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That isn't true. As you say, it implies there are $c_2$ and $N$ so that for all $n \ge N$ you have $a^n \le c_2 \log n$. Consider: $\begin{align*} \lim_{n \to \infty} \frac{a^n}{\log n} &= \lim_{n \to \infty} \frac{a^n \log a}{1 / n} = \infty \end{align*}$ Here we used l'Hôpital. But that the limit is infinite means that ...


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