Product of all real solutions of equation $\frac {2013x}{2014}=2013^{\log_x2014}$?

Question

How am I even supposed to start this task, i need some hint? I logarithm both sides and these are my steps: $$\frac{2013x}{2014}=2013^{\log_x2014}$$ $$\log_{2013}\frac{2013x}{2014}=\log_{2013}2013^{\log_x2014}$$ $$\log_{2013}\frac{2013x}{2014}=\log_x2014$$ $$\log_{2013}{2013} + \log_{2013}{x} - \log_{2013}{2014} = \log_{x}{2014}$$ $$1 + \log_{2013}{x} - \log_{2013}{2014} = \log_{x}{2014}$$ After this step I do not know what to do next.

Answer 1

From your third last step:

$$\log_{2013}\frac{2013x}{2014}=\log_x2014$$ $$\frac{\log2013+\log x-\log2014}{\log2013}=\frac{\log2014}{\log x}$$ $$(\log x)^2+(\log2013-\log2014)\log x-\log2013\log2014=0$$ $$(\log x +\log2013)(\log x-\log2014)=0$$

$$\log x=\log2014 \text{ or } -\log2013$$

$$x=2014 \text{ or } \frac1 {2013}$$

So product of solutions $=\frac{2014}{2013}$

Answer 2

It's simpler to start by taking the logarithm to the base $2014$. Then you get $$a+y-1=\frac ay,$$where $a=\log_{2014}2013$ and $y=\log_{2014}x.$ This is a quadratic equation, the sum of whose roots will give you the product of the roots (in $x$) of the original equation.