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

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?

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

• this question is a near duplicate. – lulu Nov 30 '18 at 22:25
• 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. – Pero Nov 30 '18 at 22:29
• Do I delete my post now? Or learn to solve it from the link and answer my question? Or what? – Pero Nov 30 '18 at 22:30
• Sounds fine to me! – Mostafa Ayaz Nov 30 '18 at 22:31
• Leave your question as is. We will close it as a duplicate. I have added one above. – астон вілла олоф мэллбэрг Nov 30 '18 at 23:09

We have that

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

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

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

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:

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