# The product of the two roots of $\sqrt{2014}x^{\log_{2014} x}=x^{2014}$ is an integer. Find its units digit

The product of the two roots of $$\sqrt{2014}x^{\log_{2014} x}=x^{2014}$$ is an integer. Find its units digit.

I'm quite unable to solve the problem given. I have no idea how to deal with that $$\sqrt{2014}$$ term nor the logarithms in the exponent. Is it even necessary to solve for each individual root or is there some other way to progress through the problem?

I'm really just a novice so I'd appreciate if anyone could show me the process or processes, if multiple. Thank you very much.

To simplify the notation, let $$\log$$ denote $$\log_{2014}$$.

Taking $$\log$$ of both sides then yields $$\frac 12+(\log x)^2=2014\log x$$

Letting $$z=\log x$$ we get the quadratic $$z^2-2014z+\frac 12=0$$

Now if $$r,s$$ are the roots of the original problem we note that $$\log rs=\log r+\log s$$ so the log of the product of $$r,s$$ is the sum of the two roots of the quadratic. Of course the sum of the roots of the quadratic is $$2014$$. Thus $$\log(rs)=2014\implies \boxed{r\times s=2014^{2014}}$$

It is easy to see that, if $$N$$ ends in $$4$$, any even power of $$N$$ ends in $$6$$ so the units digit of this number is $$6$$.

If we take $$\log_{2014}$$ of both sides we get $$\frac12\log_{2014}2014+(\log_{2014}x)^2=2014\log_{2014}x$$ Replace $$\log_{2014}x=y$$: $$y^2-2014y+\frac12\log_{2014}2014=0$$ Let the two solutions to the original equation be $$x_1$$ and $$x_2$$. The logarithm of their product is equivalent to the sum of the two solutions to the equation in $$y$$: $$\log_{2014}x_1x_2=y_1+y_2$$ and this is 2014 by Viète's formulas. Thus $$x_1x_2=2014^{2014}$$ and the units digit of this is $$2014^{2014}\bmod10=6$$

You can use $$a^b=\exp(b \ln(a))$$ to see that the equation is $$\sqrt{2014} \exp\left(\frac{\ln(x)^2}{\ln(2014)} \right)=\exp(2014 \ln(x)).$$ Taking the $$\ln$$ on both sides $$\frac{1}{2} \ln(2014)+\frac{1}{\ln(2014)} \ln(x)^2-2014 \ln(x)=0$$ from where you can expilictely compute the two roots.

If you want to avoid computing the two roots, denoting them by $$x_0$$ and $$x_1$$, notice that $$x_0 x_1=\exp(\ln(x_0)+\ln(x_1))$$ and by the relation betwwen coeeficient and roots of a polynomial $$\ln(x_0)+\ln(x_1)=-\frac{-2014}{1/\ln(2014)}$$ so $$x_0x_1=\exp \left(2014 \ln(2014) \right)=2014^{2014}$$.

By taking logs, we get \begin{align} \sqrt{2014}x^{\log_{2014} x}=x^{2014}\\ \frac 12+\log_{2014}^2(x)=2014\log_{2014}(x)\\ \log_{2014}^2(x)-2014\log_{2014}(x)-\frac 12=0 \end{align} hence if $$x_1,x_2$$ denote the solutions, we have $$\log_{2014}(x_1x_2)=\log_{2014}(x_1)+\log_{2014}(x_2)=2014$$, from which \begin{align} x_1x_2 &=2014^{2014}\\ &\equiv 4^{2014}\\ &\equiv 16^{1007}\\ &\equiv 6^{1007}\\ &\equiv 6\pmod{10} \end{align}