jmc

Let $$x={\log }_2(27+{\log }_2(27+{\log }_2(27+\cdots ))).$$Then $$x={\log }_2(27+x),$$so $2^x=x+27.$

To solve this equation, we plot $y=2^x$ and $y=x+27.$



By inspection, the graphs intersect at $(5,32).$ Beyond this point, the graph of $y=2^x$ increases much faster than the graph of $y=x+27,$ so the only positive solution is $x=\boxed{5}.$