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It would make sense that there's always a solution, since I'm pretty sure that equations of the form $a^{x}$ diverge faster than $x^{a}$, but does anyone have a proof?

EDIT: Whoops, sorry, I meant $P(x)-{b}^{x}+1$.

EDIT2: I really fell short on this question. I also meant $b>1$

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Since $P(0)+1\geq b^0$, this is equivalent to the question of whether there exists $x>0$ such that $P(x)+1<b^x$, and in fact the stronger statement holds that there exists $M$ such that $P(x)+1< b^x$ for all $x>M$. Related: math.stackexchange.com/q/111918 –  Jonas Meyer Dec 9 '12 at 0:00
    
Thank you! That makes sense. –  hombre Dec 9 '12 at 0:43

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up vote 2 down vote accepted

Counterexample: $P(x)=x+1$, $b=2$.

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Whoops, sorry, I made a dumb mistake. I meant $+1$ instead of $-1$. Basically, I'm trying to show that exponential functions always diverge faster than polynomials. –  hombre Dec 8 '12 at 23:50

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