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Prove that the product of the roots of $x^{\log_{2016}x}*\sqrt{2016}=x^{2016}$ is a Natural Number.

This is my solution:

by putting $log_{2016}$ on both sides we get:
$\log^2{_{2016}}x-2016\log_{2016}x=-\log_{2016}\sqrt{2016}$

then by putting $\log_{2016}x$ in front of bracket on left side, then removing $\log_{2016}$ from both sides and squaring the equation, and $t=\log_{2016}x$
$t^2-4032t+2016^2-2016=0$

solving for $t_{1/2}$ we get:
$x_1 = 2016^{12\left(168+\sqrt{14}\right)}$
$x_2 = 2016^{12\left(168-\sqrt{14}\right)}$
$x_1 * x_2 = 2016^{4032}$

Is my solution correct? (I think I might have a mistake because I haven't solved this type of problem before, and overall I just started practicing). And if correct is there a better/easier way to solve it?

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marked as duplicate by астон вілла олоф мэллбэрг, Community Nov 30 '18 at 23:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ this question is a near duplicate. $\endgroup$ – lulu Nov 30 '18 at 22:25
  • $\begingroup$ Oh it actually is the same. My bad I guess. I searched for my problem and nothing came up. That link is the exact same problem but with 2014 instead of 2016. $\endgroup$ – Pero Nov 30 '18 at 22:29
  • $\begingroup$ Do I delete my post now? Or learn to solve it from the link and answer my question? Or what? $\endgroup$ – Pero Nov 30 '18 at 22:30
  • $\begingroup$ Sounds fine to me! $\endgroup$ – Mostafa Ayaz Nov 30 '18 at 22:31
  • $\begingroup$ Leave your question as is. We will close it as a duplicate. I have added one above. $\endgroup$ – астон вілла олоф мэллбэрг Nov 30 '18 at 23:09
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We have that

$$y=\log_{2016}x \iff 2016^y=x \implies x^y=2016^{y^2}$$

$$x^{\log_{2016}x}\cdot \sqrt{2016}=x^{2016} \iff 2016^{y^2}\cdot \sqrt{2016}=2016^{2016y}$$

$$2016^{(y^2-2016y)} =\frac{1}{\sqrt{2016}} \iff y^2-2016y=\frac12$$

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Apparently, I could've solved it a lot easier. Thanks to @lulu for providing the link.

$x^{\log_{2016}x}*\sqrt{2016}=x^{2016}$

By taking $log_{2016}$ of both sides we get:

$\log^2{_{2016}}x-2016\log_{2016}x+1/2=0$
$t^2-2016t+1/2=0$

Let r, s be the roots of the original problem, we have:

$\log_{2016} rs = \log_{2016} r + \log_{2016} s = 2016 => rs = 2016^{2016}$

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