Find $n$ such that $\lim _{x \rightarrow \infty} \frac{1}{x} \ln (\frac{e^{x}+e^{2x}+\dots e^{nx}}{n})=9$

Find $n$ such that
$$\lim _{x \rightarrow \infty} \frac{1}{x} \ln (\frac{e^{x}+e^{2x}+\dots e^{nx}}{n})=9$$.

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