$[Real Analysis] Let $a>1$ and $K>0$. Show that there exists $n_0∈N$ such that $a^{n_0}>K$.$

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$[Real Analysis] Let $a>1$ and $K>0$. Show that there exists $n_0∈N$ such that $a^{n_0}>K$.$

Let $a>1$ and $K>0$. Show that there exists $n_0∈N$ such that

$a^{n_0}>K$.

Real Analysis

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If $K>0$ and $a>1$ then $\frac{\ln(K)}{\ln(a)}$ is a well defined real number. For $n_0>\frac{\ln(K)}{\ln(a)}$ you have that

$$a^{n_0} > a^{\ln(K)/\ln(a)} = e^{\ln(a) \cdot \ln(K)/\ln(a)}= e^{\ln(K)}=K$$

There is no problem finding an $n_0\in\Bbb N$ with $n_0>\frac{\ln(K)}{\ln(a)}$, this is always possible by the archimedean property of the naturals in the reals.

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$[Real Analysis] Let $a>1$ and $K>0$. Show that there exists $n_0∈N$ such that $a^{n_0}>K$.$

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