The nature of the curve $f(x) = x (\log_{2} x)^6$ is baffling me. I expect the curve looks rather like plot of $f(x) = x\log_{2} x$; however probing on Wolfram Alpha says it does not. The shape of the graph looks more like a squashed $f(x) = x^2$ which is not all that surprising but just doesn't seem right.
$x (\log_{2} x)^6$:
$x log_{2} x$:
To find out what the growth curve is really like I put in $f(x) = x^2 / x (\log_{2} x)^6$ and for comparison $f(x) = x^2 / x (\log_{2} x)$. These curves look very different from either interpretation.
$x^2 / (x\log_{2} x)^6$:
$x^2 / (x\log_{2} x)$:
My best guess is I can't get answers I understand because the range where this takes on its typical shape is much larger than $10000$. I don't care about when x is really small. $x$ is domain-restricted to be $> 1$
Incidentally this is a real function that came up at my work some time ago, and it's been on my mind on and off ever since. The original form was $f(x) = \sum_{i=0}^n tx \prod_{j=1}^i k_{j}\log_{2} x$ where $t$ is effectively constant and $k_{n}$ was a statistics table with each term being between $1$ and $0$ and $k_{7}$ turned out to be exactly $0$ so the equation terminated with $6$ dominating.
[What's the tag for interesting range of a function?]
The key fact are that
$$(\log_2x)^6\ge0$$
$$(\log_21)^6 = 0$$
$$x\to 0^+ \implies x(\log_2 x)^6\to0$$
thus there exist a maximum for $x\in(0,1)$.
The difference for $x>1$ with $x\log_2x$ can be understood taking a look at the second derivatives of the functions
second derivative $x(\log_2 x)^6$
second derivative $x\log_2x$