No: If \(\displaystyle\in\) is any positive number, then \(\displaystyle{n}^{\epsilon}\) grows faster than \(\displaystyle{\log{{n}}}\). This can be made precise in the statement

\(\displaystyle\lim_{{{n}\rightarrow\infty}}{\frac{{{\log{{n}}}}}{{{n}^{\epsilon}}}}={0}\)

for all \(\displaystyle\epsilon{>}{0}\). To prove this, just note that by L'Hospital's rule,

\(\displaystyle\lim_{{{n}\rightarrow\infty}}{\frac{{{\log{{n}}}}}{{{n}^{\epsilon}}}}=\lim_{{{n}\rightarrow\infty}}{\frac{{{\frac{{{1}}}{{{n}}}}}}{{\epsilon{n}^{{\epsilon-{1}}}}}}={\frac{{{1}}}{{\epsilon}}}\lim_{{{n}\rightarrow\infty}}{\frac{{{1}}}{{{n}^{\epsilon}}}}={0}\)

\(\displaystyle\lim_{{{n}\rightarrow\infty}}{\frac{{{\log{{n}}}}}{{{n}^{\epsilon}}}}={0}\)

for all \(\displaystyle\epsilon{>}{0}\). To prove this, just note that by L'Hospital's rule,

\(\displaystyle\lim_{{{n}\rightarrow\infty}}{\frac{{{\log{{n}}}}}{{{n}^{\epsilon}}}}=\lim_{{{n}\rightarrow\infty}}{\frac{{{\frac{{{1}}}{{{n}}}}}}{{\epsilon{n}^{{\epsilon-{1}}}}}}={\frac{{{1}}}{{\epsilon}}}\lim_{{{n}\rightarrow\infty}}{\frac{{{1}}}{{{n}^{\epsilon}}}}={0}\)