now, to get the inverse "relation", we first, do a quick switcharoo on the variables, and then solve for "y".
[tex]\bf f(x)=y=log_{2}(x+4)\qquad inverse\implies \boxed{x}=log_{2}\left(\boxed{y}+4 \right)\\\\
-------------------------------\\\\
\textit{Logarithm Cancellation Rules}\\\\
log_{{ a}}{{ a}}^x\implies x\qquad \qquad
{{ a}}^{log_{{ a}}x}=x\\
~~~\qquad \qquad \qquad \qquad \qquad \uparrow \\
~\qquad \qquad \qquad \textit{let's use this rule}\\\\
-------------------------------\\\\
[/tex]
[tex]\bf 2^x=2^{\cfrac{}{}log_{2}(y+4)}\implies 2^x=y+4\implies 2^x-4=y\impliedby f^{-1}
\\\\\\
thus\qquad 2^3-4=f^{-1}(3)\implies 8-4=f^{-1}(3)\implies 4=f^{-1}(3)[/tex]
