Question

If the product of all solution of the equation $\frac{\left(2009\right)x}{2010}=(2009{)}^{{\log }_x\left(2010\right)}$ can be expressed in the lowest form as $\frac{m}{n}$ then the value of $(m+n)$ is

Answer 1

$\frac{2009}{2010}x={\left(2009\right)}^{{\log }_x\left(2010\right)}$

Taking $\text{log}$ to the base $10$ on both sides,
$\log \left(\frac{2009}{2010}\right)+\log x={\log }_x\left(2010\right)\log \left(2009\right)$
$-[\log 2010-\log 2009]+\log x=\frac{\log 2010}{\log x}\log 2009$
${\left(\log x\right)}^2-[\log 2010-\log 2009]\log x-\log 2010\log 2009=0$

The obtained equation is an quadratic equation in $\text{log}x$
$\log x=\frac{\log 2010-\log 2009\pm \sqrt{{(\log 2010-\log 2009)}^2+4\left(\log 2010\log 2009\right)}}{2}$
$\log x=\frac{\log 2010-\log 2009\pm \sqrt{{(\log 2010+\log 2009)}^2}}{2}$
$\log x=\frac{\log 2010-\log 2009\pm (\log 2010+\log 2009)}{2}$
$\log x=\log 2010,\log x=-\log 2009$

Simplifying further,
$\Rightarrow x=2010,\log x=\log {\left(2009\right)}^{-1}$
$\Rightarrow x={\left(2009\right)}^{-1}$
$\text{Products of roots}=\frac{m}{n}=\frac{2010}{2009}$
$m+n=2010+2009$
$=4019$