Question

3^n+4=27^2n

solve using common bases?

Answer 1

`\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