If m = (3^x)^(2+3+4+6) and n = (3^y)^3, is m/n > 1 ?

\(\frac{m}{n} = {3^{15x - 3y}}\,\,\mathop > \limits^? \,\,{3^0}\,\,\,\,\mathop \Leftrightarrow \limits^{{\text{base}}3\, > \,1} 15x - 3y\,\,\mathop > \limits^? \,\,0\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\boxed{5x\,\,\mathop > \limits^? \,\,y}\)

\(\left( 1 \right)\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {x,y} \right) = \left( {0,5} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{No}}} \right\rangle \hfill \\\\
\,{\text{Take}}\,\,\left( {x,y} \right) = \left( {3,4} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{Yes}}} \right\rangle \hfill \\ \\
\end{gathered} \right.\)

\(\left( 2 \right)\,\,\,\left\{ \begin{gathered}\\
\,x = k \hfill \\\\
\,y = 6k \hfill \\ \\
\end{gathered} \right.\,\,\,\,\left( {k \ne 0} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,k = 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{No}}} \right\rangle \hfill \\\\
\,{\text{Take}}\,\,k = - 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{Yes}}} \right\rangle \hfill \\ \\
\end{gathered} \right.\)

\(\left( {1 + 2} \right)\,\,\,37{k^2} = 25\,\,\,\, \Rightarrow \,\,\,\,\left\{ \begin{gathered}\\
\,k = \sqrt {\frac{{25}}{{37}}} \,\,\,\,\, \Rightarrow \,\,\,\,\,5\sqrt {\frac{{25}}{{37}}} \,\,\mathop {\, > }\limits^? \,\,\,6\sqrt {\frac{{25}}{{37}}} \,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{No}}} \right\rangle \hfill \\\\
\,k = - \sqrt {\frac{{25}}{{37}}} \,\,\,\,\, \Rightarrow \,\,\,\,\, - 5\sqrt {\frac{{25}}{{37}}} \,\,\mathop {\, > }\limits^? \,\,\, - 6\sqrt {\frac{{25}}{{37}}} \,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{Yes}}} \right\rangle \hfill \\ \\
\end{gathered} \right.\)

The correct answer is therefore (E).


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.