Answer
$f^{-1}(x)=\log_{3}(\log_{2}x)$
Domain: $(1,\ \infty)$
Range: $(-\infty,\ \infty)$
Work Step by Step
We are given:
$f(x)=2^{3^{x}}$
To find the inverse, we switch $x$ and $y$ and solve for $y$:
$y=2^{3^{x}}$
$x=2^{3^{y}}$
$\log_{2} x=\log_{2} 2^{3^{y}}$
$\log_{2}x=3^{y}$
$\log_{3}(\log_{2}x)=y$
Therefore:
$f^{-1}(x)=\log_{3}(\log_{2}x)$
To find the domain, we keep in mind that we can only take a log of a positive number (greater than 0):
$\log_{2}x>0$
$2^01$
Thus the domain is: $(1,\ \infty)$
The range of a log function is all real numbers: $(-\infty,\ \infty)$.
