Find the product of positive roots of equation $\sqrt{2008}\,x^{\log_{2008}x}=x^2$

Question

Problem : Find the product of positive roots of equation $\sqrt{2008}\, x^{\log_{2008}x}=x^2$

Solution :

The given equation can be written as $\sqrt{2008} \, x^{\log_{2008}x}=x^2 $

$\implies\sqrt{2008} \, x^{\log_{x}{2008}^{-1}}=x^2$

$\implies\displaystyle-\sqrt{2008} \frac{1}{2008} =x^2$ [by using $a^{\log_am} =m $]

$\implies \displaystyle x^2= -\frac{1}{\sqrt{2008}}$

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

Answer 1

$\log_{2008}x = \left(\log_x2008\right)^{-1}$, but is not $\log_x(2008)^{-1}$
So when you raise $x$ to that power, you do not get $2008^{-1}$.

Answer 2

Let $t=\log_{2008}x$.

Since $\sqrt{2008}\times x^{\log_{2008}x}=x^2$, we have $$\log_{2008}\left(\sqrt{2008}\times x^{\log_{2008}x}\right)=\log_{2008}(x^2)$$ $$\iff \log_{2008}\sqrt{2008}+t^2=2t\iff \frac 12+t^2=2t\iff 2t^2-4t+1=0.$$

Here, let $t_1,t_2$ be the solution of $2t^2-4t+1=0$. Then, the answer is $$2008^{t_1}\times 2008^{t_2}=2008^{t_1\color{red}{+}t_2}=2008^{-(-4)/2}=2008^2.$$

Answer 3

Take $y=\frac {\log x}{\log 2008}$. Hence
$$\begin{align}&\therefore\sqrt{2008}x^y=x^2 \\&=>\frac 12 \log 2008+y\log x=2\log x \\&=>\frac 12*\log \frac xy+y\log x=2\log x \\&=>\log x(y+\frac 12y)=2\log x \\&=>y+\frac 12y=2 \\&=>2y^2-4y+1=0\end{align}$$ Therefore $$y_1+y_2=2 =>log(x_1*x_2)=2\log2008 =>x_1*x_2=2008^2$$