Find the product of positive roots of equation $\sqrt{2008}\phantom{\rule{thinmathspace}{0ex}}{x}^{{\mathrm{log}}_{2008}x}={x}^{2}$

Problem : Find the product of positive roots of equation $\sqrt{2008}\phantom{\rule{thinmathspace}{0ex}}{x}^{{\mathrm{log}}_{2008}x}={x}^{2}$

Solution :

The given equation can be written as $\sqrt{2008}\phantom{\rule{thinmathspace}{0ex}}{x}^{{\mathrm{log}}_{2008}x}={x}^{2}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\sqrt{2008}\phantom{\rule{thinmathspace}{0ex}}{x}^{{\mathrm{log}}_{x}{2008}^{-1}}={x}^{2}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\displaystyle -\sqrt{2008}\frac{1}{2008}={x}^{2}}$ [by using ${a}^{{\mathrm{log}}_{a}m}=m$]

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\displaystyle {x}^{2}=-\frac{1}{\sqrt{2008}}}$

Now how to find the product of positive roots please guide, thanks.

Problem : Find the product of positive roots of equation $\sqrt{2008}\phantom{\rule{thinmathspace}{0ex}}{x}^{{\mathrm{log}}_{2008}x}={x}^{2}$

Solution :

The given equation can be written as $\sqrt{2008}\phantom{\rule{thinmathspace}{0ex}}{x}^{{\mathrm{log}}_{2008}x}={x}^{2}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\sqrt{2008}\phantom{\rule{thinmathspace}{0ex}}{x}^{{\mathrm{log}}_{x}{2008}^{-1}}={x}^{2}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\displaystyle -\sqrt{2008}\frac{1}{2008}={x}^{2}}$ [by using ${a}^{{\mathrm{log}}_{a}m}=m$]

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\displaystyle {x}^{2}=-\frac{1}{\sqrt{2008}}}$

Now how to find the product of positive roots please guide, thanks.