## College Algebra 7th Edition

$f^{-1}(x)=\log_{3}(\log_{2}x)$ Domain: $(1,\ \infty)$ Range: $(-\infty,\ \infty)$
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: $(-\infty,\ \infty)$.