Is there an $a$ with $1<a<2$ such that $y=a^x$ can be bounded (or eventually dominated) by a polynomial?
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2No. Take logarithms to check it out. – saulspatz Oct 12 '18 at 13:38
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No.
We know that $2^x/x^n\to\infty$ for any $n$.
Take $1\lt a\lt 2$ and let $k=\ln2/\ln a$ so that $a^k=2$
Let $x=ky$. Then
$$a^x/x^n=a^{ky}/(ky)^n=(2^y/y^n)/k^n$$
The right-hand side increases without limit as $y$, and hence $x$, increases.
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