Proof for the product of the solutions for the equation

Question

Proof that the product of the solutions for the equation:
$ x^{\log_{2016} x} \times \sqrt{2016} = x^{2016}$ is natural number. Find the digit of the units for that number.

I started by taking log with base 2016 on both sides:
$ \log_{2016} ({x^{\log_{2016} x}}\times \sqrt{2016}) = \log_{2016}x^{2016}$
$ \log_{2016} x^{\log_{2016}x}+\log_{2016} \sqrt{2016}=2016\times log_{2016} x$
$ \log_{2016} x \times \log_{2016} x + log_{2016} 2016^ \frac{1}{2}=2016\times log_{2016} x$
$\log^{2}_{2016} x - 2016\times log_{2016} x + \log_{2016} 2016^\frac{1}{2}=2016\times \log_{2016}x$
$log^{2}_{2016} x - 2016\times log_{2016} x + \frac{1}{2}=0 /\times 2$
$2log^{2}_{2016} x - 4032\times log_{2016} x+ 1=0$
Then I did the change: $ \log_{2016} x=y$
$2y^{2} -4032\times y +1=0.$
But the solutions and their product are not a natural number and I can't find what's incorrect in this. Can you correct it?

To find the product of all $x$'s we need to find the sum of all $y$'s, not the product. — Bob Jones, Oct 14 '17 at 20:06

score 1 · Answer 1 · answered Oct 14 '17 at 20:08

1

taking the logarithm on both sides we can write $$\ln(x^{\log_{2016}x})+\ln\sqrt{2016}=2016\ln(x)$$ and simplifying further $$\frac{\ln(x)}{\ln(2016)}\ln(x)+\ln\sqrt{2016}=2016\ln(x)$$ and you will got a quadratic equation in $\ln(x)$

answered Oct 14 '17 at 20:08

Dr. Sonnhard Graubner

95,283

I have that the product of the two solutions for the quadratic equation (by x) is $2016^{2016}$. The unit of the units od that number is 6, isn't it? – Emilija S. Oct 14 '17 at 20:27
yes i think this is right – Dr. Sonnhard Graubner Oct 14 '17 at 20:31
Thank you for the help! The digit of the units for that number*..I'm sorry for the errors in the previous comment. – Emilija S. Oct 14 '17 at 20:37

Proof for the product of the solutions for the equation

1 Answers1