$\begin{align}  & 1+{{\log }_{3}}p={{\log }_{27}}q \\ & 1+{{\log }_{3}}p={{\log }_{{{3}^{3}}}}q \\ & {{\log }_{{{3}^{3}}}}q-{{\log }_{3}}p=1 \\ & \tfrac{1}{3}{{\log }_{3}}q-{{\log }_{3}}p=1\left| {{\log }_{{{x}^{n}}}}y=\tfrac{1}{n}{{\log }_{x}}y \right. \\ & {{\log }_{3}}{{q}^{\tfrac{1}{3}}}-{{\log }_{3}}p=1 \\ & {{\log }_{3}}\left( \frac{{{q}^{\tfrac{1}{3}}}}{p} \right)=1 \\ & \frac{{{q}^{\tfrac{1}{3}}}}{p}={{3}^{1}} \\ & 3p={{q}^{\tfrac{1}{3}}} \\\end{align}$