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Does $f(x)=ax^b$ grow faster than $g(x)=\ln x$ for all $a, b > 0$? Can I say that $f(x) > g(x)$ as $x$ approaches infinity?

I thought the answer is yes, but this graph appears to be telling a different story.

plot of log(x) vs 0.7x^0.1

Is the polynomial (the green curve) going to cross the log function (the red curve) and exceed in value for some large value of x?

If the answer is yes, does it mean that if I subtract the two functions and set it to zero, the resulting equation will have two roots? What are the roots?

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    $\begingroup$ The answer is yes, just for much larger values of $x$. There will be at least two roots. $\endgroup$ – David Feb 20 '16 at 5:02
  • $\begingroup$ Any suggestion on how I can go about finding the roots? $\endgroup$ – igor Feb 20 '16 at 5:15
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    $\begingroup$ Let $x=e^{100}$. Then $\log x=100$ and $(0.7)x^{0.1}=(0.7)e^{10}\gt 100$ (indeed it is quite a bit larger than $100$). But this brings up an important point. Over the range of values we are interested in, it is very possible for $\log x$ to be bigger than $(0.7)x^{0.1}$. $\endgroup$ – André Nicolas Feb 20 '16 at 5:15
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    $\begingroup$ With calculus, you can show that there are exactly two roots. The second root is between $10^{17}$ and $10^{18}$. If you want to visualize the graph, it's easiest to make a substitution $x_1 = \log_{10} x$, so you would plot $y = x_1 \log 10$ against $y = (0.7)10^{0.1x_1}$. This amounts to using a logarithmic scale on the $x$-axis, and allows you to see where the curves meet. The second intersection is between $x_1 = 17$ and $x_1 = 18$. $\endgroup$ – David Feb 20 '16 at 5:24
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    $\begingroup$ Well, I was nitpicking. I suppose if we want to get precise we could put this as..."we know that as $x \rightarrow +\infty$ that $ log_b x \rightarrow +\infty$ for $b > 1$, and we know for a polynomial P(x) of degree at least one, that $x \rightarrow +\infty$ that $P(x) \rightarrow \pm \infty$. Do we know that $\frac {\log_b x}{P(x)} \rightarrow 0$?" That'd be your question more or less (albeit very pendandtic). And the answer is YES (although it made do so very slowly). $\endgroup$ – fleablood Feb 21 '16 at 3:22
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The answer is yes, although in some cases (like the one you have given) it takes a very long time for the polynomial function to catch up to and ultimately dominate the log function.

A rigorous formation of what you are saying is: $$ \lim_{x \to \infty} \frac{\log(x)}{P(x)}=0$$

where $P(x)$ is any polynomial. The limit tending to zero just means that the bottom terms dominates as $x$ tends to $\infty$.

Here is a proof of the limit equality for the case of $P(x)=x^d$ for some $d>0$. The case of polynomials follows as an easy corollary.

$$ \lim_{x \to \infty} \frac{\log(x)}{x^d} = \lim_{x \to \inf} \frac{1/x}{dx^{d-1}} = \lim_{x \to \infty} \frac{1}{dx^d}=0$$

where the first equality follows from l'Hopital's rule.

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    $\begingroup$ Eek, $d$ as an exponent is kind of unfortunate in this case. $\endgroup$ – pjs36 Feb 20 '16 at 5:13
  • $\begingroup$ Thanks. So does the equation have two roots? What are they? $\endgroup$ – igor Feb 20 '16 at 5:13
  • $\begingroup$ Yes the equation has two roots, specifically $x \approx 2.127$ and $x \approx 4.314 \times 10^{17}$. wolframalpha.com/input/… $\endgroup$ – ASKASK Feb 20 '16 at 5:15
  • $\begingroup$ From some course, I see x^0.5 is called fraction power function, not polynomial, x^2 does. $\endgroup$ – Eric Wang Sep 2 at 8:54

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