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I am trying to solved this inequality for $k$.

$x^{2k}<\varepsilon\cdot k^k$

Here $k\in\mathbb{N}$ and $x,\varepsilon$ are fixed such that $x,\varepsilon\in\mathbb{R}$ and $\varepsilon>0$. I was thinking about taking $\log_\varepsilon$ on both sides but I am not sure about whether this will preserve the inequality or not? Moreover, I cannot really think of any other way to isolate $k$ other than to consider logarithms.

EDIT: Can someone suggest a solution?

Also if possible, could someone tell me what are the conditions for when the inequalities are preserved under logarithms and what are the properties that need to be satisfied by the base? EDIT: Got the answer for this! Thank you!

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    $\begingroup$ It's possible to show that if the function you're using is strictly monotonically increasing (which the logarithm is), it will preserve inequalities. If it's strictly monotonically decreasing, it will reverse the inequality. If it's neither, you will destroy the inequality. $\endgroup$ – Adrian Keister Nov 14 '18 at 19:41
  • $\begingroup$ If $\varepsilon>1$, the inequality will be preserved— but if $0<\varepsilon<1$, the inequality will be reversed. $\endgroup$ – Mercy King Nov 14 '18 at 19:47
  • $\begingroup$ @AdrianKeister and Mercy King : Thank you for your comments. So does that mean the value of $x$ makes no difference and does not have an impact on how the inequality will be upon taking the logarithm? EDIT : Also, what about $\varepsilon=1$? Do you just deal with that as a special case? $\endgroup$ – Bhavesh Singhal Nov 14 '18 at 19:59
  • $\begingroup$ Rather than take $\log_{\varepsilon},$ I would just do $\ln$ of both sides. $\endgroup$ – Adrian Keister Nov 14 '18 at 20:12
  • $\begingroup$ @MercyKing: I think the conditions on the variables present would make that an impossibility, right? The LHS is non-negative, and $k^k>0$ for all $k\in\mathbb{N}$, so $\varepsilon<0$ would make the inequality a contradiction. $\endgroup$ – Adrian Keister Nov 14 '18 at 20:14
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Suppose $0 < a < c$ and $b = 1$. The $b^k = 1$ for all $k$ and defining $\log_1$ makes no sense. So you can't use logarithms base $1$. (Duh....)

If $b > 1$ then for every $x > 0$ there is some $k = \log_b x$ so that $b^k = x$.

Notice that if $0 < x < 1$ then $k =\log_b x < 0$. If $x = 1$ then $k = \log_b x = 0$. And if $x > 1$ then $k=\log_b x > 0$.

Now if we are given that $a,c$ are positive then notice:

$a < c \iff 1 < \frac ca \iff \log_b \frac ca > 0 \iff \log_b c - \log_b a > 0 \iff \log_b a < \log_b c$.

So if $b > 1$ then you can "take the logs of both sides" to perserve inequality.

If $0 < b < 1$ then we still have for every $x > 0$ there is some $k = \log_b x$ so that $b^k = x$.

If this is not as intuitively obvious note if $0 < b < 1$ then $M = \frac 1b > 0$ and there is $j = \log_M x$ so that $M^j = x$. So $(\frac 1b)^{j} = b^{-j} = x$ so $k = \log_b x = -j = -\log_M x$. So it is true.

We get "double negatives" but...

If $x > 1$ then $b^k = x> 1\implies k =\log_b x < 0$. Of $x = 1$ then $b^k = x=1\implies k = \log_b x = 0$ and if $0 < x < 1$ then $0<b^k = x< 1\implies k = \log_b x > 0$.

And again if $a,c$ are positive and $a < c\iff 1 < \frac ca\iff \log_b \frac ca <0 \iff \log_b c - \log_b a <0 \iff \log_b a > \log_b c$.

So....

Upshot. You can take logs of both sides to preserve inequality if the base is more than $1$. You can take logs of both sides to reverse inequality if the base is less that $1$. And if the base is $1$ you can't do anything.

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Because the log function (to a base greater than $1$) is monotonically increasing, you can take the log of both sides. You still won't get a clean answer for $k$ given $x,\varepsilon$ If we make it an equality to find the minimum $k$ we can look at it as $$\left(\frac {x^2}k\right)^k=\varepsilon$$

If $\varepsilon$ is a small positive number as it usually is, we can see that $k$ must be greater than $x^2$. If $x^2$ is reasonably large, $k$ will not be much larger because the power will be so high. If $x=10, \varepsilon=10^{-6}$ we have $k \approx 113$

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  • $\begingroup$ Could you use the Lambert $W$ function here? Looks like you might, but I'm not sure $W$ is monotonic; you might lose the inequality. $\endgroup$ – Adrian Keister Nov 14 '18 at 22:29
  • $\begingroup$ From your equation, Mathematica yields $k=-\ln(\varepsilon)/W(-\ln(\varepsilon)/x^2).$ $\endgroup$ – Adrian Keister Nov 14 '18 at 22:32
  • $\begingroup$ @AdrianKeister: Mathematica says so, but I haven't understood the W function to say one way or the other. $\endgroup$ – Ross Millikan Nov 14 '18 at 22:47
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Choosing $k=\max \left( \lceil x^2\rceil+1,\ \lceil\frac{x^2}{\varepsilon}\rceil \right)$ works. Thanks to everyone for their suggestions.

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Since $\ln$ is increasing, we have $$ 2k\ln(|x|)<k\varepsilon\ln(k) $$ i.e $$ \frac{2k\ln(|x|)}{k\varepsilon}=2\varepsilon^{-1}\ln(|x|)<\ln(k) $$ Hence $$ k>|x|^{2/\varepsilon} $$

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