3^n+4=27^2nsolve using common bases?
\(\bf 3^{n+4}=27^{2n}\qquad 27=3^3\qquad thus \\\\ 3^{n+4}=(3^3)^{2n}\implies 3^{n+4}=3^{3\cdot 2n}\implies 3^{n+4}=3^{6n} \\\\ \textit{same bases, thus the exponent must be the same} \\\\ n+4=6n\)pretty sure you'd know what "n" is
