
x &> 1
\end{aligned}
\end{equation}
$

Thus, the domain is $(1,\infty)$. Then by substituting the domain in the function we get the range of $(-\infty,2)$