Charlie Conner

2022-09-06

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 $\sqrt{2008}\phantom{\rule{thinmathspace}{0ex}}{x}^{{\mathrm{log}}_{2008}x}={x}^{2}$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\sqrt{2008}\phantom{\rule{thinmathspace}{0ex}}{x}^{{\mathrm{log}}_{x}{2008}^{-1}}={x}^{2}$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}-\sqrt{2008}\frac{1}{2008}={x}^{2}$ [by using ${a}^{{\mathrm{log}}_{a}m}=m$]
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{x}^{2}=-\frac{1}{\sqrt{2008}}$
Now how to find the product of positive roots please guide, thanks.

odejicahfc

Take $y=\frac{\mathrm{log}x}{\mathrm{log}2008}$. Hence
$\begin{array}{rl}& \therefore \sqrt{2008}{x}^{y}={x}^{2}\\ & =>\frac{1}{2}\mathrm{log}2008+y\mathrm{log}x=2\mathrm{log}x\\ & =>\frac{1}{2}\ast \mathrm{log}\frac{x}{y}+y\mathrm{log}x=2\mathrm{log}x\\ & =>\mathrm{log}x\left(y+\frac{1}{2}y\right)=2\mathrm{log}x\\ & =>y+\frac{1}{2}y=2\\ & =>2{y}^{2}-4y+1=0\end{array}$
Therefore
${y}_{1}+{y}_{2}=2=>log\left({x}_{1}\ast {x}_{2}\right)=2\mathrm{log}2008=>{x}_{1}\ast {x}_{2}={2008}^{2}$

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