Suppose that $\alpha$ and $\beta$ are the two positive roots of the equation

$$
x^{2}-\sqrt{13} x^{\log _{13} x}=0
$$

Determine the value of $\alpha \beta$.
Using logarithm rules, the following equations are equivalent:

$$
\begin{aligned}
x^{2}-\sqrt{13} x^{\log _{13} x} & =0 \\
x^{2} & =\sqrt{13} x^{\log _{13} x} \\
\log _{13}\left(x^{2}\right) & =\log _{13}\left(\sqrt{13} x^{\log _{13} x}\right) \\
2 \log _{13} x & =\log _{13}(\sqrt{13})+\log _{13}\left(x^{\log _{13} x}\right) \\
2 \log _{13} x & =\frac{1}{2}+\log _{13} x \cdot \log _{13} x \\
0 & =2\left(\log _{13} x\right)^{2}-4 \log _{13} x+1
\end{aligned}
$$

This equation is a quadratic equation in $\log _{13} x$.

If $\alpha$ and $\beta$ are the two roots of the original equation, then

$$
\alpha \beta=13^{\log _{13}(\alpha \beta)}=13^{\log _{13} \alpha+\log _{13} \beta}
$$

But $\log _{13} \alpha+\log _{13} \beta$ is the sum of the two roots of the equation

$$
2\left(\log _{13} x\right)^{2}-4 \log _{13} x+1=0
$$

which is a quadratic equation in $\log _{13} x$.

Since the sum of the roots of the quadratic equation $a y^{2}+b y+c=0$ is $-\frac{b}{a}$, then

$$
\log _{13} \alpha+\log _{13} \beta=-\frac{-4}{2}=2
$$

Finally, this gives $\alpha \beta=13^{2}=169$.

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